Finite Group Theory-Martin Isaac

Finite Group Theory

I. Martin Isaacs

Finite Group Theory I. Martin Isaacs

Graduate Studies in Mathematics Volume 92

American Mathematical Society Providence, Rhode Island

Editorial Board

David Cox (Chair) Steven G. Krantz

Rafe Mazzeo Martin Scharlemann

2000 M a t h e m a t i c s S u b j e c t C l a s s i f i c a t i o n . P r i m a r y 2 0 B 1 5 , 2 0 B 2 0 , 2 0 D 0 6 , 2 0 D 1 0 , 2 0 D 1 5 , 2 0 D 2 0 , 2 0 D 2 5 , 2 0 D 3 5 , 2 0 D 4 5 , 2 0 E 2 2 , 2 0 E 3 6 .

F o r a d d i t i o n a l i n f o r m a t i o n a n d upda tes o n th i s b o o k , v i s i t w w w . a m s . o r g / b o o k p a g e s / g s m - 9 2

Library of Congress Cataloging-in-Publication Data Isaacs, I. Martin, 1940-

Finite group theory / I. Martin Isaacs. p. cm. — (Graduate studies in mathematics ; v. 92)

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T o D e b o r a h

Contents

Preface ix

Chapter 1. Sylow Theory 1

Chapter 2. Subnormali ty 45

Chapter 3. Spli t Extensions 65

Chapter 4. Commutators 113

Chapter 5. Transfer 147

Chapter 6. Frobenius Act ions 177

Chapter 7. The Thompson Subgroup 201

Chapter 8. Permuta t ion Groups 223

Chapter 9. More on Subnormali ty 271

Chapter 10. More Transfer Theory 295

Append ix : The Basics 325

Index 345

Preface

This book is a somewhat expanded version of a graduate course in finite group theory that I often teach at the Univers i ty of Wiscons in . I offer this course in order to share what I consider to be a beautiful subject w i t h as many people as possible, and also to provide the solid background in pure group theory that my doctoral students need to carry out their thesis work i n representation theory.

The focus of group theory research has changed profoundly in recent decades. Start ing near the beginning of the 20th century wi th the work of W . Burnside, the major problem was to find and classify the finite simple groups, and indeed, many of the most significant results i n pure group theory and i n representation theory were directly, or at least peripherally, related to this goal. The simple-group classification now appears to be complete, and current research has shifted to other aspects of finite group theory including permutat ion groups, p-groups and especially, representation theory.

It is certainly no less essential in this post-classification period that group-theory researchers, whatever their subspecialty, should have a mastery of the classical techniques and results, and so without a t tempting to be encyclopedic, I have included much of that mater ial here. B u t my choice of topics was largely determined by my pr imary goal i n wr i t ing this book, which was to convey to readers my feeling for the beauty and elegance of finite group theory.

G i v e n its origin, this book should certainly be suitable as a text for a graduate course like mine. B u t I have t r ied to write it so that readers would also be comfortable using it for independent study, and for that reason, I have tr ied to preserve some of the informal flavor of my classroom. I have tr ied to keep the proofs as short and clean as possible, but without omit t ing

ix

X P r e f a c e

details, and indeed, i n some of the more difficult material , my arguments are simpler than can be found in print elsewhere. F ina l ly , since I f irmly believe that one cannot learn mathematics without doing it, I have included a large number of problems, many of which are far from routine.

Some of the material here has rarely, if ever, appeared previously i n books. Just in the first few chapters, for example, we offer Zenkov's marvelous theorem about intersections of abelian subgroups, Wie landt ' s "zipper lemma" i n subnormali ty theory and a proof of Horosevskii 's theorem that the order of a group automorphism can never exceed the order of the group. Later chapters include many more advanced topics that are hard or impossible to find elsewhere.

Mos t of the students who attend my group-theory course are second-year graduate students, w i t h a substantial minor i ty of first-year students, and an occasional well-prepared undergraduate. Almos t al l of these people had previously been exposed to a standard first-year graduate abstract algebra course covering the basics of groups, rings and fields. I expect that most readers of this book w i l l have a similar background, and so I have decided not to begin at the beginning.

Mos t of my readers (like my students) w i l l have previously seen basic group theory, so I wanted to avoid repeating that material and to start w i th something more excit ing: Sylow theory. B u t I recognize that my audience is not homogeneous, and some readers w i l l have gaps in their preparation, so I have included an appendix that contains most of the assumed material in a fairly condensed form. O n the other hand, I expect that many i n my audience w i l l already know the Sylow theorems, but I am confident that even these well-prepared readers w i l l find material that is new to them wi th in the first few sections.

M y semester-long graduate course at Wiscons in covers most of the first seven chapters of this book, start ing wi th the Sylow theorems and culmina t ing w i t h a purely group-theoretic proof of Burnside 's famous p a q b -theorem. Some of the topics along the way are subnormali ty theory, the Schur-Zassenhaus theorem, transfer theory, coprime group actions, Frobe-nius groups, and the normal p-complement theorems of Frobenius and of Thompson. The last three chapters cover material for which I never have t ime i n class. Chapter 8 includes a proof of the s implic i ty of the groups P S L ( n , q ) , and also some graph-theoretic techniques for s tudying subdegrees of pr imi t ive and nonprimit ive permutat ion groups. Subnormali ty theory is revisited i n Chapter 9, which includes Wielandt ' s beautiful automorphism tower theorem and the Thompson-Wielandt theorem related to the Sims

P r e f a c e x i

conjecture. F ina l ly , Chapter 10 presents some advanced topics i n transfer theory, including Yoshida 's theorem and the so-called "principal ideal theorem".

F ina l ly , I thank my many students and colleagues who have contributed ideas, suggestions and corrections while this book was being wri t ten. In part icular, I mention that the comments of Yakov Berkovich and Gabr ie l Navarro were invaluable and very much appreciated.

C h a p t e r 1

Sylow Theory

1 A

It seems appropriate to begin this book w i t h a topic that underlies v i r tua l ly a l l of finite group theory: the Sylow theorems. In this chapter, we state and prove these theorems, and we present some applications and related results. A l t h o u g h much of this mater ial should be very familiar, we suspect that most readers w i l l find that at least some of the content of this chapter is new to them.

A l t h o u g h the theorem that proves Sylow subgroups always exist dates back to 1872, the existence proof that we have decided to present is that of H . Wie land t , published i n 1959. Wie land t ' s proof is slick and short, but i t does have some drawbacks. It is based on a t r ick that seems to have no other applicat ion, and the proof is not really constructive; it gives no guidance about how, i n practice, one might actual ly find a Sylow subgroup. B u t Wie land t ' s proof is beautiful, and that is the pr inc ipa l mot ivat ion for presenting i t here.

A l so , Wie landt ' s proof gives us an excuse to present a quick review of the theory of group actions, which are nearly as ubiquitous i n the study of finite groups as are the Sylow theorems themselves. We devote the rest of this section to the relevant definitions and basic facts about actions, al though we omit some details from the proofs.

Let G be a group, and let ft be a nonempty set. (We w i l l often refer to the elements of ft as "points".) Suppose we have a rule that determines a new element of ft, denoted a-g, whenever we are given a point a e ft and an element g e G. We say that this rule defines an action of G on ft i f the following two conditions hold.

1

2 1 . S y l o w T h e o r y

(1) a - l = a for a l l a G ft and

(2) { a - g ) - h = a-fo/i) for a l l a G ft and al l group elements g , h e G .

Suppose that G acts on ft. It is easy to see that i f g G G is arbitrary, then the function a g : ft -> ft defined by - a-$ has an inverse: the function C T 5 _ I . Therefore, a g is a permutat ion of the set ft, which means that og is bo th injective and surjective, and thus a g lies in the symmetric group Sym(ft) consisting of a l l permutations of ft. In fact, the map g •-> a g is easily seen to be a homomorphism from G into Sym(ft) . ( A homomorphism like this, which arises from an action of a group G on some set, is called a permutation representation of G.) The kernel of this homomorphism is, of course, a normal subgroup of G, which is referred to as the kernel of the action. The kernel is exactly the set of elements g e G that act t r iv ia l ly on ft, which means that a - g = a for a l l points a e ft.

Generally, we consider a theorem or a technique that has the power to find a normal subgroup of G to be "good", and indeed permutat ion representations can be good in this sense. (See the problems at the end of this section.) B u t our goal i n introducing group actions here is not to find normal subgroups; i t is to count things. Before we proceed i n that direction, however, it seems appropriate to mention a few examples.

Let G be arbitrary, and take ft = G. We can let G act on G by right mul t ip l ica t ion , so that x - g = x g for x , g G G. Th i s is the regular action of G, and it should be clear that it is faithful, which means that its kernel is t r iv i a l . It follows that the corresponding permutat ion representation of G is an isomorphism of G into S y m ( G ) , and this proves Cayley 's theorem: every group is isomorphic to a group of permutations on some set.

We continue to take ft = G, but this t ime, we define x - g = g ~ l x g . (The standard notat ion for g ~ x x g is x».) It is t r iv i a l to check that x l = x and that { x 9 ) h = X 9 h for a l l x , g , h G G, and thus we t ru ly have an action, which is called the conjugation action of G on itself. Note that x 9 = x i f and only i f x g = gx, and thus the kernel of the conjugation action is the set of elements g e G that commute w i t h a l l elements x e G. The kernel, therefore, is the center Z ( G ) .

A g a i n let G be arbitrary. In each of the previous examples, we took ft = G , but we also get interesting actions i f instead we take ft to be the set of a l l subsets of G. In the conjugation action of G on ft we let X - g = X 9 = { x 9 \ x e X } and i n the r ight-mult ipl icat ion action we define X - g = X g = { x g | x 6 X } . O f course, i n order to make these examples work, we do not really need ft to be a l l subsets of G. For example, since a conjugate of a subgroup is always a subgroup, the conjugation action is well defined if we take ft to be the set of a l l subgroups of G. A l so , both right mul t ip l ica t ion

1 A 3

and conjugation preserve cardinality, and so each of these actions makes sense i f we take ft to be the collection of a l l subsets of G of some fixed size. In fact, as we shall see, the t r ick in Wie landt ' s proof of the Sylow existence theorem is to use the right mul t ip l ica t ion action of G on its set of subsets w i t h a certain fixed cardinality.

We mention one other example, which is a special case of the right-mul t ip l ica t ion action on subsets that we discussed i n the previous paragraph. Let H C G be a subgroup, and let ft = { H x \ x G G } , the set of right cosets of H i n G. If X is any right coset of H , i t is easy to see that X g is also a right coset of H . (Indeed, if X = H x , then X g = H { x g ) . ) T h e n G acts on the set ft by right mul t ip l ica t ion.

In general, i f a group G acts on some set ft and a G ft, we write Ga = { g G G | a - g = a } . It is easy to check that Ga is a subgroup of G ; it is called the stabilizer of the point a . For example, in the regular action of G on itself, the stabilizer of every point (element of G ) is the t r i v i a l subgroup. In the conjugation action of G on G, the stabilizer of x G G is the centralizer C G ( x ) and in the conjugation act ion of G on subsets, the stabilizer of a subset X is the normalizer N G ( X ) . A useful general fact about point stabilizers is the following, which is easy to prove. In any action, if a - g = /?, then the stabilizers Ga and G p are conjugate i n G, and in fact, { G a ) 3 = Gp.

Now consider the action (by right mul t ipl icat ion) of G on the right cosets of H , where H C G is a subgroup. The stabilizer of the coset H x is the set of a l l group elements g such that H x g = H x . It is easy to see that g satisfies this condi t ion if and only if x g G H x . (This is because two cosets H u and H v are identical if and only if u G H v . ) It follows that g stabilizes H x if and only if g G x ~ l H x . Since x ~ l H x = H x , we see that the stabilizer of the point (coset) H x is exactly the subgroup H x , conjugate to H v i a x . It follows that the kernel of the action of G on the right cosets of H i n G is exactly f | H x . Th is subgroup is called the core of H i n G, denoted

x e G c o r e G ( H ) . The core of H is normal in G because i t is the kernel of an action, and, clearly, it is contained i n H . In fact, if AT < G is any normal subgroup that happens to be contained in H , then N = N x C H x for a l l x G G , and thus N C c o r e G ( t f ) . In other words, the core of H i n G is the unique largest normal subgroup of G contained in H . (It is "largest" i n the strong sense that it contains a l l others.)

We have digressed from our goal, which is to show how to use group actions to count things. B u t having come this far, we may as well state the results that our discussion has essentially proved. Note that the following theorem and its corollaries can be used to prove the existence of normal subgroups, and so they might be considered to be "good" results.


1.1. Theorem. L e t H C G be a s u b g r o u p , a n d l e t ft be t h e set of r i g h t cosets o f H i n G. T h e n G / c o v e G { H ) i s i s o m o r p h i c t o a s u b g r o u p o / S y m ( f t ) . I n p a r t i c u l a r , if t h e i n d e x \ G : H \ = n , t h e n G / c o r e G ( H ) i s i s o m o r p h i c t o a s u b g r o u p of Sn, t h e s y m m e t r i c g r o u p o n n s y m b o l s .

Proof. The action of G on the set ft by right mul t ip l ica t ion defines a homomorphism 6 (the permutat ion representation) from G into Sym(ft) . Since ker(0) = c o r e G ( t f ) , it follows by the homomorphism theorem that G / c o r e G { H ) ^ 0 ( G ) , which is a subgroup of S y m ( G ) . The last statement follows since if \ G : H \ = n , then (by definition of the index) |ft| = n , and thus Sym(ft) = Sn. U

1.2. Corollary. L e t G be a g r o u p , a n d suppose t h a t H C G i s a s u b g r o u p w i t h \ G : H \ = n . T h e n H c o n t a i n s a n o r m a l s u b g r o u p N of G such t h a t \ G : N \ d i v i d e s n \ .

Proof. Take N = c o r e G ( H ) . T h e n G / N is isomorphic to a subgroup of the symmetric group Sn, and so by Lagrange's theorem, \ G / N \ divides \ S n \ = n \ . •

1.3. Corollary. L e t G be s i m p l e a n d c o n t a i n a s u b g r o u p of i n d e x n > 1. T h e n \ G \ d i v i d e s n l .

Proof. The normal subgroup N of the previous corollary is contained i n H , and hence it is proper i n G because n > 1. Since G is simple, N = 1 , and thus \ G \ = \ G / N \ divides n l . U

In order to pursue our ma in goal, which is counting, we need to discuss the "orbits" of an action. Suppose that G acts on ft, and let a G ft. The set Oa = { a - g | g G G } is called the orbit of a under the given action. It is routine to check that i f ¡3 G Oa, then O p = Oa, and i t follows that distinct orbits are actually disjoint. A l so , since every point is i n at least one orbit, it follows that the orbits of the action of G on ft par t i t ion ft. In part icular, if ft is finite, we see that |ft| = where in this sum, O runs over the full set of G-orbits on ft.

We mention some examples of orbits and orbit decompositions. F i rs t , i f H C G is a subgroup, we can let H act on G by right mul t ip l ica t ion . It is easy to see that the orbits of this action are exactly the left cosets of H i n G . (We leave to the reader the problem of realizing the right cosets of H i n G as the orbits of an appropriate action of H . B u t be careful: the rule x - h - hx does n o t define an action.)

Perhaps it is more interesting to consider the conjugation action of G on itself, where the orbits are exactly the conjugacy classes of G . The fact

1 A 5

that for a finite group, the order \ G \ is the sum of the sizes of the classes is sometimes called the class equation of G.

How big is an orbit? The key result here is the following.

1.4. T h e o r e m (The Fundamental Count ing Pr inc ip le ) . L e t G a c t o n Q, a n d suppose t h a t O i s one of t h e o r b i t s . L e t a G O, a n d w r i t e H = Ga> t h e s t a b i l i z e r of a . L e t A = { H x | x G G } be t h e set of r i g h t cosets of H i n G. T h e n t h e r e i s a b i j e c t i o n 6 : A -> O such t h a t 9 { H g ) = a-g. I n p a r t i c u l a r , \ 0 \ = \ G : Ga\.

Proof. We observe first that i f H x = H y , then a-x = a-y. To see why this is so, observe that we can write y = hx for some element h s H . T h e n

a-y = a - { h x ) = { a - h ) - x = a-x ,

where the last equality holds because h G H = Ga, and so h stabilizes a .

G i v e n a coset H x G A , the point a-x lies i n O, and we know that i t is determined by the coset H x , and not just by the par t icular element x . It is therefore permissible to define the function 9 : A O by 9 { H x ) = a-x, and it remains to show that 9 is bo th injective and surjective.

The surjectivity is easy, and we do that first. If 0 G O, then by the definition of an orbit , we have 0 = a-x for some element x G G. T h e n H x G A satisfies 6 { H x ) = a-x = 0, as required.

To prove that 9 is injective, suppose that 9 { H x ) = 9 { H y ) . We have a-x = a-y, and hence

a = a-1 = { a - x ) - x - 1 = ( a - y ) - x " 1 = a - i y x ' 1 ) .

T h e n yx'1 fixes a , and so it lies i n Ga = H . It follows that y G H x , and thus H y = H x . Th is proves that 6 is injective, as required. •

It is easy to check that the bijection 9 of the previous theorem actually defines a "permutation isomorphism" between the act ion of G on A and the action of G on the orbit O. Formally, this means that 0 ( X - g ) = 9 ( X ) - g for a l l "points" X in A and group elements g G G. M o r e informally, this says that the actions of G on A and on O are "essentially the same". Since every action can be thought of as composed of the actions on the ind iv idua l orbits, and each of these actions is permutat ion isomorphic to the right-mul t ip l ica t ion action of G on the right cosets of some subgroup, we see that these actions on cosets are t ru ly fundamental: every group action can be viewed as being composed of actions on right cosets of various subgroups.

We close this section w i t h two familiar and useful applications of the fundamental counting principle.

1.5. Corol lary. L e t x € G, w h e r e G i s a finite g r o u p , a n d l e t K be t h e c o n j u g a c y class of G c o n t a i n i n g x . T h e n \ K \ = \ G : C G ( x ) \ .

Proof. The class of x is the orbit of x under the conjugation action of G on itself, and the stabilizer of x in this action is the centralizer C G { x ) . Thus \ K \ = \ G : C G { x ) \ , as required. •

1.6. Corol lary. L e t H C G be a s u b g r o u p , w h e r e G i s finite. T h e n t h e t o t a l n u m b e r of d i s t i n c t c o n j u g a t e s of H i n G, c o u n t i n g H i t s e l f , i s \ G : N G { H ) \ .

Proof. The conjugates of H form an orbit under the conjugation action of G on the set of subsets of G. The normalizer N G { H ) is the stabilizer of H i n this action, and thus the orbit size is \ G : N G ( # ) | , as wanted. •

P r o b l e m s 1 A

1A.1 . Le t H be a subgroup of prime index p i n the finite group G, and suppose that no prime smaller than p divides \ G \ . Prove that H < G.

1A.2. G i v e n subgroups H , K C G and an element g G G, the set H g K = { h g k \ h e H , k G K } is called an ( H , i^-double coset. In the case where H and K are finite, show that \ H g K \ = \ H \ \ K \ / \ K n H 9 \ .

Hint . Observe that H g K is a union of right cosets of H , and that these cosets form an orbit under the action of K .

Note. If we take g = 1 in this problem, the result is the familiar formula \ H K \ = \ H \ \ K \ / \ H n K \ .

1 A . 3 . Suppose that G is finite and that H , K C G are subgroups.

(a) Show that \ H : H n K \ < \ G : K \ , w i th equality if and only if H K = G.

(b) If \ G : H \ and \ G : K \ are coprime, show that H K = G.

Note . Proofs of these useful facts appear in the appendix, but we suggest that readers t ry to find their own arguments. A l so , recall that the product H K of subgroups H and K is not always a subgroup. In fact, H K is a subgroup if and only i f H K = K H . (This too is proved i n the appendix.) If H K = K H , we say that H and K are permutable.

1A.4. Suppose that G = H K , where H and K are subgroups. Show that also G = H x K y for a l l elements x,y G G. Deduce that if G = H H X for a subgroup H and an element x G G, then H = G.

P r o b l e m s 1 A 7

1A.5 . A n action of a group G on a set ft is transitive i f ft consists of a single orbit . Equivalently, G is transitive on ft i f for every choice of points a,/? G ft, there exists an element g G G such that a - g = 0. Now assume that a group G acts transit ively on each of two sets ft and A . Prove that the natural induced action of G on the cartesian product ft x A is transitive if and only i f G a G p = G for some choice of a G ft and 0 G A .

Hint . Show that if G a G p = G for some a G ft and 0 G A , then i n fact, this holds for a l l a G ft and 0 G A .

1A.6 . Le t G act on ft, where both G and ft are finite. For each element g G G, wri te x ( g ) = \ { a e n \ a - g = a } \ . The nonnegative-integer-valued function X is called the permutation character associated w i t h the action. Show that

J]x(0) = J2 \ G a \ = n \ G \ ,

where n is the number of orbits of G on ft.

Note. Thus the number of orbits is

1 1 g e G

which is the average value of x over the group. A l t h o u g h this orbit-counting formula is often at t r ibuted to W . Burnside, it should (according to P. Neumann) more properly be credited to Cauchy and Frobenius.

1A.7 . Let G be a finite group, and suppose that H < G is a proper subgroup. Show that the number of elements of G that do not lie in any conjugate of H is at least \ H \ .

Hint . Let X be the permutat ion character associated wi th the r ight-mul t ip l i cation action of G on the right cosets of H . T h e n £ x ( # ) = \ G \ , where the sum runs over g e G . Show that £ x W > 2| t f | , where here, the sum runs over h e H . Use this information to get an estimate on the number of elements of G where x vanishes.

1A.8. Le t G be a finite group, let n > 0 be an integer, and let C be the additive group of the integers modulo n . Let ft be the set of n-tuples [ x l , x 2 , . . . , x n ) of elements of G such that x x x 2 •••xn = l .

(a) Show that C acts on ft according to the formula

( x i , x 2 , •••, x n ) - k = ( x i + k , x 2 + k , • ••, x n + k ) ,

where k G C and the subscripts are interpreted modulo n .


(b) Now suppose that n = p is a prime number that divides \ G \ . Show that V divides the number of C-orbi ts of size 1 on ft, and deduce that the number of elements of order p i n G is congruent to - 1 m o d p .

Note . In part icular, if a prime p divides | G | , then G has at least one element of order p . Th i s is a theorem of Cauchy, and the proof i n this problem is due to J . H . M c K a y . Cauchy's theorem can also be derived as a corollary of Sylow's theorem. Alternat ively, a proof of Sylow's theorem different from Wielandt ' s can be based on Cauchy's theorem. (See P rob lem 1B.4.)

1A.9 . Suppose \ G \ = p m , where p > m and p is prime. Show that G has a unique subgroup of order p .

1A.10. Let H C G .

(a) Show that \ N G ( H ) : H \ is equal to the number of right cosets of H in G that are invariant under right mul t ip l ica t ion by H .

(b) Suppose that \ H \ is a power of the prime p and that \ G : H \ is divisible by p . Show that \ N G { H ) : H \ is divisible by p .

I B

F i x a prime number p . A finite group whose order is a power of p is called a p-group. It is often convenient, however, to use this nomenclature somewhat carelessly, and to refer to a group as a "p-group" even if there is no part icular prime p under consideration. For example, i n proving some theorem, one might say: it suffices to check that the result holds for p-groups. W h a t is meant here, of course, is that it suffices to show that the theorem holds for a l l p-groups for a l l primes p .

We mention that, al though i n this book a p-group is required to be finite, it is also possible to define infinite p-groups. The more general definition is that a (not necessarily finite) group G is a p-group if every element of G has finite p-power order. O f course, i f G is finite, then by Lagrange's theorem, every element of G has order d iv id ing \ G \ , and so i f \ G \ is a power of p, it follows that the order of every element is a power of p, and hence G is a p-group according to the more general definition. Conversely, if G is finite and has the property that the order of every element is a power of p, then clearly, G can have no element of order q for any prime q different from p. It follows by Cauchy 's theorem (Problem 1A.8) that no prime q ^ p can divide | G | , and thus \ G \ must be a power of p, and this shows that the two definitions of "p-group" are equivalent for finite groups.

A g a i n , fix a prime p. A subgroup S o f a finite group G is said to be a Sylow p-subgroup of G i f \S\ is a power of p and the index \ G : S\ is

I B 9

not divisible by p. A n alternative formulation of this definition relies on the observation that every positive integer can be (uniquely) factored as a power of the given prime p times some integer not divisible by p . In part icular, if we write \ G \ = p a m , where a > 0 and p does not divide m > 1, then a subgroup S of G is a Sylow p-subgroup of G precisely when \S\ = p a . In other words, a Sylow p-subgroup of G is a p-subgroup S whose order is as large as is permit ted by Lagrange's theorem, which requires that \S\ must divide | G | . We mention two t r i v i a l cases: i f | G | is not divisible by p, then the identi ty subgroup is a Sylow p-subgroup of G, and if G is a p-group, then G is a Sylow p-subgroup of itself. The Sylow existence theorem asserts that Sylow subgroups a l w a y s exist.

1.7. T h e o r e m (Sylow E ) . L e t G be a finite g r o u p , a n d l e t p be a p r i m e . T h e n G has a S y l o w p - s u b g r o u p .

The Sylow E-theorem can be viewed as a par t ia l converse of Lagrange's theorem. Lagrange asserts that i f i f is a subgroup of G and \ H \ = k, then k divides \ G \ . The converse, which i n general is false, would say that i f k is a positive integer that divides then G has a subgroup of order k. (The smallest example of the failure of this assertion is to take G to be the alternating group A A of order 12; this group has no subgroup of order 6.) B u t i f A; is a power of a prime, we shall see that G actual ly does have a subgroup of order k. If k is the largest power of p that divides \ G \ , the desired subgroup of order k is a Sylow p-subgroup; for smaller powers of p, we w i l l prove that a Sylow p-subgroup of G necessarily has a subgroup of order k.

We are ready now to begin work toward the proof of the Sylow E -theorem. We start w i t h a purely ari thmetic fact about b inomia l coefficients.

1.8. L e m m a . L e t p be a p r i m e n u m b e r , a n d l e t a > 0 a n d m > 1 be i n t e g e r s .

Proof. Consider the polynomial (1 + X ) p . Since p is prime, it is easy to see that the b inomia l coefficients ( p) are divisible by p for 1 < i < p - 1, and thus we can write (1 + X ) p = 1 + X p m o d p. (The assertion that these polynomials are congruent modulo p means that the coefficients of corresponding powers of X are congruent modulo p.) A p p l y i n g this fact a second t ime, we see that { 1 + X ) p 2 = ( 1 + X P ) P = 1 + X p 2 m o d p. Cont inu ing like this, we deduce that (1 + X ) p a = 1 + X p a mod p, and thus

T h e n

= m m o d p

( l + Xfam = ( 1 + X p a ) m m o d p .

10 1 . S y l o w T h e o r y

Since these polynomials are congruent, the coefficients of corresponding terms are congruent modulo p, and the result follows by considering the coefficient of X ? a on each side. •

P r o o f of the Sylow E-theorem (Wielandt) . Wr i t e \ G \ = p a m , where a > 0 and p does not divide m. Let f t be the set of a l l subsets of G having cardinal i ty p a , and observe that G acts by right mul t ip l ica t ion on f t . Because of this action, f t is part i t ioned into orbits, and consequently, \fl\ is the sum of the orbit sizes. B u t

and so |i2| is not divisible by p , and it follows that there is some orbit O such that \ 0 \ is not divisible by p .

N o w let X G O, and let H = G x be the stabilizer of X i n G. B y the fundamental counting principle, \ 0 \ = \ G \ / \ H \ , and since p does not divide \ 0 \ and p a divides \ G \ , we conclude that p a must divide \ H \ , and i n part icular p a < \ H \ .

Since H stabilizes X under right mul t ip l ica t ion, we see that if x G X , then x H C X , and thus \ H \ = \ x H \ < \X\ = p a , where the final equality holds since X G f t . We now have \ H \ = p a , and since H is a subgroup, it is a Sylow subgroup of G, as wanted. •

In P rob l em 1A.8 , we sketched a proof of Cauchy's theorem. We can now give another proof, using the Sylow E-theorem.

1.9. Corol lary (Cauchy) . L e t G be a finite g r o u p , a n d suppose t h a t p i s a p r i m e d i v i s o r of \ G \ . T h e n G has a n element of o r d e r p .

Proof. Let S be a Sylow p-subgroup of G, and note that since \S\ is the m a x i m u m power of p that divides | G | , we have \S\ > 1. Choose a non-identity element x of S, and observe that the order o { x ) divides \S\ by Lagrange's theorem, and thus 1 < o ( x ) is a power of p . In particular, we can write o { x ) = pm for some integer m > 1, and we see that o { x m ) = p , as wanted. •

We introduce the notat ion S y l p ( G ) to denote the set of a l l Sylow p-subgroups of G. The assertion of the Sylow E-theorem, therefore, is that the set Sylp(G) is nonempty for a l l finite groups G and a l l primes p . The intersection n S y l p ( G ) of a l l Sylow p-subgroups of a group G is denoted O p ( G ) , and as we shall see, this is a subgroup that plays an important role i n finite group theory.

I B 11

Perhaps this is a good place to digress to review some basic facts about characteristic subgroups. (Some of this mater ial also appears in the appendix.) F i r s t , we recall the definition: a subgroup K C G is characteristic in G i f every automorphism of G maps K onto itself.

It is often difficult to find a l l automorphisms of a given group, and so the definition of "characteristic" can be hard to apply directly, but nevertheless, i n many cases, it easy to establish that certain subgroups are characteristic. For example, the center Z ( G ) , the derived (or commutator) subgroup G', and the intersection of al l Sylow p-subgroups O p ( G ) are characteristic in G . M o r e generally, any subgroup that can be described unambiguously as " t h e something" is characteristic. It is essential that the description using the definite article be unambiguous, however. G i v e n a subgroup H Q G , for example, we cannot conclude that the normalizer N G ( H ) or the center Z ( H ) is characteristic i n G . A l though these subgroups are described using "the", the descriptions are not unambiguous because they depend on the choice of H . We can say, however, that Z ( G ' ) is characteristic in G because it is t h e center of t h e derived subgroup; it does not depend on any unspecified subgroups.

A good way to see why "the something" subgroups must be characteristic is to imagine two groups G i and G 2 , w i t h an isomorphism 9 : G i -> G 2 . Since isomorphisms preserve "group theoretic" properties, it should be clear that 9 maps the center Z ( G i ) onto Z ( G 2 ) , and indeed 9 maps each unambiguously defined subgroup of G i onto the corresponding subgroup of G 2 . Now specialize to the case where G i and G 2 happen to be the same group G , so 9 is an automorphism of G . Since i n the general case, we know that 6»(Z(GJ) = Z ( G 2 ) , we see that when G i = G = G 2 , we have 0(Z(G)) = Z ( G ) , and similarly, if we consider any "the something" subgroup in place of the center.

O f course, characteristic subgroups are automatical ly normal . T h i s is because the definition of normali ty requires only that the subgroup be mapped onto itself by i n n e r automorphisms while characteristic subgroups are mapped onto themselves by a l l automorphisms. We have seen that some characteristic subgroups are easily recognized, and it follows that these subgroups are obviously and automatical ly normal . For example, the subgroup O p ( G ) is normal in G for al l primes p .

The fact that characteristic subgroups are normal remains true i n an even more general context. The following, which we presume is already known to most readers of this book, is extremely useful. (This result also appears i n the appendix.)

1.10. L e m m a . L e t K C N C G, w h e r e G i s a g r o u p , N i s a n o r m a l s u b g r o u p of G a n d K i s a c h a r a c t e r i s t i c s u b g r o u p of N . T h e n K < G .

12 I . Sylow T h e o r y

Proof. Le t g € G. T h e n conjugation by g maps N onto itself, and it follows that the restriction of this conjugation map to N is an automorphism of N . (But note that it is not necessarily an inner automorphism of N . ) Since K is characteristic i n N , it is mapped onto itself by this automorphism of N , and thus K 9 = K , and it follows that K < G. U

P r o b l e m s I B

1B.1. Let S e S y l p ( G ) , where G is a finite group.

(a) Le t P C G be a p-subgroup. Show that PS is a subgroup if and only i f P C S.

(b) If S < G, show that S y l p ( G ) = { S } , and deduce that S is characteristic i n G.

Note . O f course, it would be "cheating" to do problems in this section using theory that we have not yet developed. In particular, you should avoid using the Sylow C-theorem, which asserts that every two Sylow p-subgroups of G are conjugate in G.

1B.2. Show that O p ( G ) is the unique largest normal p-subgroup of G. (This means that it is a normal p-subgroup of G that contains every other normal p-subgroup of G.)

1B.3. Let S £ Sy lp(G) , and write N = N G ( 5 ) . Show that N = N G ( N ) .

1B.4. Let P C G be a p-subgroup such that \ G : P \ is divisible by p. Us ing Cauchy's theorem, but without appealing to Sylow's theorem, show that there exists a subgroup Q of G containing P , and such that \ Q : P \ = p . Deduce that a max ima l p-subgroup of G (which obviously must exist) must be a Sylow p-subgroup of G.

H i n t . Use P rob l em 1A.10 and consider the group N G ( P ) / P .

Note . Once we know Cauchy's theorem, this problem yields an alternative proof of the Sylow E-theorem. O f course, to avoid circularity, we appeal to P rob lem 1A.8 for Cauchy's theorem, and not to Corol la ry 1.9.

1B.5. Let 7T be any set of prime numbers. We say that a finite group H is a vr-group i f every prime divisor of \ H \ lies in TT . A l so , a vr-subgroup H C G is a H a l l 7r-subgroup of G i f no prime d iv id ing the index \ G : H \ lies in TT. (So i f TT = {p}, a H a l l vr-subgroup is exactly a Sylow p-subgroup.)

Now let 9 : G -»• K be a surjective homomorphism of finite groups.

(a) If H is a H a l l vr-subgroup of G, prove that 9 ( H ) is a H a l l vr-subgroup of K .

P r o b l e m s I B 13

(b) Show that every Sylow p-subgroup of K has the form 9 ( H ) , where H is some Sylow p-subgroup of G.

(c) Show that | S y l p ( G ) | > | S y l p ( X ) l for every prime p.

Note. If the set vr contains more than one prime number, then a H a l l vr-subgroup can fail to exist. B u t a theorem of P . H a l l , after whom these subgroups are named, asserts that in the case where G is solvable, H a l l TT-subgroups always do exist. (See Chapter 3, Section C.) We mention also that Par t (b) of this problem would not remain true if "Sylow p-subgroup" were replaced by " H a l l 7r-subgroup".

1B.6. Let G be a finite group, and let K C G be a subgroup. Suppose that H C G is a H a l l 7r-subgroup, where TT is some set of primes. Show that i f H K is a subgroup, then H n K is a H a l l 7r-subgroup of K .

Note . In part icular, K has a H a l l 7r-subgroup if either H or K is normal in G since in that case, H K is guaranteed to be a subgroup.

1B.7. Let G be a finite group, and let vr be any set of primes.

(a) Show that G has a (necessarily unique) normal 7r-subgroup N such that N D M whenever M < G is a 7r-subgroup.

(b) Show that the subgroup N of Par t (a) is contained in every H a l l 7r-subgroup of G.

(c) Assuming that G has a H a l l 7r-subgroup, show that N is exactly the intersection of a l l of the H a l l vr-subgroups of G.

Note . The subgroup N of this problem is denoted O n ( G ) . Because of the uniqueness in (b), it follows that this subgroup is characteristic in G . F ina l ly , we note that i f p is a prime number, then, of course, 0 { p } ( G ) - O p ( G ) .

1B.8. Let G be a finite group, and let vr be any set of primes.

(a) Show that G has a (necessarily unique) normal subgroup N such that G / N is a vr-group and M D N whenever M < G and G / M is a 7r-group.

(b) Show that the subgroup N of Par t (a) is generated by the set of al l elements of G that have order not divisible by any prime in TT.

Note. The characteristic subgroup N of this problem is denoted O f f ( G ) . A l so , we recall that the subgroup generated by a subset of G is the (unique) smallest subgroup that contains that set.

14 I . S y l o w T h e o r y

1 C

We are now ready to study in greater detail the nonempty set S y l p ( G ) of Sylow p-subgroups of a finite group G .

1.11. Theorem. L e t P be a n a r b i t r a r y p - s u b g r o u p of a finite g r o u p G, a n d suppose t h a t S € S y l p ( G ) . T h e n PCS9 f o r some element g e G .

Proof. Let ft = {Sx \ x e G } , the set of right cosets of S in G , and note that |ft| = \G:S\ is not divisible by p since -S is a Sylow p-subgroup of G . We know that G acts by right mul t ip l ica t ion on ft, and thus P acts too, and ft is par t i t ioned into P-orbi t s . A l so , since |ft| is not divisible by p, there must exist some P-orb i t O such that \ 0 \ is not divisible by p.

B y the fundamental counting principle, \ 0 \ is the index in P of some subgroup. It follows that \ 0 \ divides | P | , which is a power of p. Then \ 0 \ is bo th a power of p and not divisible by p, and so the only possibil i ty is that \ 0 \ = 1. Recal l ing that a l l members of ft are right cosets of S in G , we can suppose that the unique member of O is the coset Sg.

Since Sg is alone in a P-orb i t , it follows that it is fixed under the action of P , and thus Sgu = Sg for a l l elements u e P . T h e n g u € Sg, and hence u € g-^Sg = S9. Thus PCS9, as required. •

If S is a Sylow p-subgroup of G , and g e G is arbitrary, then the conjugate S9 is a subgroup having the same order as S. Since the only requirement on a subgroup that is needed to qualify it for membership in the set S y l p ( G ) is that it have the correct order, and since S € S y l p ( G ) and \S9\ = \S\, it follows that S9 also lies in S y l p ( G ) . In fact every member of S y l p ( G ) arises this way: as a conjugate of S. Th i s is the essential content of the Sylow conjugacy theorem. P u t t i n g it another way: the conjugation action of G on S y l p ( G ) is transitive.

1.12. T h e o r e m (Sylow C ) . If S a n d T S y l o w p - s u b g r o u p s of a finite g r o u p G, t h e n T = S9 f o r some element g e G .

Proof. A p p l y i n g Theorem 1.11 w i t h T in place of P , we conclude that T C S9 for some element g e G . B u t since both S and T are Sylow p-subgroups, we have \T\ = \S\ = \S9\, and so the containment of the previous sentence must actually be an equality. •

The Sylow C-theorem yields an alternative proof of P rob lem I B . 1(b), which asserts that i f a group G has a normal Sylow p-subgroup S, then S is the only Sylow p-subgroup of G . Indeed, by the Sylow C-theorem, if T e S y l p ( G ) , then we can write T = S9 = S, where the second equality is a consequence of the normali ty of S.

1 C 15

A frequently used application of the Sylow C-theorem is the so-called "Fra t t in i argument", which we are about to present. Perhaps the reason that this result is generally referred to as an "argument" rather than as a "lemma" or "theorem" is that variations on its proof are used nearly as often as its statement.

1.13. L e m m a (Frat t in i Argument ) . L e t N < G w h e r e N i s finite, a n d s u p pose t h a t P G Sylp( iV) . T h e n G = N G { P ) N .

Proof. Let g G G, and note that P 9 C N 9 = N , and thus P 9 is a subgroup of N having the same order as the Sylow p-subgroup P . It follows that P 9 G Sylp(iV), and so by the Sylow C-theorem applied in N , we deduce that ( p g y = p , for some element n £ J V . Since P 9 n = P , we have g n G N G ( P ) , and so g G N G ( P ) n " 1 C N G ( P ) N . B u t g € G was arbitrary, and we deduce that G = N G ( P ) A T , as required. •

B y definition, a Sylow p-subgroup of a finite group G is a p-subgroup that has the largest possible order consistent w i t h Lagrange's theorem. B y the Sylow E-theorem, we can make a stronger statement: a subgroup whose order is max ima l among the orders of a l l p-subgroups of G is a Sylow p-subgroup. A n even stronger assertion of this type is that every max ima l p-subgroup of G is a Sylow p-subgroup. Here, "maximal" is to be interpreted i n the sense of containment: a subgroup H of G is max ima l w i t h some property i f there is no subgroup K > H that has the property. The t ru th of this assertion is the essential content of the Sylow "development" theorem.

1.14. T h e o r e m (Sylow D ) . L e t P be a p - s u b g r o u p of a finite g r o u p G. T h e n P i s c o n t a i n e d i n some S y l o w p - s u b g r o u p of G.

Proof. Let S G S y l p ( G ) . T h e n by Theorem 1.11, we know that PCS9 for some element g G G. Also , since \S9\ = \S\, we know that S9 is a Sylow p-subgroup of G. U

G i v e n a finite group G, we consider next the question of how many Sylow p-subgroups G has. To facilitate this discussion, we introduce the (not quite standard) notat ion n p { G ) = | S y l p ( G ) | . (Occasionally, when the group we are considering is clear from the context, we w i l l s imply write n p instead of n p { G ) . )

Fi rs t , by the Sylow C-theorem, we know that S y l p ( G ) is a single orbit under the conjugation action of G. The following is then an immediate consequence.

1.15. Corol lary. L e t S G S y l p ( G ) , w h e r e G i s a finite g r o u p . T h e n n p ( G ) -| G : N G ( 5 ) | .

16 1 . S y l o w T h e o r y

Proof. Since n p ( G ) = | S y l p ( G ) | is the to ta l number of conjugates of S i n G, the result follows by Corol la ry 1.6. •

In part icular, it follows that n p ( G ) divides \ G \ , but we can say a bit more. If 5 € S y l p ( G ) , then of course, S C N G ( S ) since S is a subgroup, and thus | G : S\ = \ G : N G ( 5 ) | | N G ( 5 ) : S\. A l so , n p { G ) = \ G : N G ( 5 ) | , and hence n p ( G ) divides \ G : S\. In other words, if we write \ G \ = p a m , where p does not divide m , we see that n p { G ) divides m. (We mention that the integer m is often referred to as the p'-part of \ G \ . )

The information that n p { G ) divides the p'-part of \ G \ becomes even more useful when it is combined wi th the fact (probably known to most readers) that n p ( G ) = 1 m o d p for a l l groups G. In fact, there is a useful stronger congruence constraint, which may not be quite so well known. Before we present our theorem, we mention that i f S,T € S y l p ( G ) , then |5| = | T | , and thus \S : SnT\ = \S\/\SnT\ = | T|/|5nr| = \ T : S n T \ . The statement of the following result, therefore, is not really as asymmetric as it may appear.

1.16. Theorem. Suppose t h a t G i s a finite g r o u p such t h a t n p { G ) > 1 , a n d choose d i s t i n c t S y l o w p - s u b g r o u p s S a n d T of G such t h a t t h e o r d e r \S D T\ i s as l a r g e as p o s s i b l e . T h e n n p { G ) = 1 m o d \S : S n T | .

1.17. Corol lary. If G i s a finite g r o u p a n d p i s a p r i m e , t h e n n p ( G ) = 1 m o d p .

Proof. If n p ( G ) = 1, there is nothing to prove. Otherwise, Theorem 1.16 applies, and there exist dist inct members S,T e S y l p ( G ) such that n p ( G ) = 1 m o d \S :SnT\, and thus it suffices to show that \S :SnT\ is divisible by p. B u t \S : S n T\ = \ T : S n T\ is certainly a power of p, and it exceeds 1 since otherwise S = S n T = T , which is not the case because S and T are distinct. •

In order to see how Theorem 1.16 can be used, consider a group G of order 21,952 = 2 6 - 7 3 . We know that n 7 must divide 2 6 = 64, and it must be congruent to 1 modulo 7. We see, therefore, that n 7 must be one of 1, 8 or 64. Suppose that G does not have a normal Sylow 7-subgroup, so that n 7 > 1. Since neither 8 nor 64 is congruent to 1 modulo 7 2 = 49, we see by Theorem 1.16 that there exist distinct Sylow 7-subgroups S and T of G such that \S : SnT\ = 7.

Let 's pursue this a bi t further. Wri te D = S ( I T i n the above si tuation, and note that since |5 : D \ = 7 is the smallest prime divisor of \S\ = 7 3 , it follows by Prob lem 1A.1 , that D < S. Similar reasoning shows that also D < T , and hence S and T are both contained in N = N G (L>) . Now S and T are distinct Sylow 7-subgroups of N , and it follows that n 7 ( N ) > 1, and hence n 7 ( N ) > 8 by Corol lary 1.17. Since n 7 ( N ) is a power of 2 that

1 C 17

divides |AT | , we deduce that 2 3 divides \ N \ . Since also 7 3 divides \ N \ , we have \ G : N \ < 8.

We can use what we have established to show that a group G of order 21,952 cannot be simple. Indeed, if n 7 ( G ) = 1, then G has a normal subgroup of order 7 3 , and so is not simple. Otherwise, our subgroup N has index at most 8, and we see that | G | does not divide \ G : N \ \ . B y the n \ -theorem (Corol lary 1.3), therefore, G cannot be simple i f N < G. F ina l ly , \{ N = G then D < G and G is not simple in this case too.

In the last case, where D < G, we see that D is contained i n al l Sylow 7-subgroups of G, and thus D is the intersection of every two distinct Sylow 7-subgroups of G. In most situations, however, Theorem 1.16 can be used to prove only the existence of some pair of dist inct Sylow subgroups wi th a "large" intersection; it does not usually follow that every such pair has a large intersection.

To prove Theorem 1.16, we need the following.

1.18. L e m m a . L e t P € S y l p ( G ) , w h e r e G i s a finite g r o u p , a n d suppose t h a t Q i s a p - s u b g r o u p o / N G ( P ) . T h e n Q C P .

Proof. We apply Sylow theory in the group N = N G ( P ) . Clearly, P is a Sylow p-subgroup of N , and since P < N , we deduce that P is the only Sylow p-subgroup of N . B y the Sylow D-theorem, however, the p-subgroup Q of N must be contained in some Sylow p-subgroup. The only possibil i ty is Q C P , as required. •

A n alternative method of proof for L e m m a 1.18 is to observe that since Q C ^ N G ( P ) , it follows that Q P = P Q . T h e n Q P is a subgroup, and it is easy to see that it is a p-subgroup that contains the Sylow p-subgroup P . It follows that P = Q P 5 Q, as wanted.

P r o o f of T h e o r e m 1.16. Let S act on the set S y l p ( G ) by conjugation. One orbit is the set { S } , of size 1, and so i f we can show that a l l other orbits have size divisible by \S : S D T\, it w i l l follow that n p ( G ) = | S y l p ( G ) | = 1 m o d \S : S D T\, as wanted. Let O be an arbi trary S-orbit in S y l p ( G ) other than { S } and let P G O, so that P ^ S. B y the fundamental counting pr inciple, \ 0 \ = \S : Q\, where Q is the stabilizer of P in S under conjugation. T h e n Q C N G ( P ) , and so Q C P by L e m m a 1.18. B u t also Q C 5 , and thus | Q | < | 5 n P | < | 5 n T | ; where the latter inequali ty is a consequence of the fact that |5 D T\ is as large as possible among intersections of two distinct Sylow p-subgroups of G . It follows that \ 0 \ = \S : Q\ > \S : S D T\. B u t since the integers \ 0 \ and \S : S n T\ are powers of p and \ 0 \ > \S : S n T\, we conclude that \ 0 \ is a mult iple of \S : S n T\. T h i s completes the proof. •

18 1 . S y l o w T h e o r y

P r o b l e m s I C

1C.1 . Let P e S y l p ( G ) , and suppose that N G ( P ) C H C G , where H is a subgroup. Prove that H = N G { H ) .

Note . Th i s generalizes P rob lem 1B.3.

1C.2. Let H C G , where G is a finite group.

(a) If P € Sy l p (P" ) , prove that P = H n 5 for some member 5 G S y l p ( G ) .

(b) Show that n p (P f ) < n p ( G ) for a l l primes p.

1C.3. Let G be a finite group, and let X be the subset of G consisting of al l elements whose order is a power of p , where p is some fixed prime.

(a) Show that X = ( J S y l p ( G ) .

1C.4. Let \ G \ = 120 = 2 3-3-5. Show that G has a subgroup of index 3 or a subgroup of index 5 (or both) .

Hint . Ana lyze separately the four possibilities for n 2 ( G ) .

1C.5. Let P e Sylp(G) , where G = A p + U the alternating group on p + 1 symbols. Show that | N G ( P ) | = p(p - l ) / 2 .

H i n t . Count the elements of order p in G .

1C.6. Let G = PTif, where P" and K are subgroups, and fix a prime p.

(a) Show that there exists P € S y l p ( G ) such that P n H € S y l p ( P ) and P H f i £ S y l p ( X ) .

(b) If P is as in (a), show that P - ( P n H ) ( P n AT).

H i n t . For (a), first choose Q € S y l p ( G ) and g £ G such that Q n H e S y l p ( P ) and Q 9 n f i e S y l p ( X ) . Wr i te 0 = Ziifc, w i th h e H and k <E K .

1C.7 . Let G be a finite group i n which every max ima l subgroup has prime index, and let p be the largest prime divisor of | G | . Show that a Sylow p-subgroup of G is normal .

H i n t . Otherwise, let M be a max ima l subgroup of G containing N G ( P ) , where P e S y l p ( G ) . Compare n p { M ) and n p ( G ) .

I D 19

1C.8 . Let P be a Sylow p-subgroup of G. Show that for every nonnegative integer a, the numbers of subgroups of order p a in P and i n G are congruent modulo p .

Note . If p a = | P | , then the number of subgroups of order p a i n P is clearly 1, and it follows that the number of such subgroups i n G is congruent to 1 modulo p . Th i s provides a somewhat different proof that n p ( G ) = 1 mod p . It is true i n general that i f p a < \P\, then the number of subgroups of order p a in P is congruent to 1 modulo p , and thus it follows that i f p a divides the order of an arbi trary finite group G, then the number of subgroups of order p a in G is congruent to 1 mod p .

I D

We now digress from our study of Sylow theory i n order to review some basic facts about p-groups and nilpotent groups. A l so , we discuss the F i t t i n g subgroup, and in the problems at the end of the section, we present some results about the Fra t t in i subgroup.

A l t h o u g h p-groups are not at a l l typ ica l of finite groups in general, they play a prominent role in group theory, and they are ubiquitous in the study of finite groups. Th i s ubiqui ty is, of course, a consequence of the Sylow theorems, and perhaps that justifies our digression.

We should mention that although their structure is a typical when compared w i t h finite groups in general, p-groups are, nevertheless, extremely abundant in comparison wi th non-p-groups. There are, for example, 2,328 isomorphism types of groups of order 128 = 2 7 ; the number of types of order 256 = 2 8 is 56,092; for 512 = 2 9 the number is 10,494,213; and there are exactly 49,487,365,422 isomorphism types of groups of order 1,024 = 2 1 0 . (These numbers were computed by a remarkable a lgori thm for counting p-groups that was developed by E . O 'Br ien . )

There is an extensive theory of finite p-groups (and also of their infinite cousins, pro-p-groups), and there are several books entirely devoted to them. Our brief presentation here w i l l be quite superficial; later, we study p-groups a bi t more deeply, but s t i l l , we shall see only a t iny part of what is known.

Perhaps the most fundamental fact about p-groups is that nontr ivia l finite p-groups have nontr ivia l centers. ( B y our definition, "p-group" means "finite p-group", but we included the redundant adjective in the previous sentence and i n what follows i n order to stress the fact that finiteness is essential here. Infinite p-groups can have t r i v i a l centers, and in fact, they can be simple groups.)

In fact, a stronger statement is true.

20 1 . S y l o w T h e o r y

1.19. Theorem. L e t P be a finite p - g r o u p a n d l e t N be a n o n i d e n t i t y n o r m a l s u b g r o u p of P . T h e n N n Z ( P ) > 1. I n p a r t i c u l a r , if P i s n o n t n v i a l , t h e n Z { P ) > 1.

Proof. Since N < P , we can let P act on N by conjugation, and we observe that N f l Z ( P ) is exactly the set of elements of N that lie in orbits of size 1. B y the fundamental counting principle, every orbit has p-power size, and so each nontr iv ia l orbit ( i . e . , orbit of size exceeding 1) has size divisible by p . Since the set N - ( N n Z ( P ) ) is a union of such orbits, we see that \ N \ — \ N n Z ( P ) | is divisible by p , and thus \ N n Z ( P ) | = \ N \ = 0 mod

p , where the second congruence follows because TV is a nontr iv ia l subgroup. N o w N n Z ( P ) contains the identity element, and so \ N n Z ( P ) | > 0. It follows that \ N n Z ( P ) | > p > 1, and hence N n Z ( P ) is nontr ivial , as required. The final assertion follows by taking N = P . U

It is now easy to show that (finite, of course) p-groups are nilpotent, and thus we can obtain addit ional information about p-groups by s tudying general nilpotent groups. B u t first, we review some definitions.

A finite collection of normal subgroups N t of a (not necessarily finite) group G is a normal series for G provided that

1 = iVo C N i C • • • C N r = G.

This normal series is a central series if in addi t ion, we have N i / N i - i C Z ( G / J V i _ i ) for 1 < i < r . F ina l ly , a group G is nilpotent if it has a central series. It is worth noting that subgroups and factor groups of nilpotent groups are themselves nilpotent, al though we omit the easy proofs of these facts.

G iven any group G, we can attempt to construct a central series as follows. (But of course, this attempt is doomed to failure unless G is nilpotent.) We start by defining Z 0 = 1 and Z x = Z ( G ) . The second center Z 2 is defined to be the unique subgroup such that Z 2 / Z 1 = Z { G / Z 1 ) . (Note that Z 2 exists and is normal in G by the correspondence theorem.) We continue like this, inductively defining Z n for n > 0 so that Z n / Z n - i = Z ( G / Z „ _ i ) . The chain of normal subgroups

1 = Z 0 C Z i C Z 2 C • • •

constructed this way is called the upper central series of G. We hasten to point out, however, that in general, the upper central series may not actually be a central series for G because it may happen that Z { < G for a l l i. In other words, the upper central series may never reach the whole group G. B u t if Z r = G for some integer r , then { Z x \ 0 < i < r } is a true central series, and G is nilpotent.

I D 21

Conversely, i f G is nilpotent, the upper central series of G really is a central series. For finite groups G, this is especially easy to prove.

1.20. L e m m a . L e t G be finite. T h e n t h e f o l l o w i n g a r e e q u i v a l e n t .

(1) G i s n i l p o t e n t .

(2) E v e r y n o n t r i v i a l h o m o m o r p h i c i m a g e of G has a n o n t r i v i a l c e n t e r .

(3) G a p p e a r s as a member of i t s u p p e r c e n t r a l s e r i e s .

Proof. We have already remarked that homomorphic images of nilpotent groups are nilpotent. A l so , since the first nont r iv ia l term of a central series for a nilpotent group is contained i n the center of the group, it follows that nontr iv ia l nilpotent groups have nontr iv ia l centers. Th i s shows that (1) implies (2).

Assuming (2) now, i t follows that i f Z { < G, where is a term i n the upper central series for G, then Z l + 1 / Z i = Z { G / Z i ) is nontr ivia l , and thus Z i < Z i + 1 . Since G is finite and the proper terms of the upper central series are s t r ic t ly increasing, we see that not every term can be proper, and this establishes (3).

F ina l ly , (3) guarantees that the upper central series for G is actually a central series, and thus G is nilpotent, proving (1). •

If P is a finite p-group, then of course, every homomorphic image of P is also a finite p-group, and thus every nontr iv ia l homomorphic image of P has a nontr iv ia l center. It follows by L e m m a 1.20, therefore, that finite p-groups are nilpotent. In fact, we shall see in Theorem 1.26 that much more is true: a finite group G is nilpotent i f and only i f every Sylow subgroup of G is normal .

Next , we show that the terms of the upper central series of a nilpotent group contain the corresponding terms of an arbi t rary central series, and this explains why the upper central series is called "upper". It also provides an alternative proof of the impl ica t ion (1) => (3) of L e m m a 1.20, without the assumption that G is finite.

1.21. Theorem. L e t G be a ( n o t n e c e s s a r i l y finite) n i l p o t e n t g r o u p w i t h c e n t r a l series

l = N Q C N 1 C . . - C N r = G,

a n d as u s u a l , l e t

1 = Z 0 C Z x C Z 2 C • • •

be t h e u p p e r c e n t r a l series f o r G. T h e n N t C Z { f o r 0 < % < r , a n d i n p a r t i c u l a r , Z r = G.

22 1 . S y l o w T h e o r y

Proof. We prove that N i C Z i by induct ion on i. Since Z 0 = 1 = A^o, we can suppose that i > 0, and by the inductive hypothesis, we can assume that N i - i C Z i - i . For notat ional simplicity, write N = and Z = Z % - X , and observe that since N C Z , there is a natural surjective homomorphism 0 : G / N G / Z , defined by 0 (W 5 ) = Z<? for elements g e G. (This map is well defined since Z g is the unique coset of Z that contains N g , and so 0 { N g ) depends only on the coset N g and not on the element g . )

Since 9 is surjective, it carries central elements of G / N to central elements of G / Z , and since N t / N is central in G/N, it is mapped by 0 into Z ( G / Z ) = Z i / Z . If x € N i } therefore, i t follows that Zx = 0 { N x ) 6 Z x / Z , and thus x E Z h as required. •

If G is an arbi trary nilpotent group, then G is a term of its upper central series, which, therefore is a true central series. We have G = Z r for some integer r > 0, and the smallest integer r for which this happens is called the nilpotence class of G . Thus nontr iv ia l abelian groups have nilpotence class 1, and the groups of nilpotence class 2 are exactly the nonabelian groups G such that G / Z ( G ) is abelian.

Since the upper central series of the nilpotent group G is really a central series, we see that i f G has nilpotence class r , then it has a central series of length r . (In other words, the series has r containments and r + 1 terms.) B y Theorem 1.21, we see that G cannot have a central series of length smaller than r, and thus the nilpotence class of a nilpotent group is exactly the length of its shortest possible central series. In some sense, the nilpotence class can be viewed as a measure of how far from being abelian a nilpotent group is.

One of the most useful facts about nilpotent groups, and thus also about finite p-groups, is that "normalizers grow".

1.22. Theorem. L e t H < G , w h e r e G i s a ( n o t n e c e s s a r i l y finite) n i l p o t e n t g r o u p . T h e n N G ( H ) > H .

Before we proceed w i t h the (not very difficult) proof of Theorem 1.22, we digress to discuss the "bar convention", which provides a handy notat ion for dealing w i t h factor groups. Suppose that N < G, where G is an arbi trary group. We write G to denote the factor group G/N, and we th ink of the overbar as the name of the canonical homomorphism from G onto G . If g e G is any element, therefore, we write g to denote the image of g i n G , and so we see that g is s imply another name for the coset N g . A l so , i f H C G is any subgroup, then H is the image of H i n G .

B y the correspondence theorem, the homomorphism "overbar" defines a bijection from the set of those subgroups of G that contain A^ onto the set of al l subgroups of G . Every subgroup of G , therefore, has the form H for

I D 23

some subgroup H Q G , and although there_are usually many subgroups of G whose image i n G is the given subgroup H , exactly o n e o f them contains N . If H Q G is arbitrary, we see that H N = H N = H since overbar is a homomorphism and N is its kernel. _It follows that H N is the unique subgroup containing N whose image i n G is H . In part icular , since indices of corresponding subgroups are equal, we have \ G : H \ = \ G : A^P"! for al l subgroups H Q G .

The correspondence theorem also yields information_about normality. If AT C H Q K C G, then < K i f and only _if H < K . In part icular, i f N C H , then since 7f C N G ( / f ) , we see that # < N G ( P ) and we have N G ( H ) C N q ( H ) . In fact, equality holds here. To see this, observe that since Ng ( f f ) is a subgroup of G, it can be wr i t ten in_the form U for some (unique) subgroup U w i t h N Q U Q G. T h e n H < U, and so H < U and U C N G ( f f ) . Th i s yields N ^ T ? ) = t7 C N G ( 7 f ) , as claimed.

We w i l l use the bar convention i n the following proof.

P r o o f of T h e o r e m 1.22. Since G is nilpotent, it has (by definition) a cent ra l series { N i | 0 < i < r } , and we have NQ = 1 C H and N r = G H . It follows that there is some subscript k w i t h 0 < k < r such that N k C H but N k + 1 % H . We w i l l show that i n fact, A ^ f c + 1 C N G ( f f ) , and it w i l l follow that N G ( f f ) > H , as required.

Wr i t e G = G/Nk and use the bar convention. Since the subgroups N i form a central series, we have

N k V i C Z ( G ) C N ^ H ) = N G W ) ,

where the equality holds because N k C H . N o w because N k C N G ( f f ) , we can remove the overbars to obtain A T f c + i C N G ( H ) . The proof is now complete. •

We return now to p-groups, w i th another appl icat ion of Theorem 1.19.

1.23. L e m m a . L e t P be a finite p - g r o u p a n d suppose t h a t N < M a r e n o r m a l s u b g r o u p s of P . Then t h e r e exists a s u b g r o u p L < P such t h a t N C L C M a n d \ L : JV| = p .

Proof. W r i t e J P - P / N and note that M is nontr iv ia l and normal i n P . Now Z ( P ) i l l is nontr ivia l by Theorem 1.19, and so this subgroup contains an element of order p . (Choose any nonidenti ty element and take an appropriate power.) Because our element is central and of order p , it generates a normal subgroup of order p , and we can write I to denote this subgroup, where N C L . Now I C M , and thus AT C L C M , as wanted. Also , as L < P , we see that L < P . F ina l ly , \ L : N \ = | Z | = p , as required. •

24 1 . S y l o w T h e o r y

1.24. Corol lary. L e t P be a p - g r o u p of o r d e r p a . T h e n f o r every i n t e g e r b w i t h 0 0 and we work by induct ion on b. B y the inductive hypothesis, there exists a subgroup N < P such that |7V| = p b ~ l and we can apply Theorem 1.22 (with M = P ) to produce a subgroup L < P w i th \ L : N \ = p . T h e n | L | = p b and the proof is complete. •

Reca l l now that the Sylow E-theorem can be viewed as a par t ia l converse to Lagrange's theorem. It asserts that for certain divisors k of an integer n , every group of order n has a subgroup of order k. (The divisors to which we refer, of course, are prime powers k such that n / k is not divisible by the relevant prime.)

We can now enlarge the set of divisors for which we know that the converse of Lagrange's theorem holds.

1.25. Corol lary. L e t G be a f i n i t e g r o u p , a n d suppose t h a t p b d i v i d e s \ G \ , w h e r e p i s p r i m e a n d b > 0 i s a n i n t e g e r . T h e n G has a s u b g r o u p of o r d e r

p b .

Proof. Let P be a Sylow p-subgroup of G and write \ P \ = p a . Since p b

divides \ G \ , we see that b < a , and the result follows by Corol la ry 1.23. •

In fact, i n the si tuat ion of Corol la ry 1.25, the number of subgroups of G having order p b is congruent to 1 modulo p . ( B y P rob lem 1C.8 and the note following it , it suffices to prove this i n the case where G is a p-group, and while this is not especially difficult, we have decided not to present a proof here.) We mention also that it does not seem to be known whether or not there are any integers n other than powers of primes such that every group of order divisible by n has a subgroup of order n .

Sylow theory is also related to the theory of nilpotent groups in another way: a finite group is nilpotent i f and only if a l l of its Sylow subgroups are normal . In fact, we can say more.

1.26. Theorem. L e t G be a finite g r o u p . T h e n t h e f o l l o w i n g a r e e q u i v a l e n t .

(1) G i s n i l p o t e n t .

(2) N G ( H ) > H f o r every p r o p e r s u b g r o u p H < G.

(3) E v e r y m a x i m a l s u b g r o u p of G i s n o r m a l .

(4) E v e r y S y l o w s u b g r o u p of G i s n o r m a l .

(5) G i s t h e ( i n t e r n a l ) d i r e c t p r o d u c t of i t s n o n t r i v i a l S y l o w s u b g r o u p s .

I D 25

Note that in statement (3), a "maximal subgroup" is max ima l among p r o p e r subgroups. B u t i n most other situations, the word "maximal" does not imp ly proper. If a group G happens to be nilpotent, for example, then the whole group is a max ima l nilpotent subgroup of G.

To help w i t h the proof that (4) implies (5), we establish the following.

1.27. L e m m a . L e t X be a c o l l e c t i o n of finite n o r m a l s u b g r o u p s of a g r o u p G, a n d assume t h a t t h e o r d e r s of t h e members of X a r e p a i r w i s e c o p r i m e . T h e n t h e p r o d u c t H = \ \ X of t h e members of X i s d i r e c t . A l s o , \ H \ = T\ \X\.

xex

Proof. Cer ta in ly \ H \ < U\x\- A l so , by Lagrange's theorem, \X\ divides \ H \ , for every member X of X , and since the orders of the members of X are pairwise coprime, i t follows that f j \X\ divides \ H \ . We conclude that \ H \ = n i ^ l , as wanted.

N o w to see that T \ X ™ direct, it suffices to show that

X n Y [ { Y e x | Y + X } = l

for every member X EX. Th is follows since by the previous paragraph, the order of n Y for Y ^ X is equal to n \Y\, and this is coprime to | X | . •

P r o o f of T h e o r e m 1.26. We saw that (1) implies (2) i n Theorem 1.22. Tha t (2) implies (3) is clear, since i f M < G is a max ima l subgroup, then N G ( M ) > M , and so we must have N G ( M ) = G.

Now assume (3), and let P E S y \ p ( G ) for some prime p . If N G ( P ) is proper i n G, it is contained i n some m a x i m a l subgroup M , and we have M < G. Since P € S y l p ( M ) , it follows by L e m m a 1.13, the F ra t t i n i argument, that G = N G ( P ) M C M , and this is a contradict ion. Thus P < G, and this proves (4).

Tha t (4) implies (5) is immediate from L e m m a 1.27. N o w assume (5). It is clear that (4) holds, and it follows that (4) also holds for every homomor-phic image of G. (See Prob lem 1B.5.) Since we know that (4) implies (5), we see that every homomorphic image of G is a direct product of p-groups for various primes p . The center of a direct product , however, is the di rect product of the centers of the factors, and since nontr iv ia l p-groups have nontr iv ia l centers, i t follows that every nonidenti ty homomorphic image of G has a nontr iv ia l center. We conclude by L e m m a 1.20 that G is nilpotent, thereby establishing (1). •

Reca l l that O p ( G ) is the unique largest normal p-subgroup of G, by which we mean that it contains every normal p-subgroup. We define the Fi t t ing subgroup of G, denoted F ( G ) , to be the product of the subgroups

26 1 . S y l o w T h e o r y

O p { G ) as p runs over the prime divisors of G. O f course, F ( G ) is characteristic i n G, and i n part icular, i t is normal .

1.28. Corol lary. L e t G be a finite g r o u p . T h e n F ( G ) i s a n o r m a l n i l p o t e n t s u b g r o u p of G. I t c o n t a i n s every n o r m a l n i l p o t e n t s u b g r o u p of G, a n d so i t i s t h e u n i q u e l a r g e s t such s u b g r o u p .

Proof. B y L e m m a 1.27, we know that | F ( G ) | is the product of the orders of the subgroups O p ( G ) as p runs over the prime divisors of G . T h e n O p ( G ) € S y l p ( F ( G ) ) , and thus F ( G ) has a normal Sylow subgroup for each prime. It follows by Theorem 1.26 that F ( G ) is nilpotent.

Now let N < G be nilpotent. If P e S y l p ( i V ) , then P < N by Theorem 1.26, and thus P is characteristic i n N and hence is normal i n G . It follows that P C O p ( G ) C F ( G ) . Since N is the product of its Sylow subgroups and each of these is contained in F ( G ) , it follows that N C F ( G ) , and the proof is complete. •

1.29. Corol lary. L e t K a n d L be n i l p o t e n t n o r m a l s u b g r o u p s of a finite g r o u p G. T h e n K L i s n i l p o t e n t .

Proof. We have K C F ( G ) and L C F ( G ) , and thus K L C F ( G ) . Since F ( G ) is nilpotent, it follows that K L is nilpotent. •

P r o b l e m s I D

1D.1. Let P 6 S y l p ( P ) , where H C G , and suppose that N G ( P ) C H . Show that p does not divide | G : H \ .

Note. In these problems, we are, as usual, dealing w i t h finite groups.

1D.2. F i x a prime p , and suppose that a subgroup H C G has the property that C G ( x ) C H for every element x E H having order p . Show that p cannot divide both | P | and | G : P | .

1D.3. Le t H C G have the property that H n P 9 = 1 for al l elements </ <E G — P .

(a) Show that N G ( K ) C P for a l l subgroups # w i t h K i C i i .

(b) Show that H is a H a l l subgroup of G . (Recall that this means that \ H \ and | G : P | are coprime.)

Note . A subgroup H < G that satisfies the hypothesis of this problem is usually referred to as a Frobenius complement i n G . In Chapter 6, we give a different definition of a "Frobenius complement", which, i n fact, is equivalent to this one. It is easy to see (as we w i l l show) that a Frobenius complement i n the sense of Chapter 6 satisfies the hypothesis of this problem.

P r o b l e m s I D 27

W h a t is much more difficult is the theorem of Frobenius, which asserts that i f H is a Frobenius complement in G in the sense denned here, then it is, i n fact, a Frobenius complement i n the sense of Chapter 6. Unfortunately, al l known proofs of Frobenius ' theorem require character theory, and so we cannot present a proof i n this book.

1D.4. Let G = N H , where I < N < G and N n H — 1 . Show that H is a Frobenius complement i n G (as denned above) i f and only i f C N ( h ) = 1 for al l nonidenti ty elements h € H .

Hint . If x and x n lie i n H , where n € N , observe that x ~ x x n € N .

1D.5. Le t H < G, and suppose that N G ( P ) C H for a l l p-subgroups P C H , for al l primes p. Show that H is a Frobenius complement i n G .

Hint . Observe that the hypothesis is satisfied by H n H 9 for 5 6 G. If this intersection is nontr ivial , consider a nontr iv ia l Sylow subgroup Q o i H n H 9

and show that Q and Q 9 " 1

a r e conjugate i n H .

1D.6. Show that a subgroup of a nilpotent group is max ima l if and only i f it has prime index.

1D.7. For any finite group G, the Fratt in i subgroup $ ( G ) is the intersection of al l max ima l subgroups. Show that $ ( G ) is exactly the set of "useless" elements of G , by which we mean the elements g e G such that if ( X U { g } } = G for some subset X of G , then ( X ) = G.

1D.8. A finite p-group is elementary abelian i f it is abelian and every nonidenti ty element has order p . If G is nilpotent, show that G / $ ( G ) is abelian, and that if G is a p-group, then G / $ ( G ) is elementary abelian.

Note. It is not hard to see that i f P is a p-group, then $(P) is the unique normal subgroup of P m in ima l w i th the property that the factor group is elementary abelian.

1D.9. If P is a noncyclic p-group, show that \ P : $(P) | > p 2 , and deduce that a group of order p 2 must be either cyclic or elementary abelian.

ID.10. Let A be max ima l among the abelian normal subgroups of a p-group P . Show that A = C P { A ) , and deduce that \ P : A \ divides ( \ A \ - 1)!.

Hint . Let C = C P { A ) . If C > A , apply L e m m a 1.23.

28 1 . S y l o w T h e o r y

1D.11. Le t n be the m a x i m u m of the orders of the abelian subgroups of a finite group G . Show that | G | divides n \ .

Hint . Show that for each prime p , the order of a Sylow p-subgroup of G divides n \ .

Note . There exist infinite groups i n which the abelian subgroups have bounded order, so finiteness is essential here.

ID.12. Let p be a prime d iv id ing the order of a group G. Show that the number of elements of order p i n G is congruent to - 1 modulo p .

ID.13 . If Z C Z ( G ) and G / Z is nilpotent, show that G is nilpotent.

ID.14. Show that the F ra t t in i subgroup $ ( G ) of a finite group G is ni lpotent.

Hint . A p p l y the F ra t t in i argument. The proof here is somewhat similar to the proof that (3) implies (4) in Theorem 1.26.

Note. Th i s problem shows that $ ( G ) C F ( G ) .

ID.15. Suppose that $ ( G ) C N < G and that N / $ ( G ) is nilpotent. Show that N is nilpotent. In part icular, if G / $ ( G ) is nilpotent, then G is ni lpotent.

Note. Th i s generalizes the previous problem, which follows by setting N = $ ( G ) . Note that this problem proves that F ( G / $ ( G ) ) = F ( G ) / $ ( G ) .

ID.16. Let N < G, where G is finite. Show that $(JV) C $ ( G ) .

H i n t . If some max ima l subgroup M of G fails to contain $ ( jV) , then $ ( W ) A f = G , and it follows that N = < P ( N ) ( N n M ) .

1D.17. Le t N < G, where AT is nilpotent and G/AT ' is nilpotent. Prove that G is nilpotent.

H i n t . The derived subgroup N ' is contained i n $(JV) by P rob lem 1D.8.

Note . If we weakened the assumption that G/N' is nilpotent and assumed instead that G / N is nilpotent, i t would not follow that G is necessarily nilpotent.

1D.18. Show that F ( G / Z ( G ) ) = F ( G ) / Z ( G ) for a l l finite groups G .

ID.19. Le t F = F ( G ) , where G is an arbi trary finite group, and let C = C G { F ) . Show that G / ( G D F ) has no nontr ivia l abelian normal subgroup.

Hint . Observe that F ( G ) < G .

I E 29

I E

F r o m the earliest days of group theory, researchers have been intr igued by the question: what are the finite simple groups? O f course, the abelian simple groups are exactly the groups of prime order, and (up to isomorphism) there is just one group of order p for each prime p : the cyclic group of that order. B u t n o n a b e l i a n simple groups are comparat ively rare. There are, for example, only five numbers less than 1,000 that occur as orders of nonabelian simple groups, and up to isomorphism, there is just one simple group of each of these orders. (These numbers are 60, 168, 360, 504 and 660.) There do exist numbers, however, such that there are two nonisomorphic simple groups of that order. (The smallest such number is 8!/2 = 20,160.) B u t no number is the order of three nonisomorphic simple groups.

Perhaps their rar i ty is one reason that nonabelian finite simple groups have inspired such intense interest over the years. It seems quite natural to collect rare objects and to attempt to acquire a complete collection. B u t a more "pract ical" explanation is that a knowledge of al l finite simple groups and their properties would be a major step i n understanding a l l finite groups. The reason for this is that, in some sense, al l finite groups are bui l t from simple groups.

To be more precise, suppose that G is any nontr iv ia l finite group. B y finiteness, G has at least one max ima l normal subgroup N . (We mean, of course, a max ima l p r o p e r normal subgroup, but as is customary i n this context, we have not made.the word "proper" explicit .) T h e n by the correspondence theorem, the group G / N is simple. (This is because the normal subgroups of G / N are in natural correspondence w i t h the normal subgroups of G that contain N , and there are just two of these: N and G.) Now i f N is nontr iv ia l , we can repeat the process by choosing a max ima l normal subgroup M of N . (In general, of course, M w i l l not be normal i n G.) Because G is finite, we see that i f we continue like this, repeatedly choosing a maximal normal subgroup of the previously selected group, we must eventually reach the identity subgroup. If we number our subgroups from the bo t tom up, we see that we have constructed (or more accurately "chosen") a chain of subgroups N such that

1 = j V 0 < N x < • • • < N r = G

such that each of the factors A ^ - i is simple for 1 < i < r . In this situation, the subgroups iV ? are said to form a composit ion series for G, and the simple groups N / N - x are the corresponding composition factors. The Jordan-Holder theorem asserts that despite the arbitrariness of the construction of the composit ion series, the set of composi t ion factors ( including mult ipl ici t ies) is uniquely determined up to isomorphism. The

30 1 . S y l o w T h e o r y

composi t ion factors of G are the simple groups from which we might say that G is constructed. (We mention that the finite groups for which al l composi t ion factors are cyclic of prime order are exactly the "solvable" finite groups, which we study i n more detail i n Chapter 3.)

The problem of finding al l nonabehan finite simple groups can be approached from two directions: construct as many simple groups as you can, and prove that every finite nonabelian simple group appears i n your list. In part icular , as part of the second of these programs, it is useful to prove "nonsimplici ty" theorems that show that groups that look different from the known simple groups cannot, i n fact, be simple. A major result of this type from early i n the 20th century is due to W . Burnside, who showed that the order of a nonabelian simple group must have at least three different prime divisors. (This is Burnside 's classic pY-theorem.)

Burnside also observed that al l of the then known nonabelian simple groups had even order. He conjectured that this holds in general: that al l odd-order simple groups are cyclic of prime order. Burnside 's conjecture, which can be paraphrased as the assertion that every group of odd order is solvable, was eventually proved by W . Feit and J . G . Thompson i n the early 1960s. (The celebrated Fei t -Thompson paper, at about 250 pages, may have been the longest published proof of a single theorem at the time.) Since around 1960, there has been dramatic progress w i t h bo th aspects of the simple-group-classification problem, and it appears that now, i n the early years of the 21st century, the classification of finite simple groups is complete.

So what are the nonabehan finite simple groups? A highly abbreviated description is this. Every finite nonabelian group is either:

(1) One of the alternating groups A n f o r n > 5,

(2) A member of one of a number of infinite families parameterized by prime-powers q and (usually) by integers n > 2, or

(3) One of 26 other "sporadic" simple groups that do not fit into types (1) or (2).

O f the simple groups i n parameterized families, the easiest to describe are the projective special linear groups P S L ( n , q ) , where q is a prime-power and n > 2. These are constructed as follows. Let F be the field of order q and construct the general linear group G L ( n , q ) consisting of al l invertible n x n matrices over F . The special linear group S L ( n , q ) is the normal subgroup of G L { n , q ) consisting of those matrices w i t h determinant 1. It is not hard to see that the center Z = Z ( S L ( n , q ) ) consists exactly of the scalar matrices of determinant 1, and by definition, P S L ( n , q ) is the factor group S L ( n , q ) / Z . It turns out that P S L ( n , q ) is simple except

I E 31

when n = 2 and q is 2 or 3. (We say more about the groups P S L ( n , q ) i n Chapter 7, and we prove their s impl ic i ty i n Chapter 8, where we also prove that the alternating groups A n are simple for n > 5.)

In fact, a l l of the simple groups w i t h order less than 1,000 are of the form P S L ( 2 , q ) , where q is one of 5, 7, 8, 9 or 11, and the corresponding group orders are 60, 168, 504, 360 and 660. The unique (up to isomorphism) simple group P S L { 2 , 5) of order 60 has two other realizations: it is isomorphic to P S L ( 2 , 4 ) and also to the al ternat ing group A 5 . The simple groups P S L ( 2 , 7) of order 168 and P S L ( 2 , 9) of order 360 also have mult iple realizations: the first of these is isomorphic to P S L ( 3 , 2) and the second is isomorphic to A 6 .

The smallest of the 26 sporadic simple groups is the smal l M a t h i e u group, denoted M n , of order 7,920; the largest is the Fischer-Griess "monster" of order

2 4 6 -3 2 0 -5 9 -7 6 - l l 2 -17-19-23-29-31-41.47-59-71 =

808,017,424,794,512,875,886,459,904,961,710,757,005,754,368,000,000,000.

For the remainder of this section, we discuss nonsimpl ic i ty theorems of the form: " i f the order of G is . . . , then G cannot be simple". (Of course, bo th Burnside 's pY-theorem and the Fe i t -Thompson odd-order theorem are of this type.) A very much more elementary result of this form is immediate from the fact that a nontr iv ia l p-group always has a nont r iv ia l center. If | G | is divisible by only one prime, it follows that G cannot be simple unless Z ( G ) = G , and i n this case, G is abelian, and so i t must be cyclic of prime order.

Burnside 's pV-theorem asserts that the order of a simple group cannot have exactly two prime divisors. His beautiful (and short) proof is not elementary since it uses character theory (which we do not discuss i n this book). It took about 50 years before a purely group-theoretic (and harder) proof of Burnside 's theorem was found by Goldschmidt , Ma t suyama and Bender, using powerful techniques of Thompson , Glauberman and others. These techniques were developed for other purposes, and eventually they led to the full classification of finite simple groups, but as a test of their power, it seemed reasonable to see if they would y ie ld a direct proof of Burnside 's theorem. Indeed they d id , and one of the goals of this book is to develop enough group theory so that we can present a somewhat simplified version of the Goldschmidt-Matsuyama-Bender proof of Burnside 's theorem. (This proof appears i n Chapter 7.)

For now, however, we use Sylow theory (and a few tricks) to prove some much more elementary nonsimplici ty theorems.

32 1 . S y l o w T h e o r y

1.30. Theorem. L e t \ G \ = p q , w h e r e q 1, we have n p > p > q. Th i s is not possible, however, since n p must divide q. We conclude that n p = 1, as wanted. A l so , since every element of G of order p generates a subgroup of order p , and there is only one of these, there are exactly p - 1 elements of order p i n G.

N o w if n , > 1, then p = n q = 1 mod q, and thus q divides p - 1 . Otherwise n q = 1, and we see that there are exactly q - 1 elements of order q i n G. Together w i th the identity, we have now accounted for p + q - 1 elements of G. B u t p + q - 1 < 2 p < p q = \ G \ , and so G must have some element g that does not have order either 1, g or p. Since o { g ) divides \ G \ = p q , the only possibil i ty is that o ( g ) = p q , and thus G = {pq) is cyclic . •

1.31. Theorem. L e t \ G \ = p 2 q , w h e r e p a n d q a r e p r i m e s . T h e n G has e i t h e r a n o r m a l S y l o w p - s u b g r o u p o r a n o r m a l S y l o w q - s u b g r o u p .

Proof. We assume that n p > 1 and n , > 1, and we note that p ^ q. Since n p = 1 m o d p , it follows that n p > p and s imilar ly n q > q.

Now n p must divide q, and hence n p = q and we have n q > q = n p > p . B u t n q divides p 2 , and we see that the only possibil i ty is that n q = p 2 , which means that G has p 2 subgroups of order q.

If Q i ^ Q2 are subgroups of order q, then since q is prime, it follows v ia Lagrange's theorem that Ql n Q2 = 1. The p 2 subgroups of order q i n G, therefore, have no nonidentity elements in common, and it follows that G has at least p 2 { q - 1) elements of order q. (Actual ly , since every element of order q i n G must lie i n some subgroup of order q, it follows that G has e x a c t l y p 2 ( q - l ) elements of order q, but we shall not need that fact.) The number of elements of G that do not have order q is then at most \ G \ - p 2 { q - 1) = p 2 . If P e Sylp(G) , then, of course, no element of P has order q, and since | P | = p 2 , it follows that P is exactly equal to the set of elements of G that do not have order q. In part icular, P is uniquely determined, and this is a contradict ion since we are assuming that n p > 1. •

B y the previous two theorems, we see that if p ^ q are primes, then a group of order p q or of order p 2 q must either have a normal Sylow p -subgroup or a normal Sylow g-subgroup. Th i s almost works for groups of order p 3 q too, but there is an exception. Th i s s i tuat ion is endemic i n finite group theory: some general fact holds w i t h a smal l number of exceptions.

I E 33

(Another example of this phenomenon is that a l l automorphisms of the symmetr ic group Sn are inner for a l l integers n > 1 except n = 6.)

1.32. T h e o r e m . L e t \ G \ = p 3 q , w h e r e p a n d q a r e p r i m e s . T h e n G has e i t h e r a n o r m a l S y l o w p - s u b g r o u p o r a n o r m a l S y l o w q - s u b g r o u p , except w h e n \ G \ = 24.

Proof. A s in the previous proof, we assume that n p > 1 and that n q > 1, and we conclude that p ^ q and that n p > p and n q > q. Cont inu ing to reason as we d id previously, we have n q > q = n p > p . Since n q divides p 3 , we are left this t ime w i t h two possibilities: either n q = p 3 or n q = p 2 .

Suppose that n q = p 3 . T h e n the p 3 subgroups of order q contain a to ta l of p 3 { q - l ) elements of order q, and thus G c&ntains at most \ G \ - p 3 { q - l ) = p 3

elements that do not have order q. It follows that i f P e S y l p ( G ) , then P is exactly the set of elements of G having order different from q, and thus P is unique, contradict ing the assumption that n p > 1.

We conclude, therefore, that n q = p 2 , and thus p 2 = 1 mod q. In other words, q divides p 2 - 1 = (p+ l ) (p - 1). Since q is prime, we see that q must divide either p + 1 or p - 1, and so i n either case, we have q < p + l . B u t we know from the first paragraph of the proof that p < q, and so the only possibi l i ty is that q = p + l . Now p and q are primes that differ by 1, and so one of them must be even. We deduce that p = 2 and q = 3, and thus \ G \ = 24. •

O f course, the previous proof does not te l l us that there actually is an exception of order 24; it only permits the possibi l i ty of such an exception. B u t an exception really does exist. Consider S4, the symmetr ic group on four symbols, which, of course, has order 4! = 24. B y examining the possible cycle structures for elements of S4, it is easy to count that there are exactly nine elements of order 2, eight elements of order 3 and six elements of order 4. (As a check, note that 9 + 8 + 6 = 23, so that together w i t h the identity, we have accounted for al l 24 elements.) B y the Sylow D-theorem, every element of 2-power order i n S4 lies in some Sylow 2-subgroup. Since each Sylow 2-subgroup has order 8, and yet there are more than eight elements of 2-power order, it follows that S4 must have more than one Sylow 2-subgroup. S imi lar reasoning shows that there must be more that one Sylow 3-subgroup. (In fact, i t is easy to see that n2(S4) = 3 and that n3(S4) = 4.)

There is s t i l l more that we can say about this si tuation. We know that among groups of order p 3 q , only groups of order 24 can fail to have a nontr iv ia l normal Sylow subgroup, and we know that at least one group of order 24 actually does fail to have such a Sylow subgroup. The remaining question is whether or not there are any other groups of order 24 that can occur as exceptions. The answer is "no".

34 1 . S y l o w T h e o r y

1.33. Theorem. L e t \ G \ = 24, a n d suppose n 2 ( G ) > 1 a n d n 3 ( G ) > 1. T h e n G^S4.

Proof. Since n 3 exceeds 1, is congruent to 1 modulo 3 and divides 8, the only possibi l i ty is that n 3 = 4, and thus \ G : N \ = 4, where N = N G ( P ) and P € S y l 3 ( G ) . N o w let K = c o r e G ( N ) , so that G / K is isomorphic to a subgroup of S4 by Theorem 1.1. It suffices to show that K = 1 since that w i l l imp ly that G is isomorphic to a subgroup of 5 4 , and this w i l l complete the proof since \ G \ = 24 = \SA\.

Reca l l that K C iV = N G ( P ) , where P G S y l 3 ( G ) . T h e n P is a normal Sylow 3-subgroup of J T P , and so P is characteristic in K P . Since P is not normal i n G, however, we conclude that K P cannot be normal in G. B u t K P / K is a Sylow 3-subgroup of G/K, and it follows that n 3 { G / K ) > 1. In part icular, G / K is not a 2-group, and so \ K \ is not divisible by 3.

Since \ G : N \ = 4, we see that |TV| = 6, and hence the only possibilities are \ K \ = 1, as wanted, or \ K \ = 2. Assuming now that \ K \ = 2, we work to obtain a contradiction. Since \ G / K \ = 12, Theorem 1.31 applies, and we deduce that either n 2 { G / K ) = 1 or n 3 { G / K ) = 1. Since we have seen that n 3 { G / K ) > 1, however, we conclude that G / K has a unique Sylow 2-subgroup S/K. T h e n S/K < G/K, and so S < G. A l so | 5 | = 8 since \ K \ = 2 and \S/K\ = 4, and thus S is a normal Sylow 2-subgroup of G. T h i s contradicts the hypothesis that n 2 { G ) > 1. •

We mention that the group S4, of order 24, is not simple, and it follows that no group of order p 3 q is simple, where p and q are primes. In fact, S4 has a normal subgroup of order 12 and one of order 4. T h e subgroup of order 12 is the alternating group A 4 (see the discussion below) and the normal subgroup of order 4 is the so-called K l e i n group consisting of the identity and the three elements of order 2 i n S4 that have no fixed points.

We can eliminate a quarter of a l l positive integers as possibilities for the order of a simple group. The argument relies on the notion of odd and even permutations, w i t h which, we assume the reader is familiar.

Briefly, the facts are these. Every permutat ion of a finite set can be wri t ten as a product of transpositions (2-cycles). A permutat ion that can be wri t ten as a product of an even number of transpositions is even and one that can be wri t ten as a product of an odd number of transpositions is odd. It is a t r iv ia l i ty that the even permutations i n the symmetric group Sn form a subgroup. (It is called the alternating group and is denoted A n . ) It is also true, but much less t r iv i a l to prove, that no permutat ion can be both even and odd, and since odd permutations certainly exist for n > 2, we see that A n is proper i n Sn in this case.

I E 35

M o r e generally, suppose that G is an arbi t rary group that acts on some finite set ft, and assume that there is at least one element x G G that induces an odd permutat ion on ft. It should be clear that the elements of G that induce even permutations on ft form a subgroup, which we cal l H . Since x i H , we see that H < G, and we c la im that \ G : H \ = 2. To see this, it suffices to show that the set G - H is contained i n a single right coset of H . We show, i n fact, that i f g G G - H , then g G E x - 1 . N o w g induces an odd permutat ion on ft, and hence gx induces an even permutat ion and we have gx G H . T h e n g G H x ~ l , as claimed. Since a subgroup of index 2 is automatical ly normal , we have established the following useful fact.

I. 34. L e m m a . L e t G a c t o n a finite set ft a n d suppose t h a t some element o f G acts " o d d l y " . T h e n G has a n o r m a l s u b g r o u p of i n d e x 2. •

1.35. Theorem. Suppose t h a t \ G \ = I n , w h e r e n i s o d d . T h e n G has a n o r m a l s u b g r o u p of i n d e x 2.

Proof. B y Cauchy's theorem, we can find an element t G G of order 2. In the regular action of G on itself by right mul t ip l ica t ion , t has no fixed points, and so the permutat ion induced by t consists entirely of 2-cycles, and there are \ G \ / 2 = n of them. Since n is odd, the permutat ion induced by t is odd, and the result follows by L e m m a 1.34. •

We have stated that 60 is the smallest number that occurs as the order of a nonabelian simple group. In fact, using what we have now proved, this is not very hard to see. F i r s t , we can eliminate a l l powers of primes since i f G is a simple p-group, then because 1 < Z ( G ) < G, we see that Z ( G ) = G and G is abelian. A l so , by Theorem 1.30, we can eliminate a l l products of two primes, and thus we can assume that | G | factors as a product of at least three primes, counting mult ipl ici ty .

Suppose now that \ G \ < 60 and write n = \ G \ . If n is odd, then, since 3 4 , 3 2 -7 and 3-5 2 a l l exceed 60, we see that the only possibilities that are products of at least three primes are n = 3 3 and n = 3 2 -5, and we have seen that neither of these numbers can be the order of a simple group.

We are left w i t h the cases where n is even. B y Theorem 1.35, we can suppose that n is divisible by 4, and we write n = 4 m , where m < 15. We have el iminated the cases where m is prime or is a power of 2, and also the cases where m = 2 p , where p is an odd prime. The the only surviving possibilities for m are 9 and 12, and so we need to show that there is no simple group of order 36 or of order 48.

If G is simple of order 36 = 2 2 - 3 2 , we see that | G : P \ = 4, where P G S y l 3 ( G ) , and yet | G : P \ does not divide 4! = 24. Th i s violates the n!-theorem, Coro l la ry 1.3. Similar ly, i f | G | = 48 = 2 4 -3 , we get a v iola t ion of

36 1 . S y l o w T h e o r y

the n!-theorem by taking P € S y l 2 ( G ) , so that | G : P \ = 3. Th i s completes the proof that 60 is the smallest possible order of a nonabelian simple group.

To prove that no number between 60 and 168 can be the order of a simple group, it is convenient to have two more nonsimplici ty results of the type we have been discussing: if G has order p 2 q 2 or order p q r , where p , q and r are primes, then G cannot be simple. (We leave the proofs of these to the exercises at the end of this section.) Us ing these two facts and the other results that we have established, a l l but about a dozen of the numbers between 60 and 168 can be el iminated, and most of those are easily dealt w i t h by a d h o c methods. For example, we need to consider the numbers 84 = 2 2-3-7, 140 = 2 2-5-7 and 156 = 2 2-3-13, and these are easily el iminated by the observation that i f p is the largest prime divisor, then n p = 1 since no divisor of n exceeding 1 is congruent to 1 modulo p .

The number 72 = 2 3 - 3 2 can be el iminated by observing that i f G is simple and \ G \ = 72, then n 3 > 1, and therefore the only possibil i ty is n 3 = 4. T h e n \ G : N \ = 4, where N is the normalizer of a Sylow 3-subgroup, and this violates the n!-theorem. More interesting is the case where \ G \ = 144 = 2 4 - 3 2 . Th i s case can be handled using Theorem 1.16, w i t h an argument analogous to the one we used i n the discussion following Corol la ry 1.17, where we showed that groups of order 2 6 - 7 3 are not simple. Another number that requires some work is 132 = 2 2 - 3 - l l . If a group of this order is simple, then n n = 12, and hence there are at least 120 elements of order 11, leaving at most 12 other elements. It follows that n 3 = 4, and this yields a contradict ion using the n!-theorem.

The number between 60 and 168 that seems to be the most difficult to eliminate is 120 = 2 3-3-5. One (rather t r icky) approach is this. If G is simple of order 120, it is easy to see that n 5 ( G ) = 6. T h e n G acts (nontrivially) on the six right cosets of the normalizer of a Sylow 5-subgroup, and this determines a nontr iv ia l homomorphism from G into the symmetric group 5 6 . Since G is simple, this homomorphism is injective, and thus there is an isomorphic copy H of G w i t h H C S 6 . B u t G has no normal subgroup of index 2, and hence by L e m m a 1.34, no element of G can act oddly on the six cosets, and thus H actually lies i n the al ternating group A 6 , and since \ A 6 \ = 360, we see that \ A 6 : H \ = 3. Recal l , however, that the alternating groups A n are simple when n > 5, and so by the n!-theorem, A 6 cannot have a subgroup of index 3. Th is contradiction shows that 120 is not the order of a simple group.

A n alternative approach for el iminat ing 120 is to quote P rob lem 1C.4, which asserts that if \ G \ = 120, then there exists a subgroup H C G w i th 1 < \ G : H \ < 5. If G is simple, this yields an isomorphism of G into the symmetric group S5. Since |55| = 120 = \ G \ , we conclude that G = S5,

P r o b l e m s I E 37

and this is a contradict ion since S5 has a normal subgroup of index 2: the alternating group A 5 .

We close this section wi th one further result which, though somewhat more general than those that we have already proved, is s t i l l only a very special case of Burnside 's pV-theorem.

1.36. Theorem. Suppose t h a t \ G \ = p a q , w h e r e p a n d q a r e p r i m e s a n d a > 0. T h e n G i s n o t s i m p l e .

Proof. We can certainly assume that p ^ q and that n p > 1, and thus n p = q. N o w choose dist inct Sylow p-subgroups S and T of G such that \S f l T\ is as large as possible, and write D = S n T.

If D = 1, then every pair of distinct Sylow p-subgroups of G intersect t r iv ia l ly , and so the Sylow p-subgroups of G account for a to ta l of q { p a - 1) nonidenti ty elements of G. A l l of these elements, of course, have orders divisible by p , and this leaves room for at most \ G \ - q { p q - 1) = q elements w i t h order not divisible by p . It follows that i f Q G S y l , ( G ) , then Q is exactly the set of these elements, and i n part icular Q is unique, and so Q < G and G is not simple, as required.

We can now assume that D > 1 , and we let N = N G ( D ) . Since D < S and D < T and we know that "normal i zes grow" i n p-groups, we can conclude that N Ci S > D and N (~]T > D .

Next , we show that N is not a p-group. Otherwise, by the Sylow D -theorem, we can write N C Re S y l p ( G ) . T h e n R n S D N n S > D , and so by the choice of D , we see that the Sylow subgroups R and S cannot be dist inct . In other words, S = R and similarly, T = R. B u t this is a contradict ion since S ^ T .

It follows that q divides | jV | , and so i f we choose Q G S y l , ( i V ) , we have | Q | = q. Now S is a p-group and Q is a g-group, and thus S n Q = 1 and we have \SQ\ = \S\\Q\ = p a q = \ G \ . T h e n SQ = G, and hence if g G G is arbitrary, we can write g = xy, where x G S and y G Q. T h e n S g = Sxy = sv Z} D V = D , where the second equality holds since x G S and the last equality follows because y e Q C N = N G ( £ > ) . We see, therefore, that D is contained in every conjugate of S i n G. T h e n D is contained in every Sylow p-subgroup of G, and we have 1 < D C O p ( G ) . Thus O p ( G ) is a nonidenti ty proper normal subgroup G , and this completes the proof that G is not simple. •

P r o b l e m s I E

1E.1. Le t | G | = p 2 q 2 , where p < q are primes. Prove that n q { G ) = 1 unless \ G \ = 36.

38 1 . S y l o w T h e o r y

Note . If | G | = 3 6 , then a Sylow 3 subgroup really can fail to be normal , but i n that case, one can show that a Sylow 2-subgroup of G is normal .

1E.2. Let \ G \ = p q r , where p < q < r are primes. Show that n r ( G ) = 1.

Hint . Otherwise, show by counting elements that a Sylow g-subgroup must be normal and consider the factor group, of order p r .

Note . In general, i f \ G \ is a product of distinct primes, then a Sylow subgroup for the largest prime divisor of \ G \ is normal . We prove this theorem of Burnside when we study transfer theory i n Chapter 5.

1E.3. Show that there is no simple group of order 315 = 3 2-5-7.

Note. Th i s , of course, is also a consequence of the Fei t -Thompson odd-order theorem, which asserts that no group of odd nonprime order can be simple.

1E.4. If \ G \ = 144 = 2 4 - 3 2 , show that G is not simple.

1E.5. If \ G \ = 336 = 2 4-3-7, show that G is not simple.

Hint . If G is simple, compute n 7 ( G ) and use Prob lem 1C.5.

1E.6. If \ G \ = 180 = 2 2 -3 2 -5 , show that G is not simple.

1E.7. If \ G \ = 240 = 2 4-3-5, show that G is not simple.

1E.8. If \ G \ = 252 = 2 2 -3 2 -7 , show that G is not simple.

I F

Reca l l that O p ( G ) is the unique largest normal p-subgroup of the finite group G, and that it can be found by taking the intersection of a l l of the Sylow p-subgroups of G. B u t do we really need al l of them? W h a t is the smallest collection of Sylow p-subgroups of G w i t h the property that their intersection is O p ( G ) ? In the case where a Sylow p-subgroup is abelian, there is a pretty answer, which was found by J . S. Brodkey. (Of course, since the Sylow p-subgroups of G are conjugate by Sylow C-theorem, they are isomorphic, and so if any one of them is abelian, they a l l are.)

1.37. T h e o r e m (Brodkey) . Suppose t h a t a S y l o w p - s u b g r o u p of a finite g r o u p G i s a b e l i a n . T h e n t h e r e e x i s t S,T e S y l p ( G ) such t h a t S n T = O p ( G ) .

O f course, every intersection of two Sylow p-subgroups of G contains O p ( G ) , so what Brodkey 's theorem really says is that if the Sylow subgroups are abelian, then O p ( G ) is the unique min ima l such intersection. In fact, what is essentially Brodkey 's argument establishes something about min ima l intersections of two Sylow p-subgroups even i f the Sylow subgroups are not abelian.

I F 39

1.38. Theorem. F i x a p r i m e p , a n d l e t G be any finite g r o u p . C h o o s e S,T e Sylp(G) such t h a t D = S n T i s m i n i m a l i n t h e set of i n t e r s e c t i o n s of t w o S y l o w p - s u b g r o u p s o f G . T h e n O p ( G ) i s t h e u n i q u e l a r g e s t s u b g r o u p of D t h a t i s n o r m a l i n b o t h S a n d T.

If we assume that S and T are abelian i n Theorem 1.38, we see that D < S and D < T , and so i n this case, the theorem guarantees that D = O p ( G ) . In other words, Brodkey 's theorem is an immediate corollary of Theorem 1.38.

P r o o f of T h e o r e m 1.38. Let K C D , where K < S and K < T. We must show that K C O p ( G ) , or equivalently, that K C P for a l l Sylow p-subgroups P of G.

Let N = N G { K ) and note that S C N , and thus S G S y l p ( iV ) . Now let P G Sylp(G) and observe that F i l i V i s a p-subgroup of N . T h e n P n iV is contained i n some Sylow p-subgroup of N , and so we can write P f ] N C Sx

for some element x € N . (We are, of course, using the Sylow D-and C -theorems i n the group N . ) N o w T C N , and thus also Tx C N and we have

p n T x = P n N r \ T x c s x n T x = D x .

T h e n D = ( D x ) x ~ 1 D ( P n ^ f ' 1 = P x ~ x n T .

Since P X ~ X and T are Sylow p-subgroups of G , and their intersection is contained in D , it follows by the min imal i ty of D that P x ~ ' n T = D . In part icular , we have J f C D C P * - 1 , and thus i f * C P . B u t K x = K since x € iV = N G ( J f ) , and thus K C P , as required. •

1.39. Corol lary. L e i P G S y l p ( G ) , w/tere G is a finite g r o u p , a n d assume t h a t P i s a b e l i a n . T h e n | G : O p ( G ) \ < \ G : P | 2 .

Proof. B y Brodkey 's theorem, we can choose S , T G S y l p ( G ) such that S H T = O p ( G ) . T h e n

| G | - | 6 T | " \snr\ - \ O p ( G ) \ •

T h i s yields | G | / | P | 2 > l / | O p ( G ) | , and the result follows by mul t ip ly ing bo th sides of this inequali ty by | G | . •

We can recast this as a "good" nonsimpl ic i ty theorem.

1.40. Corol lary. L e t P G S y l p ( G ) , w h e r e G i s finite a n d P i s a b e l i a n , a n d assume t h a t \ P \ > | G | 1 / 2 . T h e n O p ( G ) > 1 , a n d thus G i s n o t s i m p l e unless | G | = p .

40 1 . S y l o w T h e o r y

Proof. We have \ G : P \ 2 = | G | 2 / | P | 2 < \ G \ , and thus O p ( G ) > 1 by Coro l la ry 1.39. •

We mention that the conclusion of Brodkey 's theorem definitely can fail if the Sylow p-subgroups are nonabelian. A counterexample is a certain group G of order 144 = 2 4 - 3 2 for which 0 2 ( G ) = 1. Tak ing p = 2, we see that the conclusion of Corol lary 1.40 fails for this group, and thus the conclusion of Brodkey 's theorem must also fail. O f course, the Sylow 2-subgroups of G are necessarily nonabelian in this case.

Th i s counterexample of order 144 can be constructed as follows. Let E be elementary abelian of order 9. (In other words, E is the direct product of two cyclic groups of order 3.) T h e n E can be viewed as a 2-dimensional vector space over a field of order 3, and thus the full automorphism group of E is isomorphic to G L ( 2 , 3 ) , which has order 48. A Sylow 2-subgroup T of this automorphism group, therefore, has order 16, and of course, since T C Aut(£7) , we see that T acts on E and that this action is faithful.

Us ing the "semidirect product" construction, which we discuss i n Chapter 3, we can bu i ld a group G containing (isomorphic copies of) E and T , where E < G = E T , and the conjugation action of T on E w i t h i n G is identical to the original faithful action of T on E . Now 0 2 ( G ) < G and E < G , and since 0 2 ( G ) is a 2-group and E is a 3-group, we see that 0 2 ( G ) f l £ = l . It follows that 0 2 ( G ) centralizes E , and thus 0 2 ( G ) is contained in the kernel of the conjugation action of T on E . Th is action is faithful, however, and we deduce that 0 2 ( G ) = 1, and indeed we have a counterexample.

P r o b l e m s I F

1F.1. In the last paragraph of this section, we used the fact that i f M < G and N < G and M D N = 1, then M and N centralize each other. Prove this.

H i n t . Consider elements of the form m - ^ m n , where m e M and n € N .

Note. Reca l l that if x,y e G, then the element x ~ l y - l x y € G is the commutator of x and y . It is customary to write [ x , y ] to denote this element. Note that [ x , y] = 1 i f and only if xy = yx. We study commutators i n more detail in Chapter 4.

1F.2. Le t G be a group, and fix a prime p . Show that if O p ( G ) = 1, then there exist S , T e S y l p ( G ) such that Z (5 ) D Z ( T ) = 1.

1F.3. Let G = N P , where N < G , P < E S y l p ( G ) and N n P = 1, and assume that the conjugation action of P o n N is faithful. Show that P acts faithfully on at least one orbit of this action.

1 G 41

Hint . In fact, if x e N is chosen so that | P n P x \ is as smal l as possible, then P acts faithfully on the P -o rb i t containing x .

1 G

Suppose that A C G is an abelian subgroup, and that the index n = \ G : A \ is relatively small . C a n we conclude that G has a n o r m a l abelian subgroup TV such that the index \ G : N \ is under control? Yes, certainly; we know that \ G : core G ( ,4) | < n!, and of course, c o v e G ( A ) is normal , and since this subgroup is contained in A , it is abelian.

A bound much better than \ G : N \ < n \ is available i f the abelian subgroup A happens to be a Sylow subgroup of G. Assuming that A is a Sylow p-subgroup, we can take N = O p ( G ) , and then Corol la ry 1.39 tells us that \ G : N \ < n 2 . Surprisingly, it is not necessary to assume that A is a Sylow subgroup; there a l w a y s exists an abelian normal subgroup N w i t h \ G : N \ < n 2 . A l though this remarkable theorem of A . Chermak and A . Delgado does not depend on Sylow theory, i t seems appropriate to include it here since i t can be viewed as a generalization of Coro l la ry 1.39.

1.41. T h e o r e m (Chermak-Delgado). L e t G be a finite g r o u p . T h e n G has a c h a r a c t e r i s t i c a b e l i a n s u b g r o u p N such t h a t \ G : N \ < \ G : A \ 2 f o r every a b e l i a n s u b g r o u p A C G.

A s we shall see, the key idea i n the proof of the Chermak-Delgado theorem is very elementary. B u t it is powerful, and it yields even more than we stated i n Theorem 1.41. The tr ick is to define the integer m G ( H ) = \ H \ \ C G ( H ) \ for arbi t rary subgroups H of a finite group G, and to consider subgroups J i C G where this Chermak-Delgado measure is maximized.

We begin w i t h a t r iv i a l observation.

1.42. L e m m a . L e t H C G, w h e r e G i s a finite g r o u p , a n d w r i t e C = C G ( H ) . T h e n m G ( H ) < m G ( C ) a n d if e q u a l i t y o c c u r s , t h e n H = C G ( C ) .

Proof. Since H C C G ( C ) , we see that

m G { C ) = | C | | C G ( C ) | > \ C \ \ H \ = m G ( H ) ,

and the result follows. •

Next , we consider two subgroups simultaneously.

1.43. L e m m a . L e t H a n d K be s u b g r o u p s of a finite g r o u p G, a n d w r i t e D = H D K a n d J = ( H , K ) . T h e n

m G ( H ) m G ( K ) < m G ( D ) m G ( J )

a n d if e q u a l i t y h o l d s , t h e n J = H K a n d C G ( D ) = C G ( H ) C G { K ) .

42 1 . S y l o w T h e o r y

Proof. Wr i t e C H , C K , CD and C j to denote the centralizers i n G of H , K , D and J , respectively, and note that C j = C H n C K and that CH, CK C G D . T h e n

| J | > | M | = and

I w l

and it follows that ( n \ i n l i ^ i ^ ^ M l i M ^ l r n G ( H ) m G ( K )

mG(D) = \ D \ \ C D \ > — m r=rn— = > | J | \ C j \ m G { J )

which establishes the inequality. If equality occurs, then \ J\ = \ H K \ and \ C D \ = \ C H C K I and thus J = H K and G D = C H C K , as required. •

We mention that if H , K C G are subgroups, then the subgroup J = (7i , K ) is often called the join of H and K . Also , a collection £ of subgroups of a group G is said to be a lattice of subgroups i f it is closed under intersections and joins.

1.44. Theorem. G i v e n a finite g r o u p G, l e t C = C ( G ) be t h e c o l l e c t i o n of s u b g r o u p s of G f o r w h i c h t h e C h e r m a k - D e l g a d o m e a s u r e i s as l a r g e as p o s s i b l e . T h e n

(a) C i s a l a t t i c e of s u b g r o u p s o f G .

(b) I f H , K e C, t h e n ( H , K ) = H K .

(c) I f H e C , t h e n C G ( H ) e C a n d C G ( C G ( H ) ) = H .

Proof. Let m be the m a x i m u m of the Chermak-Delgado measures of the subgroups of G , and let H , K e C. T h e n m G ( H ) = m = m G ( K ) , and so by L e m m a 1.43, we have m2 = m G ( H ) m G ( K ) < m G ( D ) m G ( J ) , where D = H n K and J = ( H , K ) . B u t m G ( D ) < m and m G ( J ) < m by the max imal i ty of m, and thus m G ( D ) = m = m G ( J ) , and hence D and J lie i n C, as required. A l so , since equality holds i n L e m m a 1.43, we know that H K = J , and this proves (b). Final ly , statement (c) follows because the max imal i ty of m G { H ) forces equality i n L e m m a 1.42. •

1.45. Corol lary. E v e r y finite g r o u p G c o n t a i n s a u n i q u e s u b g r o u p M , m i n i m a l w i t h t h e p r o p e r t y t h a t m G ( M ) i s t h e m a x i m u m of t h e C h e r m a k - D e l g a d o measures of t h e s u b g r o u p s o f G . A l s o M i s a b e l i a n a n d M D Z ( G ) .

Proof. Since C = C { G ) is a lattice, the intersection of a l l of its members lies i n C, and this is the desired subgroup M . A l so , since M e C, we know that C G ( M ) e C, and thus M C C G ( M ) , and it follows that M is abelian. F ina l ly , M = C G ( C G ( M ) ) D Z ( G ) , and the proof is complete. •

P r o b l e m s I G 43

We shall refer to the subgroup M of Coro l la ry 1.39 as the Chermak-Delgado subgroup of G. It is, of course, characteristic i n G.

P r o o f of T h e o r e m 1.41. Since the Chermak-Delgado subgroup M is characteristic and abelian, it suffices to show that \ G : M \ < \ G : A \ 2 for a l l abelian subgroups A C G. B y the definition of M , we have m G ( M ) > m G ( A ) = \ A \ \ C G ( A ) \ > \ A \ 2 , where the last inequali ty holds because A is abelian. T h e n

\ G • A \ 2 = > ——— = — ——— > — = \ G • M l

as required.

Another appl icat ion of the Chermak-Delgado subgroup is the following.

1.46. Corollary. L e t H be a s u b g r o u p of a finite g r o u p G, a n d assume t h a t \ H \ \ C G ( H ) \ > \ G \ . T h e n G i s n o t a n o n a b e l i a n s i m p l e g r o u p .

Proof. Let M be the Chermak-Delgado subgroup of G and observe that m G ( M ) > m G ( H ) > \ G \ . B u t the Chermak-Delgado measure of the identity subgroup is equal to | G | , and it follows that M > 1. Since M is abelian and normal i n G, the result follows. •

P r o b l e m s 1 G

1G.1 . Let A Q G , where A is abelian, and assume that there does not exist a characteristic abelian subgroup N of G such that \ G : N \ < \ G : A \ 2 . Show that A = C G ( A ) is a member of the maximum-measure lattice C ( G ) and that \ G : Z { G ) \ = \ G : A \ 2 .

1G.2. Let C ( G ) be the maximum-measure lattice as i n Theorem 1.44, and suppose that H e C { G ) and that H < G. Show that there exists a normal subgroup M of G such that H C M < G.

Hint . Observe that a l l conjugates of H i n G lie i n C ( G ) and thus the product of any two of them is a subgroup. If H is not contained i n a proper normal subgroup of G, show that it is possible to write G = H K , where K < G and K contains a conjugate of H . Deduce a contradict ion from this.

1G.3. Let G be simple and suppose that H C G and that \ H \ \ C G ( H ) \ = \ G \ . Show that H = 1 or H = G.

Hint . If G is nonabelian, show that H £• C ( G ) .

1G.4. Le t A C G , where A is abelian and G is nonabelian. Show that there exists a normal abelian subgroup N of G such that | G : / V | < | G : , 4 | 2 .

C h a p t e r 2

Subnormality

2 A

We know that i n general, normal i ty is not a transitive relation on the set of subgroups of a group. In other words, i f K < H and H < G, we cannot usually conclude that K < G. Fol lowing Wie land t , who developed much of the theory that we discuss in this chapter, we "repair" this lack of t rans i t iv i ty by defining a new relation. A subgroup S C G is said to be subnormal i n G if there exist subgroups H i of G such that

S = H 0 < H i < • • • < H r = G,

and i n this si tuation, we write S « G. Subnormali ty, therefore, is a transitive relation that extends the notion of normality. We stress that the chain of subgroups H i extending from S up to G is required to be finite, even if G is an infinite group. Note that al though we have not required that the subgroups H i are a l l distinct, i t is always possible to delete repeated groups and to assume that H 0 < H x < • • • < H r .

O f course, if S « G, there may be a number of different chains of subgroups { H i } that realize the subnormality. The length of the shortest possible chain, or equivalent^ the smallest possible integer r , is called the subnormal depth of S. Thus the whole group G has depth 0 i n G ; proper normal subgroups have depth 1, and subnormal subgroups that are not actually normal have depth exceeding 1.

A l t h o u g h we are concerned almost exclusively w i t h finite groups i n this book, it should be noted that part of the snbnormal i ty theory that we present also works for infinite groups. Some of the theorems that we prove by induct ion on the group order can also be proved by induct ion on the subnormal

45

46 2. S u b n o r m a l i t y

depth, and most of those results hold for infinite groups too. B u t the reader should be cautioned that many of the more interesting theorems about sub-normal i ty are s imply false for general infinite groups.

We begin w i t h a s i tuat ion where subnormali ty arises naturally.

2.1. L e m m a . L e t G be finite. T h e n G i s n i l p o t e n t if a n d o n l y if every s u b g r o u p of G i s s u b n o r m a l .

Proof. B y Theorem 1.26, a finite group is nilpotent if and only if norm a l i z e s grow. In other words, G is nilpotent if and only if H < N G { H ) whenever H < G. Now if H < G and H is subnormal, we can write H = H 0 < ••• < H r = G, where r > 0, and we can assume that H i > H Q . Since H = H 0 < H i , w e have H < H i C N G ( H ) , and this shows that if every subgroup of G is subnormal, then normalizers grow, and hence G is nilpotent.

Conversely, suppose that G is nilpotent, so that normalizers grow in G. G i v e n H C G, we prove that H « G by induct ion on the index \ G : H \ . If this index is 1, then H = G, and H is certainly subnormal. Otherwise, H < G, and we have H < N G ( H ) . T h e n \ G : N G ( H ) \ < \ G : H \ , and so by the inductive hypothesis, N G ( H ) « G. The result now follows since H < N G ( H ) . U

If we wanted to prove that a subgroup S of G is subnormal by using the definition, we would need to construct a chain of subgroups running from S up to G, where each of these subgroups is normal i n the next. In the second half of the proof of L e m m a 2.1, we found such a chain by being greedy. B y this, we mean that i n going from H i to H i + i , we went as far as possible: we took H l + i to be the full normalizer of H i in G. B u t greediness does not always yie ld success. In the symmetric group S 4 , for example, consider any subgroup S of order 2 in the normal K l e i n subgroup K of order 4. (Recall that the nonidenti ty elements of K are exactly the permutations i n SA that are products of two disjoint 2-cycles.) Since K is abelian, we have S < K , and since K < S4, we see that S « S4. It is not hard to see, however, that N G ( 5 ) has order 8; it is a Sylow 2-subgroup of 5 4 . A l so , since this Sylow subgroup has index 3 and is not normal , i t follows that it is its own normalizer. In other words, we are stuck. We can not continue to construct a subnormal chain from N G ( 5 ) to G, and so greediness does not work in this case.

We pursue the connection between nilpotence and subnormali ty a bit further. Recal l that the F i t t i n g subgroup F ( G ) of a finite group G is the unique largest normal nilpotent subgroup of G. The following result shows that not only does F ( G ) contain a l l normal nilpotent subgroups of G; it also contains a l l subnormal nilpotent subgroups.

2 A 47

2.2. Theorem. L e t H C G, w h e r e G i s finite. T h e n H C F ( G ) if a n d o n l y i f H i s n i l p o t e n t a n d s u b n o r m a l i n G.

Proof. If H C F ( G ) , then of course is nilpotent since F ( G ) is nilpotent. A l so , we see by L e m m a 2.1 that iZ" « F ( G ) < G , and thus H « G , as wanted.

Conversely, assume that H « G and that H is nilpotent. We proceed by induct ion on |G| to show that H C F ( G ) . If H = G, then G is nilpotent, and so H = G = F ( G ) , and we are done i n this case. We can assume, therefore, that H < G , and we let M be the penult imate term in a subnormal chain of dist inct subgroups c l imbing from H to G . T h e n H « M < G and M < G . B y the inductive hypothesis applied i n M , therefore, we have H C F ( M ) . B u t F ( M ) is characteristic i n M , which is normal in G , and hence F ( M ) is a normal nilpotent subgroup of G . Th i s yields H C F ( M ) C F ( G ) , as required. •

We leave nilpotent subgroups for a while to study subnormali ty more generally. We begin wi th a basic (and t r iv ia l ) fact.

2.3. L e m m a . L e t S « G, w h e r e G i s a g r o u p , a n d suppose t h a t K C G i s a n a r b i t r a r y s u b g r o u p . T h e n S n K « K .

Proof. Since S is subnormal i n G , there exist subgroups H i C G for 0 < i < r , where < H i for 0 < i < r , and where 0 = 5 and H r = G. Since

< f i i , it follows that P i _ i n AT < H i n A", and so we have

5 n i ^ = p 0 n K < P i n A : ^ - - - ^ i i r n A : = G n A : = A : ,

and thus S (1 K « K , as required. •

2.4. Corol lary. L e i 5 and T be s u b n o r m a l s u b g r o u p s of a g r o u p G. T h e n SHT«G.

Proof. We have S n T « T by L e m m a 2.3, and since T <M G , it follows by the t rans i t iv i ty of subnormali ty that S n T « G , as asserted. •

If 5 and T are normal subgroups of a group G , then, of course, their intersection S n T is also normal , and so too is their jo in , ( 5 , T ) = ST. In other words, the collection of normal subgroups of an arbi t rary group forms a sublattice of the full subgroup lattice. It is perhaps surprising, and it is certainly less t r iv i a l , that the same conclusion holds for subnormal subgroups of finite groups. We have just seen that the collection of subnormal subgroups of G is closed under intersections, and that was easy. The deeper result is that this collection is also closed under joins if |G| is finite. We should mention, however, that there are counterexamples that show that

joins of subnormal subgroups can fail to be subnormal i n the general infinite case.

2.5. Theorem. L e t G be a finite g r o u p , a n d suppose t h a t S,T « G. Then (S, T ) « G.

There are several known proofs of this remarkable theorem of Wie landt , which was first established i n 1939. Wielandt ' s original argument proceeded v i a double induct ion directly from the definition of subnormality, and it is also val id for some infinite groups: those which satisfy the ascending chain condi t ion on subnormal subgroups. Our proof, on the other hand, relies on a later result of Wie land t , and it makes sense only in the finite case. B u t i t is less technical than Wielandt ' s , and it seems somewhat more conceptual.

Before we prove Theorem 2.5, we need to develop some more theory, including the following pretty result (of Wie landt , of course).

2.6. Theorem. L e t S « G, w h e r e G i s a finite g r o u p , a n d l e t M be a m i n i m a l n o r m a l s u b g r o u p o f G . Then M C N G ( S ) .

Reca l l that a minimal normal subgroup of G is a nonidentity normal subgroup M of G such that the only normal subgroup of G that is properly contained in M is the identity. For example, if G is simple, then G is a min ima l normal subgroup of itself.

To prove Theorem 2.6, we must digress briefly to discuss the socle of a finite group G, which is the subgroup generated by a l l min ima l normal subgroups of G, and which is denoted Soc(G) . (Of course, the subgroup generated by a collection of normal subgroups is just the product of those subgroups.)

The key observation here is that since G is finite, every nonidenti ty normal subgroup N of G contains a min ima l normal subgroup of G, and thus N n Soc(G) > 1. In part icular, if G > 1, then Soc(G) > 1.

We remind the reader of a useful general fact, which is needed i n the proof of Theorem 2.6.

2.7. L e m m a . L e t M a n d N be n o r m a l s u b g r o u p s of a g r o u p G, a n d assume M n N = 1. Then every element of M commutes w i t h every element of N .

Proof. Let m € M and n E N and consider the commutator c = [ m , n ] , which, by definition, is the element m ^ n ^ m n . Since c = m _ 1 m n and M is normal , we see that c G M . B u t also, c = ( n _ 1 ) m n , and thus c G N because TV is normal . T h e n c G M n N = 1, and so 1 = c = m _ 1 m n , and we deduce that mn = m. It follows that m and n commute, as asserted. •

P r o o f of T h e o r e m 2.6. We proceed by induct ion on | G | . There is nothing to prove i f S = G, and so we can assume that S < G. T h e n since S « G, we

2 A 49

can reason as we d id previously to find a subgroup N < G w i t h S <¡< N < G. (Simply take TV to be the penult imate term of a subnormal series w i t h dist inct terms c l imbing from S to G.)

There are now two cases. If M n N = 1, then by L e m m a 2.7, we have M C C G ( N ) C CG(S) C N G ( S ) , as wanted. We can suppose, therefore, that 1 < M n N . T h e n since M n N C M , it follows from the fact that M is m in ima l normal i n G that M C\ N = M , and we have M C N .

Since S «¡ N < G, the inductive hypothesis guarantees that every min ima l normal subgroup of N normalizes S. Now M < N , but we do not know that M is m in ima l normal i n N , and so we cannot conclude directly that M normalizes S. B u t we do know that Soc(TV) normalizes S, and so it w i l l suffice to show that M C Soc( iV).

Fi rs t , observe that M n Soc( iV) > 1 since 1 < M < N . A l so , Soc(TV) is characteristic i n N , and hence it is normal i n G, and thus M n Soc(TV) < G . B u t 1 < M n Soc( iV) C M , and it follows from the fact that M is m in ima l normal i n G that M n Soc(TV) = M . T h e n M C Soc(TV) C N G ( 5 ) , and the proof is complete. •

F ina l ly , we are ready to prove that i n finite groups, joins of subnormal subgroups are subnormal.

P r o o f of T h e o r e m 2.5. We prove by induct ion on \ G \ that if S,T «i G, then (S, T ) « G. Since we can certainly assume that G > 1, we can choose a m in ima l normal subgroup M of G. Let G = G / M , and observe that the images S and T of S and T are subnormal i n G. B u t \ G \ < \ G \ , and so the inductive hypothesis applies, and we conclude that ( S , T ) « G. Since "overbar" is a homomorphism wi th kernel M , it follows that

( S , T ) = ( S , T ) = ( S , T ) M ,

and thus { S , T ) M « G.

The correspondence theorem yields a bijection between the set of a l l subgroups of G and the set of a l l of those subgroups of G that contain M . Furthermore, this bijection preserves normality, and hence it also preserves subnormality, and we conclude from this that ( S , T ) M « G. B u t M is m in ima l normal in G, and thus by Theorem 2.6, each of the subgroups S and T is normalized by M . It follows that M normalizes ( S , T ) , and we see that ( S , T ) < ( S , T ) M « G, and thus ( S , T ) is subnormal in G, as wanted. •

T w o subgroups H and K oí a group G are said to be permutable if H K = K H . (Recall that this is equivalent to H K being a subgroup.) We know that each normal subgroup of G is permutable w i t h a l l subgroups,

and it is of some interest to ask if the converse is true. Is a subgroup that is permutable w i t h a l l subgroups necessarily normal? The answer is "no", but it is true (in a finite group) that such a subgroup must be subnormal. ( A subgroup permutable w i t h a l l subgroups i n some group G is said to be quasinormal i n G.)

Actua l ly , a result stronger than "quasinormal implies subnormal" is true. G i v e n S C G , it is enough to know that S is permutable w i t h a l l of its own conjugates to deduce that S is subnormal; one need not assume that S is permutable wi th every subgroup.

2.8. Theorem. L e t S C G, w h e r e G i s a finite g r o u p , a n d assume t h a t SSX = SXS f o r a l l x e G . T h e n S « G.

Since it is certainly not obvious how to construct a subnormal series from S to G in the s i tuat ion of Theorem 2.8, we must find some more subtle method of proof. T ry ing induct ion on | G | , we see that i f S C H < G, then the hypothesis on S is automatical ly satisfied i n H , and so S « H by the inductive hypothesis. To see what good that might be, imagine what the s i tuat ion would be i f we were t ry ing to prove that S is n o r m a l i n G . We would then know that every proper subgroup of G that contains S is contained i n N G ( S ) , and so i f S is not actually normal in G , then N G ( S ) would be a max ima l subgroup of G . In fact, it would be the unique max ima l subgroup containing S.

B y the following "zipper lemma" of Wielandt , a s imilar conclusion holds when we are t ry ing to prove that S is subnormal, and this is the key to the proof of Theorem 2.8.

2.9. T h e o r e m (Zipper L e m m a ) . Suppose t h a t S C G , w h e r e G i s a finite g r o u p , a n d assume t h a t S « H f o r every p r o p e r s u b g r o u p H of G t h a t c o n t a i n s S. If S i s n o t s u b n o r m a l i n G, t h e n t h e r e i s a u n i q u e m a x i m a l s u b g r o u p of G t h a t c o n t a i n s S.

A s we have seen, the analog of the zipper lemma is true (and t r ivial) if "normal" replaces "subnormal", and i n that si tuation, the unique max ima l subgroup of G that contains S is the normalizer of S. B u t i n general, there is no such th ing as a subnormalizer, and so we must work harder to prove Theorem 2.9. W h a t we mean by saying that subnormalizers do not exist is that if S C G , there need not be a unique subgroup max ima l w i t h the property that it contains S subnormally. For example, a Sylow 2-subgroup P of the simple group G = P S L ( 2 , 7) of order 168 is contained i n two maximal subgroups of G , each of order 24. It is not hard to check that the subgroup Z ( P ) , which has order 2, is subnormal in each of these max ima l subgroups, but of course, Z ( P ) cannot be subnormal i n G because G is simple.

2 A 51

Before we prove the zipper lemma, we demonstrate how it can be used by proving Theorem 2.8. We need an easy pre l iminary result.

2.10. L e m m a . L e t H C G, w h e r e G i s a n a r b i t r a r y g r o u p , a n d suppose t h a t H H X = G. T h e n H = G.

Proof. Wr i te x = uv, where u e H and v G H x . T h e n xv~l = u and we have

where the first equality holds because v 6 H x and the th i rd holds because u e H . T h e n G = H H X = H H — H and the proof is complete. •

We also need a bit of notation. If H is an arbi trary subgroup of G, we write H G = ( H x \ x <E G ) . Th i s is the normal closure of H i n G, and we see that H G < G because conjugation by elements of G s imply permutes the generating subgroups of H G . In fact, H G is the unique smallest normal subgroup of G containing H since i f I f C N < G, then N contains a l l conjugates of H , and thus N contains the subgroup H G generated by these conjugates.

P r o o f of T h e o r e m 2.8. Work ing by induct ion on \ G \ , we can assume that S is subnormal i n every subgroup H such that S C H < G. Assuming that S is not subnormal in G, we have S < G, and we derive a contradiction.

B y the zipper lemma, S is contained i n a unique max ima l subgroup M of G. Now consider a conjugate Sx of S. Since S < G, L e m m a 2.10 guarantees that SSX < G, and since SSX is a subgroup because SSX = SXS by hypothesis, we conclude that SSX is contained i n some max ima l subgroup of G. B u t the only max ima l subgroup containing S is M , and so we must have SSX C M . T h e n Sx C M for a l l elements x e G.

Now consider the normal closure SG = (Sx | x € G ) . B y the result of the previous paragraph, we know that SG C M , and thus SG < G. It follows that S « S G < G, and thus S « G. Th i s is a contradict ion, and the proof is complete. •

We now prove the Wie landt zipper lemma.

P r o o f of T h e o r e m 2.9. Work by induct ion on \ G : 5 | . Since we are as. suming that S is not subnormal in G, we certainly have S < G, and thus the case \ G : S\ = 1 of the theorem is vacuously satisfied and our induct ion starts.

A s S is not normal , we have N G ( S ) < G, and thus N G { S ) C M for some max ima l subgroup M of G. O f course S C M , and our task is to show

that M is the only m a x i m a l subgroup of G that contains S. Suppose then, that also S C i f , where i f is max ima l i n G. We work to show that K — M .

Now S Q K < G, and so by hypothesis, 5 « i f . Suppose that i n fact S < K . T h e n i f C N G ( S ) C M , and since i f is max ima l i n G, it follows that i f = M , as wanted. We can suppose, therefore, that S is not normal i n i f .

Sinee S « i f , there is some shortest subnormal series

S = Ho < H i o • • • o H r = i f ,

and we see that r > 2. A l so , S is not normal i n H 2 since otherwise we could delete H i to obtain a shorter subnormal series for S in i f . Let x G I f 2 w i t h Sx + S and write T = (5, S*). Note that T C i f since 5 C i f and x e i f . A l s o , S x C ( H i ) x = H i C N G (5) , and thus T C N G (5) C M . Furthermore, we see that S < T < G.

Observe that Sx satisfies the hypotheses imposed on S i n the statement of the theorem. (This , of course, is because conjugation by x defines an automorphism of G.) We c la im that the subgroup T = (S, Sx) also satisfies these hypotheses. Specifically, we need to show that if T C H < G, then T oo H and that T is not subnormal i n G.

Fi rs t , if T C H < G, then clearly S Q H , and thus S oo H , and similarly, oo H . T h e n T is the jo in of two subnormal subgroups of H , and we conclude by Theorem 2.5 that r oo H as wanted. Also , S o T , and so if T oo G, it would follow that S oo G, which is not the case. Thus T is not subnormal, as claimed.

N o w T > S since Sx is contained i n T, but it is not contained in S. T h e n | G : T\ < \ G : 5|, and so by the inductive hypothesis, T is contained in a unique m a x i m a l subgroup of G. O n the other hand, we know that T C M and r C i f , and thus M = i f . The proof is now complete. •

Perhaps we should explain the name "zipper lemma" . The key idea of the proof of Theorem 2.9 is that if S is contained in two different max ima l subgroups, then some larger subgroup T is also contained i n two different m a x i m a l subgroups. Repeat ing this argument, we would get a s t i l l larger subgroup contained i n two different max ima l subgroups, and so on. B u t this cannot go on indefinitely i n a finite group. A s we cl imb higher, the two max ima l subgroups are forced to be the same. Th i s is analogous to z ipping up an open zipper. A s we pu l l up on the zipper pu l l , the two top parts of the zipper are forced together.

We conclude this section w i t h another appl icat ion of the zipper lemma.

2.11. Theorem. L e t A be a n a b e l i a n s u b g r o u p of a finite g r o u p G, a n d assume t h a t f o r every s u b g r o u p H w i t h A C H C G, we have \ H : A \ 2 < \ H : Z { H ) \ . T h e n A C F { G ) .

The condi t ion that \ H : A \ 2 < \ H : Z ( H ) \ for a l l subgroups H such that A C H C G may seem somewhat artificial , but i n fact, it arises natural ly in character theory. If A is abelian and some character of A induces irreducibly to G, then this condi t ion is automatical ly satisfied. A l so , we observe that if A is abelian and satisfies this condit ion, then A = C G ( A ) . To see this, take H = C G ( A ) and note that \ H : Z { H ) \ < \ H : A \ , and so the assumed inequali ty forces \ H : A \ = 1.

P r o o f of T h e o r e m 2.11. Work ing by induct ion on | G | , we can assume that A C F { H ) whenever A C H < G, and thus A « P . A l so , by Theorem 2.2, we are done if A « G , and so we can assume that A is not subnormal in G, and the zipper lemma applies. We conclude, therefore, that there is a unique max ima l subgroup M containing A .

Now let g <E G. If ( A , A 9 ) < G, then { A , A 9 ) is contained i n some m a x i m a l subgroup of G, and hence { A , A 9 ) C M , and i n part icular, ,4« C M . If this happens for a l l g € G, then A G C M , and thus ^4G < G and we have A < M A g < G, which is a contradict ion. We conclude that G = ( 4 , 4 » ) for some element g £ G.

Since A and are abelian, we have A r \ A 9 C Z ( G ) , and it follows that

| G | > = J J > ' ' \ A C \ A 9 \ ~ \ Z { G ) \ '

where the strict inequali ty follows by L e m m a 2.10 since A < G. M u l t i p l y i n g , , we conclude that \ G : A \ 2 > | G :

the hypothesis. • ' bo th sides by | G | / | A | 2 , we conclude that \ G : A \ 2 > \ G : Z ( G ) | , contrary to

P r o b l e m s 2 A

2A.1 . Let 7t be a set of prime numbers, and recall that O f f (G) is the unique largest normal rr-subgroup of G . Show that O w ( G ) contains every subnormal rr-subgroup of G . Conclude that the subgroup generated by two subnormal 7r-subgroups of G is itself a vr-subgroup.

2A.2 . A g a i n let vr be a set of prime numbers, and recall that 0 * ( G ) is the unique smallest normal subgroup of G whose factor group is a rr-group. If K « G and | G : K \ is a rr-number, show that i f D 0 * ( G ) .


2A.3 . Let H , i f C G and suppose that \ G : H \ and \K\ are relatively prime.

(a) If H « G, show K Q H .

(b) If i f « G, show i f C H .

2A.4 . Let S C. G, where G is a finite group and S is simple, and suppose that S H = H S for a l l subnormal subgroups H of G. Show that 5 C N G ( i / ) for a l l I f « G .

Note. The intersection of the normalizers of a l l subnormal subgroups of a group G is the Wie landt subgroup of G .

2A.5 . Let X be any collection of min ima l normal subgroups of G , and let N = Y i x .

(a) Show that N is the direct product of some of the members of X .

(b) Show that every min ima l normal subgroup of N is simple.

(c) Show that TV is a direct product of simple groups.

H i n t . For (b), show that Soc(TV) = N .

Note. Th i s problem shows that min ima l normal subgroups and socles of finite groups are direct products of simple groups.

2A.6 . In this s i tuat ion of the previous problem, show that every nonabelian normal subgroup of G contained i n N contains a member of X .

2A.7 . Let S « G , where S is nonabelian and simple. Show that SG is a min ima l normal subgroup of G .

Hint . Work by induct ion on \ G \ to conclude that S C Soc(Jf) whenever S C H . Deduce that each conjugate of S i n G is a min ima l normal subgroup of SG. T h e n apply the previous problem to the group SG, where X is the set of a l l G-conjugates of S.

2A.8 . Let S and T be different nonabelian subnormal simple subgroups of G . Show that S and T commute elementwise.

2A.9 . Let H«G and assume that H = 0 * { H ) , where vr is a set of primes. Show that C v ( G ) normalizes H .

Hint . It is no loss to assume that G = H O n ( G ) . Show i n this case that H = 0 * ( H ) .

2A.10. We say that subgroups H , i f C G are strongly conjugate if they are conjugate i n the group ( H , i f ) . Show that H « G if and only if the only subgroup of G that is strongly conjugate to H is H itself.

2 B 55

2 B

In this section, we present a useful result that is essentially due to R . Baer , and which we derive from the Wie landt zipper lemma. Versions of Baer 's theorem have also been proved by M . Suzuki and by J . A l p e r i n and R . Lyons.

2.12. T h e o r e m (Baer) . L e t H be a s u b g r o u p of a finite g r o u p G. T h e n f f C F ( G ) if a n d o n l y if (ff, H x ) i s n i l p o t e n t f o r a l l x € G.

Proof. One direction is t r iv i a l . If f f C F ( G ) , then also H x C F ( G ) since F ( G ) < G. T h e n ( H , H X ) is a subgroup of the nilpotent group F ( G ) , and hence it is nilpotent.

Conversely, assume that (ff, H x ) is nilpotent for a l l x G G . In part icular, f f is nilpotent, and so to prove that f f C F ( G ) , it suffices by Theorem 2.2 to prove that f f « G . We establish this by induct ion on | G | . Assume that f f is not subnormal in G , and observe that if f f is contained in a proper subgroup f f , then ( i f , H x ) is nilpotent for a l l x € K , and thus f f « AT by the inductive hypothesis. The zipper l emma now applies, and we deduce that f f is contained in a unique max ima l subgroup M .

N o w consider a conjugate H x of f f . If ( i f , H x ) = G , then G is nilpotent, and hence f f « G , which is not the case. T h e n ( i f , f f x ) < G , and so this subgroup must be contained in some m a x i m a l subgroup of G. B u t (ff, H x ) contains f f , and the only max ima l subgroup of G that contains f f is M . It follows that (ff, f f x ) C M , and we conclude that M contains a l l conjugates of f f i n G , and so H G C M . The normal closure i f G is thus proper i n G , and we have f f C H G < G. Then , H « H G < G , and so i f « G . Th i s contradict ion shows that f f « G , as required. •

The following is a useful consequence of Baer 's theorem. Theorem 2.13 is the key to Matsuyama's extension to groups of even order of Goldschmidt ' s non-character proof of the Burnside pV-theorem. (Goldschmidt required that bo th primes were odd, and Mat suyama removed that restriction.)

Reca l l that an involution i n a group G is any element of order 2.

2.13. Theorem. L e t t be a n i n v o l u t i o n i n a finite g r o u p G, a n d assume t h a t t # 0 2 ( G ) . T h e n t h e r e exists a n element x € G of o d d p r i m e o r d e r such t h a t x l = x ~ l .

Since 0 2 ( G ) is the unique largest normal 2-subgroup of G , the condit ion that t 0 0 2 ( G ) says that t is not contained in any normal 2-subgroup of G .

In order to prove Theorem 2.13, we need to digress briefly. A group D is dihedral i f it contains a nontr iv ia l cyclic subgroup C of index 2 such that every element of D - C is an involut ion. O f course, \ D \ = 2 | G | , and so dihedral groups have even order at least 4, and it is not hard to show

that a dihedral group is determined up to isomorphism by its order. Group theorists usually refer to the dihedral group of order 2 n as D 2 n , and that is the notat ion we use here. (But this notat ion is not completely standard, and this group is often referred to as D n . ) The dihedral group D 4 is isomorphic to the K l e i n group C2 x C2, and except for this degenerate case, dihedral groups are always nonabelian. It is easy to see that D 2 n exists for n > 2: it is the group of rotat ional symmetries i n 3-space of a regular n-gon. (In Chapter 3, we shall see how to construct dihedral groups as semidirect products, thereby establishing their existence without relying on geometric reasoning.)

T h e following presents some of the basic facts about dihedral groups.

2.14. L e m m a . L e t D be a g r o u p .

(a) Suppose t h a t C = (c) i s a n o n t r i v i a l c y c l i c s u b g r o u p of i n d e x 2 i n D , a n d t h a t t <E D - C i s a n i n v o l u t i o n . T h e n every element of D - C i s a n i n v o l u t i o n if a n d o n l y if cl = c~l. I n t h i s case, D i s d i h e d r a l , a n d x y = x ' 1 f o r a l l x € C a n d y e D - C. A l s o , D i s g e n e r a t e d by t h e d i s t i n c t i n v o l u t i o n s c t a n d t.

(b) Suppose t h a t D i s g e n e r a t e d by d i s t i n c t i n v o l u t i o n s s a n d t. T h e n t h e c y c l i c s u b g r o u p C = ( s t ) i s n o n t r i v i a l , a n d i t f a i l s t o c o n t a i n s a n d t. A l s o \ D : C\ = 2, a n d t i n v e r t s t h e g e n e r a t o r st of C. I n p a r t i c u l a r , w e a r e i n t h e a b o v e s i t u a t i o n , a n d D i s d i h e d r a l .

P r o o f . For (a), observe that D - C is exactly the coset C t . If a e C then ( a t ) 2 = a t a t = a a \ and thus a} = a ' 1 i f and only if ( a t ) 2 = 1. In part icular, if every element of D - C is an involut ion, then ( c t ) 2 = 1, and so c 1 = c " 1 . Conversely, suppose that c* = c " 1 . Since every element of C is a power of c, we have a* = a " 1 for a l l a € C, and hence ( a t ) 2 = 1, and every element of D - C = C t is an involut ion. In other words, D is dihedral. Let x E C and y € D - C. T h e n y = a t for some element a G C, and we have x y = x a t = x * = x ~ l . A l so , c t is an involution, and the group ( c t , t ) properly contains C = (c), and thus it is the whole group D . F ina l ly , c t and t are dist inct because c ^ 1.

Now assume the si tuat ion of (b) and observe that C is nontr iv ia l since s + t. We have (si)* = t ( s t ) t = ts = ( s t ) - \ and thus t inverts the generator st of C, as wanted. T h e n C * = ( s t ) 1 = ((si) ') = ( ( s i ) " 1 ) = C, and so t e N D ( C ) . It follows that C ( t ) is a subgroup containing both i and ( s i ) i = s, and thus C ( i ) is a l l of D . Because st <E C, we see that i f either s or i lies i n C, then both s and t lie i n C, and this is impossible since a cyclic group can contain at most one involut ion. Thus C contains neither s or t, and i n part icular, C n (i) = 1. Since D = C ( t ) , we have \ D : C\ = | ( i ) | = 2, and this completes the proof of (b). •

P r o b l e m s 2 B 57

The fact that two elements of order 2 i n an arbi t rary group always generate a dihedral subgroup does not seem to generalize. There appears to be almost nothing that can be said about the structure of a subgroup generated by two elements of given orders m > 1 and n > 1 i n any case other than m = 2 = n .

P r o o f of T h e o r e m 2.13. Let T = ( t ) . If T C F ( G ) , then T is contained i n the unique Sylow 2-subgroup of F ( G ) , which is 0 2 ( G ) , contrary to hypothesis. B y Baer 's theorem (2.12) therefore, there exists g € G such that ( T 9 , T ) is not nilpotent. In particular, ( t 9 , t ) is not a 2-group.

B y L e m m a 2.14, it follows that D = ( t 9 , t ) is dihedral of order 2 | G | , where C is a cyclic subgroup whose elements are inverted by t. Since D is not a 2-group, neither is G , and thus there exists an element x € C of odd prime order. Since x t = a T 1 , the proof is complete. •

P r o b l e m s 2 B

2B.1. Let H C G and assume that for each element g e G , either ( H , H 9 ) is nilpotent or H H 9 = H 9 H . Show that H « G .

2B.2. Let L> be the dihedral group of order 2n.

(a) If n is odd, show that D contains exactly n involutions, and that they a l l lie i n a single conjugacy class.

(b) If n is even, show that D contains exactly n + 1 involutions, and these lie i n exactly three conjugacy classes, w i t h sizes 1, n / 2 and n / 2 , respectively.

2B.3. Let s and t be involutions i n a group G . If s and t are not conjugate i n G , show that there exists an involution, z e G, different from s and t, and commut ing w i t h both of them.

2B.4. Suppose that G has more than one Sylow 2-subgroup and that every two distinct Sylow 2-subgroups of G intersect t r iv ia l ly . Show that G contains exactly one conjugacy class of involutions.

2B.5. Let B Q G , w i t h | G : B \ = 2. Show that the following are equivalent.

(1) G - B contains an involut ion t such that 6* = 6 " 1 for a l l elements b e B .

(2) G-B consists entirely of involutions.

(3) Every element t of G - B is an involut ion such that 6* = b~l for a l l elements b G B .

Show also that if these conditions hold, then B must be abelian.

Note. In this si tuation, G is said to be generalized dihedral.

2B.6. Show that G has a normal Sylow p-subgroup if and only i f every subgroup of the form ( x . y ) has a normal Sylow p-subgroup, where x and y are conjugate elements of G having p-power order.

2 C

A l t h o u g h there really is no connection wi th subnormality, we digress i n this section to discuss the terms "local" and "global" as they are used i n finite group theory. ( A n d then we w i l l present an applicat ion of Theorem 2.13 i n this context.) B y definition, a subgroup i f of a group G is p-local, where p is prime, if i f is of the form i f = N G ( P ) , where P is some nonidentity p-subgroup of G. More generally, a subgroup is local if it is p-local for some prime p. The idea here is that in some sense, the local subgroups are those that are easiest to find: choose a Sylow subgroup, pick a nontr ivia l subgroup of it , and then take the normalizer.

For example, suppose we want to prove that up to isomorphism, the only simple group of order 60 is the alternating group A 5 . To establish this, it suffices to show that an arbi trary simple group G of order 60 must have a subgroup H of index 5. T h e n the action of G on the set of right cosets of H yields an isomorphism from G into A 5 , and because | G | = 60, this map must be surjective, and G = A $ , as required.

G i v e n a simple group G of order 60, we can find the required subgroup ff , as follows. The number n 2 of Sylow 2-subgroups of G must be either 5 or 15. If n 2 = 5, take H to be the normalizer of a Sylow 2-subgroup of G. If n 2 = 15, then since 1 5 - 1 is not divisible by 4, it follows by Theorem 1.16 that there exist distinct Sylow 2-subgroups S and T that intersect nontrivially. Let D = S n T > 1 and H = N G ( D ) . T h e n f f contains both S and T, and thus | i f | is a proper mult iple of 4. Since G is simple, D is not normal , and so i f < G. A l so , \ G : i f | + 3 by the n!-theorem, and the only remaining possibi l i ty is that i f has index 5, as wanted.

In bo th cases, where n 2 = 5 and where n 2 = 15, we found a 2-local subgroup of index 5, and as we explained, this shows that G A 5 . (Actually, since n 2 ( A 5 ) = 5, we see that the case n 2 = 15 does not occur, but nevertheless, it had to be considered.)

N o w consider the analogous problem for A 6 . It is true that every simple group of order 360 is isomorphic to A 6 , but this is considerably more difficult to prove. It would suffice to find a subgroup of index 6 i n an arbi trary simple group of order 360, but this is hard to do because no l o c a l subgroup of A 6

has index 6, and so no argument such as we used for A 5 can possibly work. In fact, a l l subgroups of index 6 i n A 6 are isomorphic to A 5 , and hence

2 C 59

they are simple. N o subgroup of index 6, therefore, can be the normalizer of anything i n a simple group of order 360. A l t h o u g h it is possible, using character theory, to show that a simple group of order 360 really does have a subgroup of index 6, the usual proof that up to isomorphism, there is a unique simple group of order 360 proceeds by showing that every such group is isomorphic to P S L { 2 , 9). (Of course, it follows that A 6 and P S L { 2 , 9) are isomorphic.)

Properties of groups that do not direct ly refer to local subgroups are often referred to as global properties. W h a t is remarkable, is how often it happens that global information can be deduced from appropriate data about local subgroups. For example, here is a "local-to-global" result that is not terr ib ly difficult to prove.

2.15. Theorem. Suppose t h a t f o r every o d d p r i m e p , every p - l o c a l s u b g r o u p of a finite g r o u p G has a n o r m a l Sylow 2 - s u b g r o u p . Then G has a n o r m a l Sylow 2 - s u b g r o u p .

If a group G has a normal Sylow 2-subgroup, it is easy to see that every subgroup of G also has a normal Sylow 2-subgroup. Theorem 2.15 can thus be viewed as a strong converse to this remark. Its proof relies on Theorem 2.13 together w i th the general observation that local subgroups of homomorphic images are images of local subgroups. (We state this a bi t more precisely, as follows.)

2.16. L e m m a . L e t N < G, a n d w r i t e G = G/N, u s i n g t h e s t a n d a r d " b a r c o n v e n t i o n " , w h e r e o v e r b a r i s t h e c a n o n i c a l h o m o m o r p h i s m G ^ G . Then

f o r a l l p r i m e s p , every p - l o c a l s u b g r o u p of G has t h e f o r m L , w h e r e L i s some p - l o c a l s u b g r o u p of G.

Proof. B y the correspondence theorem, every subgroup of G has the form H , for some subgroup H of G w i th H D N . Now let N C M C G, where M is p-local i n G. A l though M may not be p- local i n G, we w i l l complete the proof by showing that M = L for some p-local subgroup L of G.

Since M is p-locat, it is the n o r m a l i z e r j n G of some nontr iv ia l p-subgroup, which we can write m the form U, where N < U C G and \ U : A | is a power of p. T h e n M = N^(C7), and since U 2 N , it follows that M = N Q ( U ) .

Let P G Sylp(C/), and let L = N G ( P ) . N o w U = N P because N and P have coprime indices in U, and since N < U, it follows that P > 1, and thus L is p-local i n G. Since /^normalizes both N and P , we have L C N G ( N P ) = N G ( U ) = M , and thus I C M . To obtain the reverse containment, observe that since U < M and P G S y l p ( [ / ) , the F ra t t i n i argument yields

M = U N M ( P ) = N P N M ( P ) = N N M ( P ) C N L .

T h e n M C N L = L , where the equality holds because N is the kernel of the overbar homomorphism. It follows that M = L , as wanted. •

P r o o f of T h e o r e m 2.15. F i r s t , we suppose that 0 2 ( G ) = 1, and we show i n this case that \ G \ is odd. If G has even order, we can choose an involut ion t e G, and we observe that t & 0 2 ( G ) . T h e n by Theorem 2.13, there exists an element x of odd prime order p such that x t = x ~ l . Let X = ( x ) , and note that t G N G { X ) and that N G ( A ) is p-local . B y hypothesis, N G ( A ) has a normal Sylow 2-subgroup S, which necessarily contains t. A s X and S are normal i n N G ( A ) and XnS = 1, it follows by L e m m a 2.7 that X and 5 centralize each other, and i n particular, t centralizes x . Thus x = x t = x " 1 , and this is a contradict ion since x has order p > 2. T h i s shows that \ G \ is odd, as wanted.

In the general case, let N = O ^ G ) , and observe that by the previous lemma, the p-local subgroups of G = G / N are homomorphic images of p-local subgroups of G for a l l odd primes p . B y hypothesis, the p-local subgroups of G have normal Sylow 2-subgroups, and thus the same is true for their homomorphic images, and it follows that G satisfies the hypotheses of the theorem. B u t 0 2 ( G ) is t r iv ia l , and by the first part of the proof it follows that that \ G \ is odd. T h e n N = 0 2 ( G ) is a normal Sylow 2-subgroup of G , and the proof is complete. •

We close this section w i t h a l emma that w i l l be needed later. A l t h o u g h this result has nothing to do w i t h subnormality, it is related to L e m m a 2.16, and so it seems appropriate to present it here. The assertion of L e m m a 2.16 is that every p-local subgroup of a homomorphic image of G is the image of a p- local subgroup of G. T h i s suggests the question of whether or not the image of a p- local subgroup must be p-local . The answer is "not always". For example, if O p ( G ) is nontr iv ia l , then G is p-local i n itself, but this cannot be true for G / O p ( G ) , which has no nonidenti ty normal p-subgroup. B u t p-local subgroups do map to p-local subgroups if the kernel of the homomorphism is a p'-group.

2.17. L e m m a . L e t G be a finite g r o u p , a n d l e t N < G. W r i t e G = G/N, a n d l e t p be a p n m e t h a t does n o t d i v i d e \ N \ . If P i s a n o n t r i v i a l p - s u b g r o u p of G, t h e n P i s n o n t r i v i a l , a n d N ^ ( P ) = N G ( P ) . I n p a r t i c u l a r , if L i s p - l o c a l i n G, t h e n L i s p - l o c a l i n G.

Proof. Since the p-group P is nontr ivial , and the order of N is not divisible by p, we have_P % N , and thus P is nontr ivia l . Wr i t e L = N G ( P ) , and observe that P < L because overbar is a homomorphism. Thus I C I % ( P ) and it suffices to prove the reverse containment.

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B y the correspondence theorem, every subgroup of G has the form X for some (unique) jsubgroup X C G w i t h X D TV, and so i n part icular, we can wri te I % ( P ) = M , where TV C M C G. N o w P ~ N = P < M , and thus P / V < M by the correspondence theorem. Al so , P € S y l p ( P / V ) since p does not divide \ N \ , and thus by the F ra t t i n i argument, wej iave M = N M ( P ) ( P / V ) = N M ( P ) N C N G ( P ) / V = L / V . Th i s yields N ^ ( P ) = M C L , as wanted. The last statement should now be clear. •

P r o b l e m s 2 C

2C.1 . A finite group G is said to be an N-group i f every local subgroup is solvable.

(a) Show that every proper homomorphic image of an N-group is solvable.

(b) Le t G be a nonsolvable N-group. Show that G has a unique min ima l normal subgroup S, and that S is nonabelian simple.

Hint . The F ra t t i n i argument is relevant. A l s o , recall that subgroups and homomorphic images of solvable groups are solvable and that i f K < L w i t h K and L / K bo th solvable, then L is solvable.

Note . A major step leading toward the classification of finite simple groups was J . Thompson 's classification of nonsolvable N-groups. A nonabelian finite simple group is minimal simple i f every proper subgroup is solvable, and since min ima l simple groups are clearly N-groups, Thompson's work provided a classification of a l l m in ima l simple groups.

2 D

Next , we present an amazing result of Zenkov. It is another appl icat ion of Baer 's theorem.

2.18. T h e o r e m (Zenkov). L e t A a n d B be a b e l i a n s u b g r o u p s of a f i n i t e g r o u p G, a n d l e t M be a m i n i m a l member of t h e set { A f l B 9 \ g e G } . T h e n M C F ( G ) .

T h e assumption i n Theorem 2.18 that M is a "min imal member" of some set of subgroups means, as usual, that no member of the set is properly contained in M . O f course, if M is chosen to have m i n i m a l order among members of the set, then M w i l l necessarily be m i n i m a l in this sense, but not conversely since i n general, a min ima l member of some collection of subgroups need not have min ima l order.

Suppose that G has an abelian Sylow p-subgroup P . If we apply Zenkov's theorem w i t h A = P = B , we deduce that P n F C F ( G ) for some element g E G, and thus P n F C O p ( G ) , which is the unique Sylow p-subgroup of F ( G ) . We know, however, that in general, O p ( G ) is contained in every Sylow p-subgroup of G , and it follows i n this case that P n F = O p ( G ) . In other words, if G has abelian Sylow p-subgroups, then O p ( G ) is the intersection of two of them. Th i s is exactly Brodkey 's theorem, which appeared i n the previous chapter as Theorem 1.37. We can view Zenkov's theorem, therefore, as a far-reaching generalization of Brodkey 's result.

P r o o f of T h e o r e m 2.18. The set { A n B 9 \ g e G } is unchanged i f we replace B by an arbi trary G-conjugate B 9 , and thus it is no loss to assume that M = A n B . We show by induct ion on | G | that M C F ( G ) . F i r s t , suppose that G = ( A , B 9 ) for some element g € G . Since A and B 9 are abelian, we have A n B 9 C Z ( G ) , and thus A n B 9 = ( A n B 9 ) 9 ' 1 C P . It follows that A n B 9 C A n B = M , and by the min imal i ty of M we have M = A n B 9 C Z ( G ) C F ( G ) , as required.

We can now assume that ( A , B 9 ) < G for a l l g G G . To show that M C F ( G ) , it is enough to prove for al l primes p that a full Sylow p-subgroup P of M is contained in F ( G ) . (This suffices, of course, because M is generated by its Sylow subgroups.) B y Baer 's theorem (2.12), therefore, it is enough to show that (P, P 9 ) is nilpotent for a l l elements g € G .

F i x g e G and let H = ( A , B 9 ) , so that H < G. Wr i t e C = B n i f , and observe that i f /* e i f , then ^ n G / l = , 4 n ( P n P ) / l = A n P / l n P = A n P \ where the final equality holds because A C P . In part icular, M = i n B = 4 n G is m in ima l i n the set { A n C h \he H } . B y the inductive hypothesis

applied i n P , and w i t h C replacing B , we conclude that P C M C F ( P ) , and thus P C O p ( P ) since O p ( P ) is the unique Sylow p-subgroup of F ( P ) . Also , P 9 C B 9 C H , and thus P 9 normalizes O p ( i f ) and O p ( H ) P 9 is a p-group containing (P, P 9 ) . In particular, (P, P 9 ) is a p-group, and hence it is nilpotent, as required. •

2.19. Corollary. L e t A C G, w h e r e A i s a b e h a n a n d G i s a n o n t r w i a l finite g r o u p , a n d assume t h a t \ A \ > \ G : A \ . T h e n A n F ( G ) > 1.

Proof. If g £ G, we compute that \ A \ \ A 9 \ = \ A \ 2 > \ A \ \ G : A \ = | G | . A l so , we can assume that A < G , and thus A A 9 < G by L e m m a 2.10. Th i s yields

G | > \ A A 9 \ = \ A \ \ A 9

> G

A n A 9 A n A 9

and thus A n A 9 > 1 . Since this holds for a l l g E G, we can apply Zenkov's theorem wi th B = A to deduce that A n F ( G ) > 1, as wanted. •

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B y Corol la ry 2.19, we see that if G is nontr iv ia l and A is an abelian subgroup of G w i th \ A \ > \ G : A \ , then A contains a nontr iv ia l subnormal subgroup of G. In fact, if A is cyclic, we get more: A contains a nontr ivia l n o r m a l subgroup of G. Th i s is a pretty result of A . Lucch in i . (We use Lucchin i ' s theorem in Chapter 3 to prove that every automorphism of a finite group G has order less than |G | . )

2.20. T h e o r e m (Lucchini ) . L e t A be a c y c l i c p r o p e r s u b g r o u p of a f i n i t e g r o u p G, a n d l e t K = c o r e G ( A ) . T h e n \ A : K \ < \ G : A \ , a n d i n p a r t i c u l a r , i f \ A \ > \ G : A \ , t h e n K > 1.

Proof. We proceed by induct ion on | G | . N o w A / K is a proper cyclic subgroup of G/K, and the core of A / K i n G / K is t r iv i a l . If K > 1, we can apply the inductive hypothesis i n the group G / K to deduce that \ A / K \ < \ ( G / K ) : [ A / K ) \ = \ G : A \ , and there is nothing further to prove. We can assume, therefore, that K = 1 , and we work to show that \ A \ < \ G : A \ .

Assume that \ A \ > \ G : A \ . Since G > A , we know that G is nontr ivia l , and hence A n F ( G ) > 1 by Corol la ry 2.19. In part icular , F ( G ) > 1, so we can choose a min ima l normal subgroup E of G w i t h E C F ( G ) . T h e n J5l~lZ(F(G)) > 1, and since E is m in ima l normal i n G, we have E C Z ( F ( G ) ) , and i n part icular, E is abelian. A l so , and again using the min imal i ty of E , it follows that E is an elementary abelian p-group for some prime p . (In other words, x p = 1 for al l x e E . )

Since E C Z ( F ( G ) ) , we see that E normalizes the nontr iv ia l group A n F ( G ) , and of course A normalizes this subgroup too. T h e n A n F ( G ) < A E ,

and since core G ( ,4) = 1, it follows that A E < G, as is indicated in the diagram.

Wri te G = G / E . Let M = c o r e ^ ) , w i th M D E , and note that M < G and A M = A E _ A l s o ^ Z _ < G since A E < G, and so by the inductive hypothesis, \ A : M \ < \ G : A \ , and we have \ A E : M | < \ G : A E \ .

Let B = A n M , so that 73 is cyclic. We have

\ A E : A \ = \ A M : A \ = \ M : A n M \ = \ M : B \ ,

and hence \ A E : M \ = \ A : B \ . We conclude that

Suppose that M is abelian, and let ip : M -»• M be the homomorphism defined by ^ ( m ) = m p . T h e n i? C ker(v>), and since M = E B by Dedekind's lemma, it follows that <p(M) = <p{B) C B C A . N o w M < G, and hence <p(M) < G, and we conclude that i p ( M ) = 1 since core G ( ,4) = 1. T h e n f ( B ) = 1, and since B is cyclic, it follows that \ B \ < p . T h e n \ M : B \ < \ B \ < p , and since M / B is ap-group, we deduce that M = B . B u t M < G and M C A , and thus M = 1, and i n part icular, E = 1, which is a contradiction.

It follows that M is nonabelian, and since M / E is cyclic, we conclude that E is not central i n M , and so E n Z ( M ) 1. Now F ( M ) C F ( G ) , and so £ centralizes F ( M ) , and thus B n F ( M ) is a nontr iv ia l central subgroup of B E = M . Since Z ( M ) is cycl ic , we see that 7 3 n F ( M ) is characteristic in Z ( M ) < G . Thus B n F ( M ) is a nontr iv ia l normal subgroup of G contained in A , and this is our final contradiction. •

P r o b l e m s 2 D

2D.1. Let G = N A , where N < G , C A ( N ) — 1 and A is abelian.

(a) If F(JV) = 1, show that \ A \ < \ N \ .

(b) If \ N \ and \ A \ are coprime, show that \ A \ < \ N \ .

2D.2. Let G = N A , where N < G, C A { N ) = 1 and A n N = 1. If TV is nontr iv ia l and ,4 is cyclic, show that \ A \ < \ N \ .

C h a p t e r 3

Split Extensions

3 A

G i v e n TV < G , a subgroup H C G is a complement for TV i n G i f / V P = G and N n P = 1, and i f /V has a complement i n G , we say that G splits over N . Note that i f P complements N i n G , then H = H / { N n P ) = 7 V P / / V = G / / V , and thus a l l complements for N i n G are isomorphic to G / i V , and hence they are isomorphic to each other. A n easy example where a group fails to split over some normal subgroup is G = G 4 , the cyclic group of order 4, where N is the unique subgroup of order 2. It is clear that N is not complemented i n G because if a complement existed, it would be isomorphic to G/N, which has order 2. B u t N is the only subgroup of order 2 i n G , and of course, N is not a complement for itself.

If P complements N i n G , then each element g € G is of the form g = n h , w i t h n e N a n d h e H , and this factorization is unique. (The uniqueness follows v i a an elementary computat ion, but i f G is finite, an easier argument is to observe that \ G \ = \ N \ \ H \ and so the surjective map from pairs (n, h ) to elements n h G G must also be injective.) It is also true, and for similar reasons, that each element of G is uniquely of the form h n , w i t h h G H and n e J V . (To be specific, suppose that g = n h , and we wish to write g = h ' n ' , w i t h ri e N and h ' G P . We have g e N h = h N since J V ^ G , and thus we must have /»' = h . The equation / i n ' = n h then yields n ' = n h . )

If P complements N i n G , then clearly, every conjugate of H i n G also complements / V . In general, however, a normal subgroup N of G may have nonconjugate complements, al though as we have seen, a l l complements are isomorphic. A n example where complements are not conjugate occurs is in the K l e i n group G = G 2 x G 2 . If TV is any one of the three subgroups of

65

66 3. S p l i t E x t e n s i o n s

order 2, then each of the other two subgroups of order 2 complements N , but these complements are not conjugate in G since G is abelian.

Our goal in this section is the following: given groups N and H , construct a l l (up to isomorphism) groups G having a normal subgroup N 0 complemented by a subgroup H 0 , where N = N 0 and H = H 0 . (We refer to such a group as a split extension of H by N , but unfortunately, the literature is inconsistent on this point, and this group is sometimes also referred to as a split extension of N by H . ) One such split extension, of course, is the (external) direct product G = N x H , which is the set of ordered pairs (n, h ) w i t h n € N and h € H , and where mul t ip l ica t ion is componentwise. Here, we take N 0 = {(n, 1) | n € N } and H 0 = { ( l , h ) \ h e H } , and it is clear that N 0 = N , HQ = H , N o < G and H 0 is a complement for N 0 i n G. In this case, the complement H Q also happens to be normal , and in general, it is easy to see that i f the normal subgroup 7V0 has a n o r m a l complement H 0 i n G, then G is the (internal) direct product of iV 0 and H 0 , and so G = N x H . Our task, therefore, is to generalize the direct product construction sufficiently so that we get complements H 0 for iV 0 in G, where H 0 is not necessarily normal i n G. The desired construction is called the "semidirect product".

Before we proceed, however, we digress briefly to discuss extensions that are not necessarily split . G i v e n groups N and H , a group G is said to be an extension of H by iV i f there exists N 0 < G such that N 0 = N and G/N0 = H . Note that in our definition, the group preceded by the word "of" corresponds to the factor group, and the group preceded by "by" corresponds to the normal subgroup. A s we mentioned, however, this use of prepositions is sometimes reversed in the literature, so readers should attempt to determine the precise meaning from the context.

There is a more fundamental ambiguity here. If G is an extension of H by N , then the normal subgroup N 0 of G such that N 0 = N and G/N0 = H is not, in general, uniquely determined. In other words, G can be an extension of H by N i n more than one way. In fact, it can happen that w i th one choice of the normal subgroup N 0 the extension is split, and w i t h another it is not. Proper ly speaking, therefore, we really should say something like " G is an extension of H by N w i th respect to the normal subgroup 7V 0 " , and then it would make sense to ask i f the extension is split .

Imagine that the "extension problem" could be completely solved. Suppose, i n other words, that we knew how to construct (up to isomorphism) a l l extensions of i f by A for given finite groups H and N . (We stress that this seems to be far from a realistic assumption.) Then , given that we know al l simple groups, we could recursively construct a l l finite groups. To do this, assume that we have already constructed all groups of order less than n , where n > 1. E a c h group G of order n has some max ima l normal subgroup,

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say N . Since the factor group G / N is simple, every group of order n can be constructed as an extension of some (known) simple group S by some (already constructed) group N of order n/\S\. B u t even if we could do this, we would s t i l l be faced w i t h the very difficult problem of deciding whether or not two groups of order n constructed in this way are isomorphic.

Re turn ing to split extensions now, suppose that G is split over a normal subgroup N , and let H be a complement. Observe that H acts on N v i a conjugat ion in G, and i n fact, conjugation by h £ H induces an automorphism of N . U H < G, then since iV n H = 1, we know that H centralizes N , and so this conjugation action is t r iv ia l . If we wish to construct split extensions that are not merely direct products, therefore, we must consider nontr iv ia l conjugation actions of H on N . Before proceeding w i t h our construction, however, we show that no further information is needed; the subgroups N and H and the conjugation action of H on N completely determine G up to isomorphism.

To state the following uniqueness theorem, we introduce some convenient, but nonstandard notation. Consider isomorphic groups N and 7V 0 )

where some specific isomorphism is given. For n £ N , we w i l l write n 0 e iVo to denote the image of n under the given isomorphism, so that we can th ink of ( ) 0 as the name of the isomorphism. We w i l l use the same name ( ) 0 to denote a given isomorphism from H to H 0 .

3.1. L e m m a . L e t G a n d G0 be g r o u p s , a n d suppose t h a t N < G i s c o m p l e m e n t e d by H , a n d N 0 < G0 i s c o m p l e m e n t e d by H 0 . Assume t h a t N = N 0

a n d H H 0 , w h e r e each of these i s o m o r p h i s m s i s d e n o t e d by ( ) 0 , a n d suppose t h a t

( n h ) 0 = (no)' 1 0

f o r a l l elements n £ N a n d h <E H . T h e n t h e r e i s a u n i q u e i s o m o r p h i s m f r o m G t o G0 t h a t extends t h e g i v e n i s o m o r p h i s m s N -> JV0 a n d H - > H 0 .

P r o o f . Every element g £ G is uniquely of the form h n , where h £ H and n £ N , and so if we seek an isomorphism 9 : G G0 that extends the given isomorphisms N - > N 0 and H ->• H 0 , we have no choice but to define 9 ( g ) = 9 ( h n ) = 9 { h ) 9 { n ) = h 0 n 0 . Since the decomposit ion g = h n is unique, 9 is well defined, and since every element g 0 £ G0 has the form h 0 n 0 w i t h h 0 £ H o and n 0 £ N 0 , it is clear that 9 is surjective. It is injective because the decomposit ion g 0 = h 0 n 0 is unique and h and n are the unique elements of H and N that map to h o and n 0 respectively.

It remains to show that 0 is a homomorphism, and so we must compute 9 ( ( h n ) ( k m ) ) , where h , k £ H and n , m £ N . We have

( h n ) { k m ) = h k k - l n k m = ( h k ) { n k m ) ,


and so by definition, appl icat ion of 9 to this element yields

{ h k ) o ( n k m ) 0 = h 0 k o { n k ) o m 0 = hQk0{nQ)k°mQ = h 0 n 0 k o m o ,

and this is exactly 9 { h n ) 9 { k m ) , as required. •

In order to construct a l l possible split extensions, we need to examine a l i t t le more abstractly the conjugation action of a group H o n a group N . Reca l l that i f i f acts on a set ft, then each element h€H induces a map o~}i '. ft —> ft, defined by a ^ a - h . Furthermore, a h is a permutat ion of ft, and the map h ^ a h is a homomorphism from i f into the symmetric group Sym(ft) . It should be clear that the given action of i f on ft is completely determined by this homomorphism from i f into Sym(ft) .

Suppose now that the set ft of the previous paragraph happens to be a group, which we now call N . In this case, it is natural to require that for a l l elements h e H , the map a h : iV -> N is actually an automorphism (and not just a permutation) of N . O f course, since this map is automatical ly bo th injective and surjective, it suffices to check that it is a homomorphism, or equivalents , that { x y ) - h = { x - h ) { y - h ) for a l l x,y € N .

Given groups i f and N , we say that i f acts via automorphisms on N i f i f acts on N as a set, and i n addit ion, ( x y ) - h = { x - h ) ( y - h ) for a l l x,y £ N a n d h e H . Just as an action of i f on ft determines and is determined by a homomorphism i f -> Sym(ft) , so too, an action v i a automorphisms of i f on N determines and is determined by a homomorphism i f -> A u t ( i V ) .

Perhaps the most natural examples of actions v i a automorphisms are where i f is actually a subgroup of A u t ( i V ) and n-h is s imply ( n ) h . Other examples of actions v i a automorphisms occur when both i f and N are subgroups of some group G w i th i f C N G ( J V ) , and where the action is given by conjugation wi th in G. In particular, the action of an arbi trary group G on itself by conjugation is an action v i a automorphisms.

It is common to use exponential notat ion instead of dots for actions v i a automorphisms, and so we write n h instead of n - h , where n G N and h € i f . Th i s is natural for conjugation actions, of course, but it is potential ly confusing for actions v i a automorphisms that are not defined by conjugation inside some given group. The following theorem, however, asserts that every action v i a automorphisms is essentially a conjugation action wi th in an appropriate group. To some extent, therefore, this justifies the exponential notat ion for actions v i a automorphisms.

In order to state our result, we again use the notat ion ( ) 0 to describe isomorphisms N N 0 and i f -> i f 0 . Technically, this is ambiguous, however, because we w i l l not assume that N and i f are disjoint, and so i f x lies i n bo th of these groups, it may not be clear which isomorphism to apply

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i n order to compute x 0 . Nevertheless, we trust that this ambiguity w i l l not cause confusion.

3.2. Theorem. L e t H a n d N be g r o u p s , a n d suppose t h a t H acts o n N v i a a u t o m o r p h i s m s . T h e n t h e r e exists a g r o u p G c o n t a i n i n g a n o r m a l s u b g r o u p N 0 = N , c o m p l e m e n t e d by a s u b g r o u p H 0 * H , a n d such t h a t f o r a l l n € N a n d h < E H , w e h a v e

( n h ) o = ( n o ) h o .

H e r e , n h G N i s t h e r e s u l t of l e t t i n g h a c t o n n i n t h e g i v e n a c t i o n , w h i l e ( n o ) h o € N 0 i s t h e r e s u l t of c o n j u g a t i n g n 0 by h 0 i n G.

It follows by L e m m a 3.1 that the group G of Theorem 3.2 is uniquely determined by N and H and the given action v i a automorphisms of H on TV; it is called the semidirect product of N by H w i t h respect to the given action. A l so , by L e m m a 3.1, every group G w i t h a normal subgroup N and complement H is isomorphic to a semidirect product of N by H , and so once we prove Theorem 3.2, it w i l l be fair to say that we have constructed al l possible split extensions (up to isomorphism).

A common notat ion for a semidirect product G of N by H is G = N x > H . (This notat ion is incomplete, of course, because it fails to mention the specific act ion v i a automorphisms that is essential i n the definition. W h e n we construct a group G = N x < H , we should always specify the action.) Note that the semidirect product is a direct product if and only if H 0 < G, and this happens precisely when the conjugation action of H 0 on JV 0 is t r i v i a l , or equivalent^, when the original action of H on AT is t r iv i a l .

We offer a mnemonic for the correct use of the symbol ' V , which looks like " x " w i t h an extra l i t t le vertical line on the right. If G = N x > H , the l i t t le line reminds us that from the point of view of H , the semidirect product can be different from a direct product since H may not be normal . B u t N is normal i n G, and so from the point of view of N , the semidirect product resembles a direct product.

We often identify N w i th N 0 and H w i t h H 0 v i a the isomorphisms x x Q , so that the semidirect product G — N x > H can be viewed as a split extension wi th N < G, N H = G and N n H = 1, and where the original action v i a automorphisms of H on N is just the conjugation action wi th in G. We stress, however, that that this identification is not always a good idea, and it can lead to confusion. Consider, for example, an arbi trary group G and let T = G x G w i t h respect to the natural conjugation action of G on itself. (Somewhat remarkably, it turns out that T = G x G for a l l groups G.)

It is seldom necessary or appropriate to refer to the proof of Theorem 3.2. In other words, the specific construction of the semidirect product is largely

irrelevant. The point here is that given the three ingredients: a group N , a group H and an action v i a automorphisms of H on N , there is a group G, which is unique up to isomorphism and satisfies the conclusions of Theorem 3.2. It is usually unnecessary to know exactly how G is constructed. (Indeed, there is more than one way to prove Theorem 3.2, which means that there is more than one possible construction for the semidirect product.) In order to emphasize this point, we give a few applications of semidirect products before we present the construction.

Reca l l that in Chapter 2, we defined the dihedral group D = D 2 n as a group having a nontr iv ia l cyclic subgroup C of order n such that a l l elements of D - C are involutions. We argued using geometric reasoning that such a group actually exists when n > 3. Us ing semidirect products, we can give a cleaner construction of the dihedral group D 2 n and some related groups.

Let A be an arbi trary abelian group, and let T = ( t ) be cyclic of order 2. Since t is the only nonidenti ty element of T , we can define an action of T on A by setting x t = x ~ l for a l l x e A . (Since t 2 = 1 and ( a ; " 1 ) " 1 = x , this is clearly a genuine action.) Because A is abelian, we have ( x y ) ~ l = x ~ l y ~ l , and so our action of T on A is an action v i a automorphisms. N o w let G = A x T w i t h respect to this action, and identify (as we may) A and T wi th the corresponding subgroups of the semidirect product G. T h e n A < G and A is complemented by T i n G, and thus \ G : A \ = \T\ = 2, and we have \ G \ = 2 \ A \ . A l so , the equation x* = x ' 1 , can be viewed as a statement about conjugation i n the semidirect product G. Now every element g of G - A lies in the coset A t , and so g = a t for some element a G A . In part icular, g

2 = a t a t = aa* = a a ' 1 = 1, and so a l l elements of G - A are involutions. In other words, G is a generalized dihedral group and i n part icular, i f A is cyclic of order n , then G is dihedral of order 2n.

We can also use the semidirect product construction to prove things. A t first glance, the following may seem almost obvious, but its proof depends on nontr iv ia l facts from Chapter 2.

3.3. Corol lary (Horosevskii) . L e t a G A u t ( G ) , w h e r e G i s a n o n t r i v i a l finite g r o u p . T h e n t h e o r d e r o { a ) of a i s less t h a n \ G \ .

Proof. Le t A = ( a ) , so that A is a cyclic subgroup of A u t ( G ) and \ A \ = o ( a ) . Since A C A u t ( G ) , there is a natural action of A on G by automorphisms, and we construct the semidirect product Y = G x > A w i th respect to this action. A s usual, we identify G and A w i th subgroups of T, so that G < T and A is a complement for G. A l so , the conjugation action of A on G i n T is the original action. Furthermore, since by definition, nonidentity automorphisms of G act nontrivial ly, it follows that no nonidenti ty element of A acts t r iv ia l ly on G by conjugation i n T. In other words, A C \ C V { G ) = 1.

3 A 71

Since G is nontr iv ia l and A is cycl ic , Lucchin i ' s theorem (2.20) applies i n T, and we deduce that \ A : K \ < \T : A \ , where K = c o r e r ( A ) . B u t K n G C A n G = 1, and bo th AT and G are normal i n T, and thus K C A n C r ( G ) = 1. Thus o(<r) = \ A \ = \ A : K \ < \ T : A \ = | G | , as required. •

Somewhat similarly, we have the following.

3.4. Corol lary. L e t P be a n a b e l i a n p - s u b g r o u p of A u t ( G ) , w h e r e G i s a finite g r o u p of o r d e r n o t d i v i s i b l e by t h e p r i m e p . T h e n P has a r e g u l a r o r b i t o n G. I n p a r t i c u l a r , if G i s n o n t r i v i a l , t h e n \ P \ < | G | .

Reca l l that if a group P acts on a set f l , then a P -o rb i t A i n f l is regular if | A | = | P | , or equiva len ts (by the fundamental counting principle) the stabilizer i n P of an element of A is t r iv i a l . A l s o , i n the si tuat ion of the corollary, if P > 1 and A is a regular P -o rb i t , then A does not contain the identi ty of G , and so | G | > 1 + | A | > | P | . Th i s shows that the last sentence of Corol la ry 3.4 w i l l follow once the existence of a regular P -o rb i t is established.

P r o o f of Corol lary 3.4. Since P C A u t ( G ) , there is a natural action of P on G v i a automorphisms. Let T = G x P w i t h respect to this action, and as usual, identify P and G wi th subgroups of T. T h e n |T : P | = | G | is not divisible by p , and so P € S y l p ( r ) . A l so , the conjugation act ion of P on G i n r is the original action, and since P consists of automorphisms, only the identi ty i n P acts t r ivial ly . It follows that P n C r ( G ) = 1.

N o w O p ( r ) C P , and so O p ( r ) D G = 1. A l s o , bo th G and O p ( r ) are normal i n T, and hence they centralize each other, and we have O p ( r ) C P n C r ( G ) = 1. Since the Sylow subgroup P is abelian, it follows by Brodkey 's theorem (1.37) that O p ( T ) is an intersection of two Sylow p -subgroups. Replacing them by conjugates i f necessary, we can assume that one of these Sylow subgroups is P , and the other is P x for some element x € T. We thus have P n P x = O p ( T ) = 1.

We can write x = u g , w i th u G P and g € G , and thus P x = P u 3 = P » , and we have We c la im that the P -o rb i t i n G that contains g is regular, and this w i l l complete the proof. It suffices to show that the stabilizer i n P of g is t r iv ia l , but since the act ion of P on G is conjugation in r , what we want to establish is that C P ( g ) = 1. We have C P ( g ) = C P ( g ) 9 , and hence C P ( g ) C P n P 9 = 1, as wanted. •

A comparison of the previous two corollaries suggests that perhaps whenever A i s & cyclic subgroup of A u t ( G ) , there is always a regular A-orb i t i n G . If that were true, it would be a strong form of Coro l la ry 3.3, but alas, it is false.


Fina l ly , we are ready to prove Theorem 3.2 by actually constructing the semidirect product T = N x H . A common way to do this, by analogy w i t h the construction of the external direct product, is to define an appropriate mul t ip l ica t ion on the set of ordered pairs ( h , n ) w i t h h £ H and n £ N . One tedious, but necessary, step using this construction is to check that the denned mul t ip l ica t ion is associative. We choose to use a different approach, which avoids checking associativity. Th i s is accomplished by working i n an appropriate symmetric group, where, of course, associativity is automatic.

P r o o f of T h e o r e m 3.2. Let ft be the set of ordered pairs ( k , m ) , where k € H and m € N . It is t r i v i a l to check that actions of N and of H on ft are denned by the following:

( k , m ) - n = ( k , m n ) and { k , m ) - h = { k h , m h ) ,

where ( k , ra) £ ft and n e N and h £ H . Here, of course, mh is the result of let t ing h act on m i n the given action v i a automorphisms of H on N .

Since N acts on ft, we have a natural homomorphism n i-> n 0 from N into the symmetric group Sym(ft) , where n 0 is the permutat ion a ^ a-n for a £ ft. Wr i te N 0 = { n 0 \ n £ A/'}, so that N 0 C Sym(ft) is a subgroup and ( )o is a surjective homomorphism from N to A ^ . If n 0 = 1, then ( k , m ) - n = ( k , m ) for a l l k £ i f and m £ TV, and in part icular, (1,1) = (1, l ) - n = (1, n) . It follows that n = l , and so the homomorphism ( ) 0 has t r i v i a l kernel. It is thus an isomorphism from N onto N 0 .

Similar ly , i f h £ H , write h 0 £ Sym(ft) to denote the map a H * a - h , and let H 0 = { h 0 \ h £ H } . T h e n i T 0 C Sym(ft) is a subgroup and ( ) 0 is a homomorphism from H onto 7J 0 - (As before, we are using the notat ion ( ) 0

for two different maps.) If h o = 1, then (1,1) = (1, l ) - h = ( h , l h ) , and thus h = 1 and ( ) 0 is an isomorphism from H onto H 0 .

Next , we argue that ( n h ) 0 = (n0)h° for a l l n £ N and h £ H . Since this is a statement about elements of Sym(ft) , it suffices to check that both { n h ) o and {n0)h° have the same effect on each element of ft. We have

( k , m ) ( n h ) o = { k , m ) - n h = ( k , m n h ) .

Also ,

( k , m ) ( n 0 ) h o = { k , m ) { h o ) - l n o h o = { { { k , m ) - h - l ) - n ) - h

= ( O ^ - V ^ V H = ( k h - \ m h ~ \ ) - h

= ( k , m n h ) , where the last equality holds because ( m h ~ 1 n ) h = m n h . (This is val id because the action of i f on TV is an action v i a automorphisms.) We conclude

that ( n h ) 0 = ( n 0 ) h o as claimed, and i n part icular , H 0 C N S y m ( Q ) ( 7 V 0 ) . We can now define G = N 0 H 0 , so that G is a subgroup of Sym(ft) , and N 0 < G. A l l that remains is to show that H 0 is a complement for N 0 i n G. Since by definition, G = N 0 H 0 , it suffices to check that N 0 n H 0 = 1. To this end, suppose that n 0 = h 0 w i t h n G TV and / i G H . T h e n

(/i, 1) = (1, l ) - h = (1, l ) / i 0 = (1, l ) n 0 = (1, l ) - n = ( l , n ) ,

and it follows that / i = 1. Thus n 0 = h 0 = l , and the proof is complete. •

We close this section w i t h a description of the wreath product, which is useful for constructing examples and counterexamples. The ingredients here are two gronps G and H , and a set ft, acted on by G. Let B be the set of a l l functions from ft into H , and make B into a group by defining mul t ip l ica t ion pointwise. Thus i f / , g G B , then f g is defined by the formula ( f g ) ( a ) = f ( a ) g ( a ) for a G ft. It is t r i v i a l to check that B is a group, and i n fact, B is really just the external direct product of |ft| copies of H .

The act ion of G on ft induces an action v i a automorphisms of G on B . If we view B as a direct product, the act ion of G can be described s imply as a permutat ion of coordinates, but it is useful to be more precise. G i v e n / G B and x G G, the function f x on ft is defined so that f x ( a - x ) = / (a) , or equivalents , f x ( a ) = f { a - x ~ l ) . (It is routine to show that this defines an act ion v i a automorphisms, and we omit the proof.)

N o w let W = B x < G w i th respect to this action. T h e n W is the wreath product of H w i t h G, and B , viewed as a subgroup of W, is called the base group of the wreath product. It is common to write W = H I G, but of course, this notat ion for the wreath product is defective i n that it does not mention the set ft or the action of G on ft. If G is given as an abstract group, and no set ft is mentioned, then it is understood that H I G is constructed using the "regular" act ion of G. T h i s means that we take ft = G, w i t h G acting by right mul t ip l ica t ion . Sometimes, this is referred to as the regular wreath product of H by G.

P r o b l e m s 3 A

3A.1 . Let C be a cyclic group of order n divisible by 8, and let z be the unique involut ion i n C.

(a) Show that C has a unique automorphism a such that ca = c~lz for every generator c of C, and show that a has order 2.

(b) Let S = Cx(o- ) , so that |5| = 2 \ C \ . Show that half of the elements o i S - C have order 2 and that the other half have order 4.

(c) Show that the elements of order 2 i n S - C form a single conjugacy class of S, and s imilar ly for the elements of order 4.


Note . The group S is the semidihedral group S D 2 n , al though i n the literature, the word "semidihedral" is usually reserved for the case where n is a power of 2, and there seems to be no standard name for other members of this family of groups.

3A.2 . Let S and C be as i n the previous problem, and let B be the subgroup of index 2 i n G . Show that the elements of order 4 i n S - C form a coset of B , and let Q be the union of this coset and B . Show that Q is a subgroup of order n .

Note . Since a l l elements of Q - B have order 4, the involut ion i n B is the unique involut ion i n Q. The group Q is the generalized quaternion group Qn. (The phrase "generalized quaternion" is often restricted to the case where the order n is a power of 2. The undecorated word "quaternion" is usually reserved for Q 8 . )

3A .3 . Let p be a prime, and suppose that m is a divisor of p-1 w i t h m > 1. Show that there exists a group G of order pm w i t h a normal subgroup P of order p , and such that G/P is cyclic and Z ( G ) = 1.

3A.4 . Let q be a power of a prime p . Show that there exists a group G of order q ( q - 1) w i t h a normal elementary abelian subgroup of order q and such that a l l elements of order p i n G are conjugate.

Hint . Let F be a field of order q and observe that the mult ipl icat ive group of F acts v i a automorphisms on the additive group of F .

3A.5 . Let G be an arbi t rary finite group. Show that G x G = G x G , where the semidirect product is constructed using the natural act ion of G on itself by conjugation.

3A.6 . Reca l l that a group action is faithful if the only group element that fixes a l l points is the identity. Let P be a p-group acting faithfully v i a automorphisms on a group G w i t h order not divisible by p . Show that P acts faithfully on some P-orb i t A i n G .

Hint . Use Theorem 1.38, the generalized Brodkey theorem.

3A.7 . Let a € A u t ( G ) , and suppose that at most two prime numbers divide o { a ) . Show that ( a ) has a regular orbit on G .

3A.8 . Let C be cyclic of order p q r , where p , q and r are distinct odd primes.

(a) Let s £ { p , q , r } . Show that A u t ( G ) contains a unique involut ion a fixing elements of G of order s and inverting elements of prime orders different from s.

3 B 75

(b) A p p l y i n g (a) three times, w i t h s = p , s = q and s = r, we get three involutions i n A u t ( C ) . Show that these together w i t h the identity form a subgroup K C A u t ( C ) of order 4.

(c) Le t G = C x i K w i th the natural action, and let a be the inner automorphism of G induced by a generator of C. Show that ( a ) has no regular orbit on G.

3A.9 . Let W = H l G be a wreath product constructed w i t h respect to a transitive action of G on some set Q. Le t B be the corresponding base group.

(a) Show that C B ( G ) is the set of constant functions from fi into H .

(b) N o w assume that W is the regular wreath product, so that i ) = G. Let C C G be an arbi trary subgroup. Show that there exists beB such that C G ( 6 ) = C.

3 A . 10. G i v e n a finite group H and a prime p , show that there exists a group G having a normal abelian p-subgroup A such that G splits over A , where G/A = H and A = C G { A ) .

3 B

The Schur-Zassenhaus theorem is a powerful result that guarantees that a finite group G always splits over a normal H a l l subgroup N . (Recall that by definition, N is a H a l l subgroup of G i f | / V | and \ G : N \ are relatively prime.)

The Schur-Zassenhaus theorem says even more: i f N is a normal H a l l subgroup of G, then a l l complements for N i n G are conjugate. Unfor tunately, however, the proof of the conjugacy part of the Schur-Zassenhaus theorem requires the addi t ional hypothesis that either N or G / N is solvable. (We review the definition and basic properties of solvable groups later i n this section.) If we are wi l l ing to apply the very deep Fei t -Thompson odd-order theorem, however, we see that this solvabil i ty requirement is not really a l imi ta t ion . Th is is because the assumption that N and G / N have coprime orders guarantees that at least one of these groups has odd order, and hence is solvable by the Fe i t -Thompson theorem. The conjugacy part of the Schur-Zassenhaus theorem is therefore true uncondit ionally, al though no direct or elementary proof of this is known. For the purposes of this book, we w i l l assume solvabili ty when appropriate, and we w i l l generally not appeal to the Fei t -Thompson theorem.

O f course, i f H is a complement for a normal subgroup N i n G, then \ H \ = \ G : N \ . It is useful to observe that i n the s i tuat ion of the Schur-Zassenhaus theorem, where |JV| and \ G : N \ are coprime, the converse of this

statement is also true. If H C G and \ H \ = \ G : N \ , then i f is a complement for TV i n G. To see this, observe that \ H \ = \ G : N \ is coprime to \ N \ , and thus TV n H = 1. F r o m this, we deduce that \ N H \ = \ N \ \ H \ = \ G \ , and hence N H = G, and indeed H is a complement for TV as claimed.

A l t h o u g h several proofs of the Schur-Zassenhaus theorem are known, the key step i n a l l of them is to establish the result i n the case where N is abelian. W h a t are essentially routine tools can then be used to reduce the general s i tuat ion to this case. (Of course, the assumption for the conjugacy part that either N or G / N is solvable is used i n the reduction.) Our first major goal, therefore, is the following.

3.5. Theorem. L e t N < G, w h e r e G i s a finite g r o u p , a n d assume t h a t N i s a b e l i a n a n d t h a t \ N \ a n d \ G : N \ a r e c o p r i m e . T h e n N i s c o m p l e m e n t e d i n G, a n d a l l c o m p l e m e n t s a r e c o n j u g a t e .

If our goal were to find a certain n o r m a l subgroup i n G, we might proceed by constructing an appropriate homomorphism from G, and then considering its kernel. B u t we need a complement for TV, which w i l l not generally be normal . We w i l l construct it as the "kernel" of a certain k ind of distorted homomorphism.

Let G act v i a automorphisms on N , and consider a map TV. T h e n ip is a crossed homomorphism i f <p(xy) = <p(x)*<p(y) for a l l elements x and y i n G. (Of course, the exponent y here refers to the given action v i a automorphisms of G on N . )

If the action of G on TV is t r iv ia l , which means that n 9 = n for a l l n G N and g G G, then a crossed homomorphism from G to N is s imply an ordinary homomorphism. B u t we w i l l see that there are also interesting examples when the action is nontr ivia l . For example, let G be an arbi trary group and suppose TV < G, so that G acts on N by conjugation. N o w fix an element n G TV, and define the map N by setting i p ( x ) = [ x , n ] . (Recal l that [ x , n] is the commutator of x and n , which, by definition, is the element x ^ n ^ x n = { n ~ l ) x n . ) Since n G N < G, it is clear that <p(x) = [ x , n ] lies i n N , as claimed. Now ( f ( x y ) = [ x y , n ] , and an easy computat ion, which we omit, establishes the identity [ x y , n ] = [ x , n ] 9 [ y , n ] . Thus ( p ( x y ) = <p(x)v<p(y), and so N to be the subset ker^) = { i £ G | (f(x) = 1}. The following establishes a few basic facts about crossed homomorphisms and their kernels.

3.6. L e m m a . L e t G a n d N be g r o u p s , a n d suppose t h a t G acts v i a a u t o m o r p h i s m s o n N . L e t ip : G - > N be a c r o s s e d h o m o m o r p h i s m w i t h k e r n e l K . T h e f o l l o w i n g t h e n h o l d .

3 B 77

(a) <p(l) = 1.

(b) i f i s a s u b g r o u p of G.

(c) I f x , y e G, t h e n <p{x) = <p(y) if a n d o n l y if K x = K y .

(d) \ G : K \ = \ < p ( G ) \ .

Proof. We have

<P(I) = < p ( i - i ) = <p(i) V ( i ) = v ( i ) 2 ,

and it follows that 1 = < p ( l ) . Th i s establishes (a) and shows that i f is nonempty. If k € K , we have

i = ^,(1) = ^ ( a r 1 ) = w i k f ' ^ i k - 1 ) = i f i k ' 1 ) ,

where the last equality holds because <p{k) — 1 and l f f = 1 for a l l elements g G G. Thus fc"1 G i f , and i f is closed under inverses. A l so , i f x , y G If , then

<p(xy) = < p ( x ) M y ) - l y l = 1,

and so xy G K , and hence i f is a subgroup, proving (b).

N o w i f x , y G G and K x = K y , we can write y = kx w i t h k G K . T h e n

p(y) = i f ( k x ) = < f ( k ) x < p ( x ) = < p { x ) ,

where the last equality holds because <p(k) = 1.

Conversely, i f <p(x) = <^(y), we compute that

p ( x y - 1 ) = ^ ( x ) y ~ ^ ( y ) = ^ ( y ) y ~ ^ ( y ) = ^ ( y y " 1 ) = p ( l ) = 1,

and so xy'1 G iT . Thus x G i f y and i f x = i f y , as wanted. F ina l ly , since <p is constant on right cosets of i f and takes dist inct values on distinct right cosets, it follows that \ i p ( G ) \ is exactly the number of such cosets. Th is proves (d). •

Our next task is to construct an appropriate crossed homomorphism from a finite group G into an abelian normal subgroup N . To do this, we consider the transversals for N in G. These are the subsets T of G constructed by choosing exactly one member from each coset of N i n G. (Note that since N is normal , left cosets and right cosets are the same, so we need not specify "left transversal" or "right transversal".) Since right mul t ip l ica t ion by an element g G G permutes the right cosets of an arbi trary subgroup of G, it follows that if T is a transversal for N , then T g is also a transversal for N . We write T to denote the set of a l l transversals for N i n G.

If x , y G G, write x = y i f x and y lie i n the same coset of N . Thus x = y i f and only if N x = Ny, or equivalently, x N = y N . If S, T G T , the relat ion s = t for s G S and t G T defines a natural bijection between S and


T, and we use this bijection to define an element of TV that we cal l d ( S , T ) . (In some sense, d ( S , T ) measures the difference between S and T.) We set

d ( S , T ) = Y[s-H, s=t

where the product runs over a l l s E S and i E T such that s = t. O f course, if s = t, then s N = t N , and so s ~ H £ TV, and hence each factor i n the product lies i n TV. Also , because we are assuming that TV is abelian, the order i n which the factors appear is irrelevant, and thus d(S, T ) is a well defined element of TV.

Some elementary properties of the function d are given i n the following.

3.7. L e m m a . L e t TV be a b e l i a n a n d n o r m a l i n a finite g r o u p G, a n d l e t S, T a n d U be t r a n s v e r s a l s f o r TV. Then u s i n g t h e n o t a t i o n d e f i n e d a b o v e , t h e f o l l o w i n g h o l d .

(a) d ( S , T ) d ( T , U ) = d ( S , U ) .

(b) d { S g , T g ) = d { S , T ) 9 f o r a l l g E G .

(c) d { S , Sn) = n l G : 7 V l f o r a l l n E TV.

Proof. If s = t and t = u for s E S, t E T and u E U, then TVs — N t = N u , and so s = u. Since ( s " 1 i ) ( r 1 u ) = s~lu and the order of factors i n the abelian group TV is irrelevant, (a) follows.

N o w let g E G. If s = t, then TVs = N t , and so TVs 5 = N t g , and sg = t g . The factor { s g ) ~ l { t g ) of the product defining d { S g , T g ) is equal to ( s ^ i ) 9 , and assertion (b) follows.

F ina l ly , i f n E TV, we have sTV = (sn)TV, and so s = sn. Each factor S - l { s n ) of the product defining d { S , Sn) equals n , and so (c) follows because | 5 | = \ G : TV|. •

We are now ready to prove the case of the Schur-Zassenhaus theorem where TV is an abelian normal H a l l subgroup of G.

P r o o f of T h e o r e m 3.5. F i x an arbi trary transversal T for TV in G, and define 9 : G -> TV by setting 9 { g ) = d { T , T g ) . We argue first that 9 is a crossed homomorphism w i t h respect to the conjugation action of G on TV. If x,y E G, then using L e m m a 3.7(a,b), we compute that

d(T, T x y ) = d(T, T y ) d { T y , T x y ) = d { T , T y ) d ( T , T x ) y ,

and so 9 { x y ) = B ( y ) 9 ( x ) v . Th i s , of course, is equal to 9{x)«9{y) since TV is abelian, and so 9 is a crossed homomorphism, as claimed.

For n E TV, we have 9 ( n ) = d(T, T n ) = J G : 7 V I by L e m m a 3.7(c). Since we are assuming that |TV| and \ G : TV| are coprime, however, the map x i-> X \ G : N \ is a permutat ion of the elements of TV. (One way to see this is to observe

3 B 79

that this map has an inverse: the map x i-> x k , where the integer k is chosen so that k \ G : 7V| == 1 mod \ N \ . ) It follows that 9 maps N onto N , and i n part icular, 9 ( G ) = N . Now let H = ker(0), so that i f is a subgroup by L e m m a 3.6(b). Also , \ N \ = \ 9 { G ) \ = \ G : H \ by L e m m a 3.6(d), and hence \ N \ \ H \ = \ G \ and \ H \ = \ G : N \ . We have already observed that since \ N \ and \ G : N \ are coprime, this is sufficient to guarantee that i f is a complement for N , as wanted.

N o w suppose that K is an arbi trary complement for N i n G. Since N K = G, we see that every coset of N i n G contains an element of K , and i n fact, each coset of N contains a unique element of K because N C \ K = \. Thus K is a transversal for N , and we let m = d{K, T ) € N . We saw that the restriction of 9 to N is the map x x l G : A f l , which is surjective, and so we can find n € N such that 9 ( n ) = m, and thus 9 ( n ~ 1 ) = TO"1.

To complete the proof, we show that K n — H . Since \K\ = \ G : N \ = \ H \ , i t suffices to show that K n C i f , or equivalently, that 9 ( k n ) = 1 for a l l keK. (Recall that H = ker(0).) F i r s t , observe that i f G AT, then

m f c = d ( K , T ) k = d ( K k , T k ) = d ( K , T k ) = d ( K , T ) d ( T , T k ) = m 9 ( k ) ,

and so 1 = { m k ) - 1 m 9 { k ) = ( m k ) - 1 9 { k ) m . We have

9 { k n ) = 9 { n - 1 k n ) = 9 ( n - 1 k ) n 9 ( n ) = 0 ( n _ 1 f c ) m

= ( m _ 1 ) f c 0 ( f c ) m ,

where we were able to drop the exponent n i n the th i rd equality because N is abelian. Thus 9 { k n ) = 1, and we are done. •

Next , we prove the existence part of the Schur-Zassenhaus theorem wi th out assuming that the normal H a l l subgroup is abelian.

3.8. Theorem. L e t N < G, w h e r e G i s a finite g r o u p a n d \ N \ a n d \ G : N \ a r e c o p r i m e . T h e n N i s c o m p l e m e n t e d i n G.

Proof. We proceed by induct ion on \ G \ . Suppose that there exists a subgroup K < G such that N K = G. T h e n \ K : K C \ N \ = \ G : N \ is coprime to | N | , and hence it is also coprime to \ K n N \ . Since K n N < K , the inductive hypothesis guarantees that K n N is complemented i n AT, and we choose a complement H . T h e n | i f | = \ K : K n JVj = |G : 7V| and as we have seen, this guarantees that i f is a complement for N i n G.

We may thus suppose that there is no proper subgroup K of G such that N K = G, and so i n part icular, N must be contained in every max ima l subgroup of G. In other words, N C $ ( G ) , the F ra t t i n i subgroup, and so

N is nilpotent. A l so , we can assume that N > 1 or else G is the desired complement for N i n G.

_ N o w let Z = Z( /V),_and^bserve that 1 < Z < G. Work ing i n the group G = G / Z , we_see that | G : N \ = \ G : 7V| is coprime to |TV| and hence is also coprime t o \ N \ . It follows by the inductive hypothesis that G splits over N , and we let K be a complement. T h e n G = N K = N K , and hence G = N K . It follows by the ear l ie rpar t of the proof t h a t K cannot be proper, and we have K = G. Because K is a complement for N , however, their intersection must be the t r i v i a l subgroup Z . Thus

Z = N n K = N n G = N ,

and hence TV = Z is abelian. The result now follows v i a Theorem 3.5. •

A s we said, it is necessary to assume that either N or G / N is solvable i n order to to prove the conjugacy part of the Schur-Zassenhaus theorem. Perhaps, therefore, a review of some basic facts about solvable groups would be appropriate, and so we begin w i t h the definition. A (not necessarily finite) group G is solvable if there exist normal subgroups N t such that

1 = N 0 C N i C N 2 C • • • C N r = G,

where N / N ^ is abelian for 1 < i < r . (In part icular, a l l abelian groups are solvable since i f G is abelian, we can take Ao = 1 and TVi = G.) A finite chain of nested normal subgroups of G , extending from the t r i v i a l subgroup to the whole group, is called a normal series for G , and thus G is solvable precisely when it has a normal series w i t h abelian factors.

Reca l l that the commutator subgroup (or derived subgroup) of an arbi t rary group G is the subgroup G' generated by a l l commutators [ x , y ] = x ~ x y ~ x x y , w i t h x,y G G . Since [ x , y ] = 1 i f and only i f x and y commute, we see that G is abelian i f and only if G' = 1. A l so , if tp : G -> H is a surjective homomorphism, then ip maps the set of commutators i n G onto the set of commutators in H , and thus <p(G') = H ' . It follows that H is abelian if and only i f G' C ker(</?), and i n particular, i f N <3 G , then G / / Y is abelian precisely when G' C AT. In other words, the derived subgroup of G is the unique smallest normal subgroup of G having an abelian factor group.

The subgroup ( G ' ) ' is usually denoted G", and its derived subgroup, is G'". Th i s notat ion rapidly gets unwieldy, however, and so we write G ^ for the n t h derived subgroup. In other words, G ^ = G and for n > 0, we define G^ = (G^" 1))'. The subgroups G ^ are characteristic, and they constitute the derived series of G .

The connection w i t h solvabili ty is the following.

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3.9. L e m m a . A g r o u p G i s s o l v a b l e if a n d o n l y i f G ^ = 1 f o r some i n t e g e r m > 0.

P r o o f . If G is solvable, choose normal subgroups N i such that

1 = N 0 C • • • C N r = G,

where N i / N i - i is abelian for 1 2) that G " C (7V r _ i ) ' C i V r _ 2 . Cont inu ing like this, we get G ( m ) C / V r _ m for m < r, and in part icular, G W C N o = 1.

Conversely, suppose that G ( m ) = 1. Since the subgroups G « are characteristic i n G , they are certainly normal , and we have the normal series

1 = G ( m ) c G ( m _ 1 ) c • • • c G ( 0 ) = G .

The factors here are abelian, and so G is solvable. •

The characterization of solvabil i ty i n terms of the derived series allows us to prove a number of useful facts.

3.10. L e m m a . L e t G be a n a r b i t r a r y g r o u p .

(a) I f H Q G , t h e n C G ^ f o r a l l m > 0 .

(b) I f H i s a s u r j e c t i v e h o m o m o r p h i s m , t h e n 0.

(c) If G i s s o l v a b l e a n d H C G, t h e n H i s s o l v a b l e .

(d) I f G i s s o l v a b l e a n d N < G, t h e n G / N i s s o l v a b l e .

(e) L e t N < G a n d suppose t h a t N a n d G / N a r e s o l v a b l e . T h e n G i s s o l v a b l e .

P r o o f . If H C G , then H ' C G' and (a) follows by repeating this observation. Similar ly , (b) follows by repeated appl icat ion of the fact that <p(G') = H ' i f ip : G - > H is a surjective homomorphism.

N o w suppose that G is solvable, so that G^ = 1 for some integer n . If H C G , then C G ( n ) = 1 by (a), and thus H is solvable, proving (c). If i f is a surjective homomorphism, then i f ( n) = ( ^ ( G ^ ) = ^(1) = 1, and H is solvable. In part icular , if N < G and we take i f be the canonical homomorphism. Since i f is solvable, i p ( G ^ ) = i f (") = 1 for some integer n , and thus G ( n ) C ker(y>) = TV. A l so , since /V is solvable, we have 7V"(m) = 1 for some integer m, and thus G ( m + n ) = ( G ^ ) ^ C / V ( m ) = 1, and thus G is solvable. •

Reca l l that an abelian p - g m u p P is elementary abelian i f x p = 1 for a l l x £ P . A useful fact about solvable groups is that a min ima l normal subgroup of a finite solvable group must be an elementary abelian p-group for some prime p . The following is a slightly more general version of this.

3.11. L e m m a . L e t M be a s o l v a b l e m i n i m a l n o r m a l s u b g r o u p of a n a r b i t r a r y g r o u p G. T h e n M i s a b e l i a n , a n d if M i s f i n i t e , i t i s a n e l e m e n t a r y a b e l i a n p - g r o u p f o r some p r i m e p .

Proof. Since M is solvable, its derived series must eventually reach 1, and thus since M > 1, we conclude that M ' < M . B u t M ' is characteristic i n M , and hence it is nermal in G. It follows by the min imal i ty of M that M ' = 1, and thus M is abelian, as wanted. N o w assume that M is finite, and let u £ M be an element of prime order, say p . Since M is abelian, the set S = { x £ M | x p = 1} is a characteristic subgroup of M and hence it is normal i n G. A l so , S > 1 since u £ S, and thus S = M by the min imal i ty of M . T h i s completes the proof. •

F ina l ly , we mention that the derived length of a solvable group G is the smallest integer n such that G^ = 1. (Thus, for example, a solvable group has derived length 1 precisely when it is nont r iv ia l and abelian.) The proof of L e m m a 3.10 shows that subgroups and homomorphic images of a solvable group w i t h derived length n have derived lengths at most n . Also , if N < G, where N is solvable w i t h derived length m and G / N is solvable w i t h derived length n , then the derived length of G is at most m + n .

We are now ready to complete our proof of the Schur-Zassenhaus theorem.

3.12. Theorem. L e t N < G, w h e r e G i s finite a n d \ N \ a n d \ G : N \ a r e c o p r i m e , a n d assume e i t h e r t h a t N i s s o l v a b l e o r t h a t G / N i s s o l v a b l e . T h e n a l l c o m p l e m e n t s f o r N i n G a r e c o n j u g a t e .

Proof. We proceed by induct ion on \ G \ . Let U C G be arbitrary. We w i l l show that U satisfies the hypotheses of the theorem w i t h respect to its normal subgroup U n N , and also that i f U contains a complement H for N i n G, then H is also a complement for U P i N i n U. It w i l l then follow v i a the inductive hypothesis that i f U is proper in G and contains two complements for N , these complements are conjugate in U and hence in G, as wanted.

Now \ U n N \ divides \ N \ and \ U : U n N \ = \ U N : N \ divides \ G : N \ , and thus \ U C \ N \ and \ U : U n N \ are coprime. Also , either G / N is solvable, in which case U / { U n N ) = U N / N C G / N is solvable, or else N is solvable, and so U n N is solvable. If U contains a complement H for N in G, then \ H \ divides |J7| and is coprime to \ U n N \ , and hence \ H \ divides \ U : U n N \ . B u t \ U : U n TV| divides \ G : N \ = \ H \ , and so we have \ H \ = \U : U n I f |,

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and therefore i f complements U n N i n U. T h i s establishes the assertions of the previous paragraph.

Next^we_consider factor groups. Suppose L < G , and write G = G / L , so that N < G . Since N is a homomorphic image of N , it is solvable if N is, and its order divides \ N \ . A l so G / N = G / N L is a homomorphic image of G/N, and_so it is solvable i f G / N is, and its order divides \ G : N \ . In other words, G satisfies the hypotheses w i t h respect to the normal_subgroup N . If_H is a complement for N in G, then \ H \ _ \ s coprime to \ N \ , and so AT n i f = 1. A l so , N H = ~NH = G, and thus H is a complement for N in G .

Let i f and K be complements for N i n G . If L_> 1, then by the inductive hypothesis applied i n G, we deduce that i f and K are conjugate in G , and thus H L and ATL are conjugate i n G. We can thus write i f f , = ( K L ) 9 for some element g e G, and hence H L contains both H and K 9 , each of which is a complement for N in G. If i f L < G, therefore, it follows by the result of the first paragraph of the proof that i f and K 9 are conjugate, and thus i f and K are conjugate, as required. We can thus assume that H L = G for every nonidenti ty normal subgroup L of G.

Suppose that N is solvable, and note that we can assume that N > 1, or else i f = G = K and there is nothing to prove. Let L C N be a min ima l normal subgroup of G, so that L is abelian by L e m m a 3.11. B y the result of the previous paragraph, i f L = G, and since L C N and \ N \ = \ G : i f |, it follows that N = L , and so N is abelian and we are done by Theorem 3.5.

F ina l ly , assume that G / N is solvable. We can assume that N < G, or else i f = 1 = K and there is nothing to prove. Let M / N be min ima l normal in G/N, and apply L e m m a 3.11 to conclude that M / N is a p-group for some prime p . A l so , since p divides \ G : 7V|, we see that p does not divide |7V|.

Since N H = G and N C M , Dedekind 's l emma yields M = N ( M n i f ) , and it follows that M n i f complements N i n M . In part icular | M n i f | = | M : 7V|, and so M n i f is ap-group. Also , | M : M n i f | = \ N \ is not divisible by p , and we conclude that M n i f is a Sylow p-subgroup of M . Similar ly, M n K is a Sylow p-subgroup of M , and thus by the Sylow C-Theorem, M n i f = ( M n A T ) m = M n A T m for some element m G M .

N o w write L — M n i f = M C \ K m , and note that L < i f and L o i f m since M < G. Thus i f and A T m are complements for N i n G that are contained i n N G ( L ) . If N G ( L ) < G , we are done by the first part of the proof. We can assume, therefore, that L < G , and since L > 1, we have L H = G. B u t L C i f , and so i f = G and thus also K = G and we are done i n this case too. •

Perhaps it is of interest to note that the conjugacy part of Theorem 3.5 was used i n the proof of Theorem 3.12 only i n the case where N is solvable. Conjugacy i n the case where G / N is solvable ul t imately depends on the conjugacy of Sylow subgroups, but not on Theorem 3.5.

P r o b l e m s 3 B

3B.1 . Show that a max ima l subgroup of a finite solvable group must have prime power index.

H i n t . Look at a min ima l normal subgroup and work by induct ion.

3B.2. Suppose 1 = N 0 < N x < • • • < N r = G, where the factors N i / N ^ are abelian for 1 < i < r . Show that G is solvable. (Note that we are not assuming that the subgroups iVj are normal i n G.)

3B.3. Let G be finite and let 1 = 7V0 < N x < • • • < N r = G, where the factors ZVi/TVi- i are simple for 1 < i < r . Show that G is solvable i f and only i f these factors a l l have prime order.

N o t e . Reca l l that a subnormal series such as { N } , where the factors are simple, is a composi t ion series for G, and note that if G is finite, such a series necessarily exists. A l though G may have several different composit ion series, the Jordan-Holder theorem asserts that the factors N i / N ^ are uniquely determined by G (except for the order in which they occur). Th is problem says that solvable finite groups are exactly the groups for which every composi t ion factor has prime order.

3B.4. Let N < G , where \ N \ and \ G : N \ are coprime. Le t U C G, where \ U \ divides \ G : N \ , and assume that either N or U is solvable. Show that U is contained i n some complement H for N i n G.

3B.5. Suppose that P e S y l p ( G ) and that P C Z(G). Show that the set X of elements of G w i t h order not divisible by p is a subgroup of G, and that G = X x P .

3B.6. Let N < G and g € G, and suppose that when N g is viewed as an element of G/N, it has has order TO.

(a) Show that there exists h € N g such that every prime divisor of o { h ) divides m.

(b) If TO and \ N \ are coprime, show that the element h of part (a) must have order TO.

(c) Now assume that N g is conjugate to its inverse i n G / N and that TO and \ N \ are coprime. Let h € N g be as i n (a). Show that h is conjugate to its inverse in G.

P r o b l e m s 3 B 85

H i n t . For (a), find a cyclic rr-subgroup C such that N C = N ( g ) , where rr is the set of pr ime divisors of m. For (c), let x € G , where ( N g ) N x = ( N g ) - \ Observe that x normalizes N ( h ) . Consider the subgroup ( h ) x .

3B.7. A group G is supersolvable i f there exist normal subgroups N i w i t h

1 = 7V0 C TVi C • • • C N r = G,

and where each factor N i / N ^ is cyclic for 1 < i < r. Clearly, supersolvable groups are solvable, and it is routine to check that subgroups and factor groups of supersolvable groups are supersolvable. Suppose now that G is finite and supersolvable.

(a) Show that the order of every m i n i m a l normal subgroup of G is prime.

(b) Show that the index of every m a x i m a l subgroup of G is prime.

H i n t . For (b), look at a min ima l normal subgroup and work by induct ion.

3B.8. Le t G be finite, and assume that every m a x i m a l subgroup of G has prime index. Show that G is solvable. Show also that G has a normal Sylow p-subgroup, where p is the largest pr ime divisor of \ G \ , and that G has a normal g-complement, where q is the smallest prime divisor of \ G \ .

Note . In fact, G must be supersolvable, but this theorem of B . Hupper t is much harder to prove. Reca l l that a g-complement i n a group G is subgroup w i t h g-power index and having order not divisible by q. In other words, it is a subgroup whose index is the order of a Sylow g-subgroup.

3B.9 . Let G be finite and supersolvable. If n is a divisor of \ G \ , show that G has a subgroup of order n .


3B.10. Let G be finite and supersolvable, and suppose that p is the largest prime divisor of \ G \ . Show that G has a normal Sylow p-subgroup.


3B.11. A s usual, let $ ( G ) be the F ra t t i n i subgroup of the finite group G. Show that every pr ime divisor of | $ ( G ) | also divides | G : $ ( G ) | .

3B.12. Let G be solvable, and assume that $ ( G ) = 1. Let M be a max ima l subgroup of G , and suppose that H C M . Show that G has a subgroup w i t h index equal to | M : H \ .

3B.13. Show that every finite group contains a unique largest solvable normal subgroup.

3B.14. Let F — F ( G ) , where G is finite, and let C = C G ( F ) . Show that Z ( F ) is the unique largest solvable normal subgroup of C.

H i n t . P rob lem I D . 19 is relevant.

Note . If G is solvable, it follows that C = Z ( F ) . Th i s shows that for solvable groups, F ( G ) 5 C G ( F ( G ) ) .

3B.15. (Berkovich) Let G be solvable, and let i f < G be a proper subgroup having the smallest possible index i n G . Show that H < G .

3 C

There exists an extensive and well developed theory of finite solvable groups to which we give only superficial mention here. (Readers interested in pursuing this might wish to consult the comprehensive book on solvable groups by Doerk and Hawkes.) Under ly ing this theory are some results of P h i l i p H a l l analogous to the Sylow E - , C - and D-theorems, but where sets of primes replace ind iv idua l primes. Hal l ' s results are not especially difficult to prove directly, but they also follow easily from the Schur-Zassenhaus theorem, and that is why we present them here.

Reca l l that i f vr is a set of primes, then a vr-group is a finite group such that a l l primes d iv id ing its order lie i n TT , and a vr-subgroup of a group G is s imply a subgroup that happens to be a vr-group. A H a l l vr-subgroup of a finite group G is a rr-subgroup w i t h index involving no prime of vr.

For convenience in discussing Hal l ' s theorems and related topics, we say that an integer m is a vr-number if every prime divisor of m lies i n vr. Also , we write vr' to denote the complement of vr i n the set of a l l prime numbers, so that a vr'-number is an integer w i t h no divisors i n vr. (In the case where TT consists of a single prime p , we generally write p ' in place of { p } ' . ) A subgroup i f of G , therefore, is a H a l l rr-subgroup of G precisely when \ H \ is a vr-number and | G : i f | is a vr'-number, or equivalently, \ H \ is the largest vr-number d iv id ing | G | .

O f course, if vr consists of just one prime, so that vr = { p } , then a H a l l vr-subgroup is just a Sylow p-subgroup, and we know by the Sylow E-theorem that such a subgroup actually exists for every finite group. B u t if |vr| > 1, then an arbi t rary finite group G can fail to have a H a l l vr-subgroup. The si tuat ion is different, however, if G is solvable.

3.13. T h e o r e m ( H a l l - E ) . Suppose t h a t G i s a finite s o l v a b l e g r o u p , a n d l e t TT be a n a r b i t r a r y set of p r i m e s . T h e n G has a H a l l t r - s u b g r o u p .

Proof. We can certainly assume that G > 1, and we proceed by induct ion on | G | . Let M be a min ima l normal subgroup of G a n d let J f be a H a l l vr-subgroup of G = G / M , where H D M . (Of course, i f exists by the inductive

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hypothesis.) T h e n \ H : M \ = \ H \ is a Tr-number and | G : H \ = \ G : H \ is a vr'-number.

B y Theorem 3.11, we know that M is a p-group for some prime p . If p G vr, then since \ H \ = \ M \ \ H : M | , it follows that \ H \ is a vr-number, and thus H is a H a l l vr-subgroup of G, as wanted. If p & TT , on the other hand, then | M | and \ H : M \ are coprime, and hence by the Schur-Zassenhaus theorem, M has a complement K in H . T h e n \ K \ = \ H : M \ is a vr-number and | G : AT| = | G : H \ \ H : K \ = \ G : H \ \ M \ is a rr '-number. In this case, K is the desired H a l l vr-subgroup of G . •

We used the Schur-Zassenhaus theorem to prove Theorem 3.13, but nothing that deep is really needed. In fact, the H a l l E-theorem can be proved using a fairly elementary argument.

Next , we present (with a somewhat sketchy proof) the H a l l C-theorem, but we defer the H a l l D-theorem to the problems at the end of this section.

3.14. T h e o r e m ( H a l l - C ) . Suppose t h a t G i s a finite s o l v a b l e g r o u p , a n d l e t TT be a n a r b i t r a r y set of p r i m e s . Then a l l H a l l T x - s u b g r o u p s of G a r e c o n j u g a t e .

Proof. A s i n the previous proof, we proceed by induct ion on | G | . Assuming that G > 1, choose a min ima l normal subgroup M of G , which must be a p-group for some prime p . Let H and K be H a l l vr-subgroups of G , and observe that H and K _ a v e Ha]l_ vr-subgroups of G = G / M . B y the inductive hypothesis, therefore, H and K are conjugate in G , and it follows that H M and K M are conjugate in G . We can thus write H M = { K M ) 9 for some element g G G .

If p G vr, then \ H M \ and \ K M \ are vr-numbers. B u t \ H \ = \ K \ is the largest vr-number d iv id ing | G | , and it follows that H = H M and K = K M , and so H = K 9 , as wanted. If p vr, then H and K 9 are complements i n H M for the normal H a l l subgroup M . T h e conjugacy part of the Schur-Zassenhaus theorem then yields the result. •

One might wonder how essential solvabil i ty is i n the H a l l E-theorem. Certainly, some nonsolvable groups have H a l l vr-subgroups for some interesting sets vr of primes. (By "interesting" here, we mean cases where vr contains more than one, but not a l l of the pr ime divisors of the group order. In other cases, the existence of a H a l l vr-subgroup is either a t r ivia l i ty , or is guaranteed by Sylow's theorem.) For example, the simple group A 5 contains the H a l l {2, 3}-subgroup A 4 , and the simple group of order 168 contains a H a l l {3, 7}-subgroup and a H a l l {2, 3}-subgroup.

We say that G satisfies E T i f i t contains a H a l l vr-subgroup. We know that solvable groups satisfy E T for a l l sets vr of primes, and perhaps surprisingly,

the converse is true: if G satisfies E v for a l l IT, then G must be solvable. In fact, G is solvable i f it satisfies E T for a l l prime sets of the form TT = p ' , where p is an arbi t rary prime. (Recall that p ' is the complement of { p } in the set of a l l primes.) A H a l l p'-subgroup of G is called a p-complement i n G, and of course, i f \ G \ is not divisible by p , then G is a p-complement of itself. The condi t ion E p > has no content i n this case, and so we can state the converse of the H a l l E-theorem as follows.

3.15. Theorem. Suppose t h a t a finite g r o u p G has a p - c o m p l e m e n t f o r a l l p r i m e d i v i s o r s p of \ G \ . T h e n G i s s o l v a b l e .

W h a t does this converse of Hal l ' s E-theorem say in the case where \ G \ involves just two (or fewer) primes? If \ G \ = p a q b , where p and q are prime and a and b are nonnegative integers, then a Sylow g-subgroup of G is a p-complement and a Sylow p-subgroup of G is a g-complement. In this si tuation, Sylow's theorem guarantees that the hypotheses of Theorem 3.15 are satisfied, and thus the theorem says that every group of order p a q b is solvable. T h i s is Burnside 's famous pY-theorem, which we are certainly not prepared to prove at this point. (We present a proof i n Chapter 7, however.) B u t i f we are wi l l ing to assume Burnside 's theorem, it is not hard to establish Theorem 3.15 i n its full generality.

We need a few prel iminary results. The first of these is fairly standard, and it also appears i n the gronp-theory review in the appendix.

3.16. L e m m a . L e t H , K C G, w h e r e G i s a finite g r o u p , a n d suppose t h a t \ G : H \ a n d \ G : K \ a r e c o p r i m e . T h e n G = H K a n d \ H : H n K \ = \ G : K \ .

Proof. Let D = H n K , and observe that since \ G : D \ = \ G : H \ \ H : D \ , the index \ G : D \ is divisible by \ G : H | . Similar ly, \ G : D \ is divisible by \ G : K \ , and since we are assuming that \ G : H \ and \ G : K \ are coprime, we conclude that their product divides \ G : D \ . In particular, \ G : H \ \ G : K \ < \ G : D \ , and hence \ G \ < \ H \ \ K \ / \ D \ = \ H K \ . It follows that H K = G, as wanted. A l so , since equality holds here, we obtain \ G : K \ = \ G \ / \ K \ = \ H \ / \ D \ = \ H : H n K \ , as wanted. •

We mention that this proof of L e m m a 3.16 is val id only i f \ G \ is finite, but i n fact, the lemma is true even if G is infinite, assuming, of course, that H and K have finite coprime indices.

The following is due to H . Wie landt .

3.17. Theorem. L e t H , K , L C G, w h e r e G i s a finite g r o u p , a n d suppose t h a t t h e i n d i c e s \ G : H \ , \ G : K \ a n d \ G : L \ a r e p a i r w i s e c o p r i m e . If each o f H , K a n d L i s s o l v a b l e , t h e n G i s s o l v a b l e .

3 C 89

Proof. Since the result is t r iv i a l i f | G | = 1, we can proceed by induct ion on \ G \ . Suppose that N < G, and consider the subgroups H , K and L of G = G/N. Since these are homomorphic images of the solvable groups H , K and L , respectively, each of them is s o l v a b l e ^ A l s c J G : H | = | G : N ~ H \ = | G : N H \ , which divides | G : H \ . S imilar ly , | G _ ^ / ^ d i v i d e s | G : K \ and | G : L \ divides | G : L | , a n d ^ o the indices of H , K and L are pairwise coprime. If N > 1, therefore, G is solvable by the induct ive hypothesis.

If H = 1, then | G | = | G : i f | is relatively pr ime to | G : K \ , and thus G = K is solvable. We can assume, therefore, that H > 1, and we choose a m in ima l normal subgroup M of H . B y L e m m a 3.11, we know that M is a p-group for some prime p, and since p cannot divide bo th | G : Af| and | G : L | , we can assume that p does not divide | G : AT|. In other words, K contains a full Sylow p-subgroup of G , and by Sylow theory, some conjugate of that Sylow subgroup contains the p-subgroup M . Thus M C K 9 for some element g G G .

N o w | G : K 9 \ = \ G : K \ is relatively pr ime to | G : i f | , and hence by L e m m a 3.16, we have G = H K 9 . Let s € G be arbitrary, and write x = w , where u e H a n d v e K 9 . T h e n M * = M U 1 ; = M v C K 9 , where the second equality holds because u £ H and M < H . A l l conjugates of M , therefore, lie i n K 9 , and hence the subgroup M G that they generate is contained i n K 9 , which is solvable. T h e n M G is a nonidenti ty solvable normal subgroup of G , and also, G / M G is solvable by the first paragraph of the proof. It follows by L e m m a 3.10(e) that G is solvable, as required. •

P r o o f of T h e o r e m 3.15 (assuming Burnside) . We proceed by induct ion on the number n of different primes that divide | G | . If n = 1, then G is a p-group, and hence is solvable, and i f n = 2, then G is solvable by Burnside 's theorem. We can assume, therefore, that at least three primes p, q and r divide | G | , and we choose ap-complement H , a g-complement K and an r-complement L i n G . The indices of these subgroups are respectively, a power of p, a power of q and a power of r , and so they are pairwise coprime. If we can show that H , K and L are solvable, it w i l l follow by Theorem 3.17 that G is solvable, and we w i l l be done.

The prime divisors o f \ H \ are just the prime divisors of | G | other than p, and so there are exactly n - 1 of them. Let s be one of these prime divisors of \ H \ , and let Q be an s-complement i n G . T h e n | G : H \ is a power of p and | G : Q\ is a power of s, and so these indices are coprime, and it follows by L e m m a 3.16 that \ H : H f ] Q \ = \ G : Q\, which is the full power of s d iv id ing | G | . Since this is also the full power of s d iv id ing \ H \ , it follows that H n Q is an s-complement in H . Th i s shows that H has an s-complement for each of the n - 1 primes s d iv id ing its order, and so H is solvable by the inductive hypothesis. Similar ly, K and L are solvable, and the proof is complete. •


We have seen that for every nonsolvable finite group G, there is a prime set rr for which G fails to have a H a l l rr-subgroup. How about conjugacy? Since H a l l rr-subgroups do sometimes exist i n nonsolvable groups, we can ask whether or not (for a fixed prime set rr) a l l H a l l rr-subgroups of a non-solvable group G must be conjugate. The answer is "no"; they need not even be isomorphic. In the simple group PSL(2,11) of order 660 = 2 2 - 3 - 5 - l l , for example, there exists a H a l l {2, 3}-subgroup isomorphic to the alternating group A A , and there is another H a l l {2, 3}-subgroup isomorphic to the d i hedral group D 1 2 . These groups of order 12 are certainly not isomorphic. We close this discussion by mentioning (without proof) a lovely theorem of Wie land t which asserts that if G has a n i l p o t e n t H a l l rr-subgroup, then a l l H a l l rr-subgroups of G are conjugate.

P r o b l e m s 3 C

3C.1 . Let U C G be a rr-subgroup, where rr is a set of primes and G is solvable and finite. Show that U is contained i n some H a l l rr-subgroup of G.

H i n t . Use P rob lem 3B.4 .

Note . Th i s , of course, is the H a l l D-theorem.

3C.2 . Let 7t be a set of primes, and suppose that G has a p-complement H p for each prime p e r t . Show that C\pen H v is a H a l l rr '-subgroup of G.

3C.3 . A Sylow system i n a group G is a set S of Sylow subgroups of G, one chosen from S y l p ( G ) for each prime divisor p of and such that P Q = Q P for a l l P , Q € 5.

(a) Show that if G has a Sylow system, then it has a H a l l rr-subgroup for every set rr of primes.

(b) If G is solvable, prove that it has a Sylow system.

Hint . For (b), choose a p-complement H p of G for each prime divisor of \ G \ , and consider intersections of a l l but one of these p-complements.

3C.4 . Le t M be m i n i m a l normal i n a finite solvable group G, and assume that M = C G ( M ) . Show that G splits over M and that a l l complements for M i n G are conjugate.

Hint . Let L / M be min ima l normal i n G / M , and note that L / M is a q-group for some prime q. Show that q does not divide | M | , and consider a Sylow q subgroup of L .

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3C.5 . Le t i f be a max ima l subgroup of G , where G is solvable, and assume that c o r e G ( i f ) = 1. Show that G has a unique m i n i m a l normal subgroup M , that H complements M and that M = C G { M ) . Deduce that i f K is also m a x i m a l i n G w i t h t r iv i a l core, then H and K are conjugate i n G.

3C.6 . Le t G be finite and solvable, and suppose that x , y , z G G have orders that are pairwise relatively prime. If xyz = 1, show that x = 1, y = 1 and z = l .

H i n t . W o r k by induct ion on the derived length of G.

3C.7 . A Carter subgroup of a solvable group G is a nilpotent subgroup C s u c h t h a t C = N G ( C ) .

(a) Show that every solvable group has a Car ter subgroup.

(b) Show that a l l Car ter subgroups of the solvable group G are conjugate i n G.

(c) If C is a Carter subgroup of G and G / N is nilpotent for some normal subgroup N of G, show that N C = G.

Note. T h i s problem and the next one seem quite a bit harder than most of the problems i n this book. We present them as challenges to the reader, and to give a taste of some of the theory of solvable groups that we have omit ted.

3C.8 . Le t G be solvable. A nilpotent injector of G is a nilpotent subgroup / containing the F i t t i n g subgroup F ( G ) , and max ima l w i t h this property. Prove that a l l nilpotent injectors of G are conjugate i n G.

3 D

A finite group G is said to be rr-separable, where rr is some set of primes, if there exists a normal series

1 = N 0 C TVi C • • • c 7Vr = G

such that each factor N . / N ^ x is either a rr-group or a rr'-group. (Observe that a rr-separable group is the same th ing as a vr'-separable group.) It is easy to prove, as we shall see, that finite solvable groups are vr-separable for al l prime sets rr, and so rr-separability can be viewed as a generalization of solvability.

F i r s t , we mention that i f G is rr-separable, then so too is every subgroup and every homomorphic image. (This fact is easy to establish, and we trust that the reader can supply a proof.) Next , we recall that C v ( G ) is the unique largest normal vr-subgroup in G , which may, of course, be the t r i v i a l subgroup. (The uniqueness is an immediate consequence of the fact that i f N

92 3. S p l i t Extensions

and M are normal rr-subgroups of G, then J V M is also a normal 7r-subgroup.) Because O n ( G ) is uniquely determined, it is, of course, characteristic i n G.

In the definition of rr-separability, we can renumber the normal subgroups N i i f necessary, i n order to eliminate repeats among them. If G is rr-separable and nontr ivia l , therefore, we can assume that N i > 1. Since N i is either a normal rr-subgroup or a normal rr '-subgroup of G, we see that either C v ( G ) > 1 or C v ( G ) > 1 (or both). In particular, G has a nontr iv ia l characteristic subgroup N that is either a rr-group or a rr '-group. Also , i f N < G, then G / N is rr-separable, and so we can find a nontr iv ia l characteristic subgroup M / N of G / N that is either a rr-group or a rr'-group. Cont inu ing like this, we can construct a series of characteristic subgroups of G whose factors are rr-groups and rr'-groups.

The discussion i n the previous paragraph shows that i f we strengthen the condi t ion on the normal subgroups N i n the definition of rr-separability, and we require them to be characteristic, and not merely normal , then the same class of groups would result. Somewhat paradoxically, this observation allows us to prove that a weakened version of the definition also yields exactly the same class of groups. It suffices to consider subnormal series.

3.18. L e m m a . Suppose t h a t G i s a finite g r o u p a n d t h a t

1 = N Q < N i < • • • < N r = G

i s a series of subgroups such t h a t each f a c t o r A ^ / A / j - i i s e i t h e r a i t - g r o u p o r a i t ' - g r o u p . Then G i s i t - s e p a r a b l e .

Proof. We can assume that r > 0, and we proceed by induct ion on r . Since N T - x is rr-separable by the inductive hypothesis, the preceding remarks guarantee that there exists a characteristic series

1 = M 0 C • • • C M s = N r - i

having rr and rr'-factors. Since 7 V r _ i < G, the characteristic subgroups M i of 7 V r _ i are normal i n G, and hence G has the normal series

1 = M 0 C • • • C M s = N r - x C N r = G

w i t h 7T and rr'-factors. Thus G is rr-separable, as desired. •

3.19. Corol lary. L e t G be finite a n d s o l v a b l e . Then G i s i t - s e p a r a b l e f o r every set i t of p r i m e s .

Proof. Consider a composit ion series for G. (Recall that this is a chain of subgroups

1 = 7V0 < N x < • • • < 7Vr = G

such that each of the factors N / N - i is simple.) Since we are assuming that G is solvable, its composit ion factors are solvable too, and thus each of

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the simple groups N t / N t - x has prime order. B u t every pr ime p is either a vr-number or a vr'-number, depending on whether p G vr or p 0 vr, and thus G is vr-separable by L e m m a 3.18. •

A somewhat stronger condit ion than "vr-separability" is "vr-solvability". A finite group G is vr-solvable if it has a normal series where each factor is either a vr'-group or is a solvable vr-group. (Of course, a solvable group is automatical ly vr-solvable for every set vr of primes.) We shal l not have much to say about vr-solvable groups except in the case where vr = { p } consists of a single prime. Since ^ g r o u p s are solvable, it follows that {p}-separability and {^ - so lvab i l i ty are the same thing, and groups i n this class are generally described as being > s o l v a b l e " . Note that a finite group is solvable i f and only i f it is p-solvable for every prime p .

A key result about vr-separable groups is the following, which is a generalizat ion of Ha l l ' s E-theorem, wi th essentially the same proof.

3 . 2 0 . Theorem. L e t G be a I T - s e p a r a b l e g r o u p . Then G has a H a l l vr-s u b g r o u p .

Proof. We proceed by induct ion on \ G \ . Suppose first that O n ( G ) > 1. B y the inductive hypothesis, the group G / O t ( G ) has a H a l l vr-subgroup H / O n ( G ) , and it is immediate that i f is a H a l l vr-subgroup of G . If C v ( G ) = 1, then since we can certainly assume that G is nontr iv ia l , we have C v ( G ) > 1. We can thus apply the inductive hypothesis to G / C v ( G ) to produce a H a l l rr-subgroup H / 0 ^ { G ) . B y the Schur-Zassenhaus theorem, C v ( G ) has a complement in H , and it is easy to see that such a complement is the desired H a l l vr-subgroup of G . •

T h e reader might guess that analogs of the C-theorem and D-theorem also hold in the context of vr-separable groups, and indeed they do. The i r proofs rely on the conjugacy part of the Schur-Zassenhaus theorem, however, and thus they require either some solvabil i ty assumption or else an appeal to the Fe i t -Thompson theorem.

Some of the basic definitions and results about vr-separability go back to the work of S. A . C u n i h i n i n 1948, but it was the historic paper of P h i l i p H a l l and G r a h a m H i g m a n i n 1956 that established a number of deeper properties, at least for p-solvable groups. A basic and frequently used result i n the H a l l -H i g m a n paper is their L e m m a 1.2.3, which we state here somewhat more generally: for vr-separable groups.

3 . 2 1 . T h e o r e m (Ha l l -Higman 1.2.3). L e t G be a i r - s e p a r a b l e g r o u p , a n d assume t h a t O t / ( G ) = 1. Then O f f ( G ) D C g ( O t ( G ) ) .

Proof. Wr i t e C = C G ( C v ( G ) ) and let B = C n C v ( G ) . Our goal is to show that 5 = 0 , and so we assume that B < C, and we work to derive a contradiction. Note that B and C are normal (and i n fact, characteristic) i n G and that B is a vr-group.

Since C/B is a nontr iv ia l vr-separable group, it has a nontr iv ia l characteristic subgroup K / B , where K / B is either a vr-group or a vr'-group. Also , since K / B is characteristic in C/B < G / B , we see that K / B < G/B, and hence K < G . Since £? is a vr-group, it follows that i f K / B is also a vr-group, then AT is a normal vr-subgroup of G , and hence K C C v ( G ) . Th i s is not the case, however, since £? < K C G and 5 = C n C v ( G ) . We can thus conclude that K / B is a vr'-group. B y the Schur-Zassenhaus theorem, therefore, there is a complement H for B in K , and we see that H > 1.

We have H C G = C G ( C v ( G ) ) C C G ( B ) , where the final containment holds because 5 C C v ( G ) . Thus £? centralizes and since B H = K , we see that H < K . Thus 1 < H C O^(AT) < G , and this is a contradict ion since by hypothesis, O t / ( G ) = 1, and hence G has no nontr iv ia l normal rr '-subgroup. •

In a group G , a normal subgroup N that contains its own centralizer can be thought of as a "large" subgroup, and indeed, i n this si tuation, the index | G : N \ , is bounded in terms of N . To see this, recall that for an arbi t rary normal subgroup N , the factor group G / C G ( N ) is isomorphically embedded i n A u t ( / V ) , and so \ G : C G ( / V ) | < |Aut(7V) | . If C G ( N ) C N , therefore, we have | G : 7V| < | A u t ( / V ) | . L e m m a 1.2.3 asserts that in a TT-separable group G for which C v ( G ) = 1, the normal subgroup C v ( G ) is large i n this sense. Another example of a characteristic subgroup that is guaranteed to be "large" is the F i t t i n g subgroup F ( G ) of a solvable group G . (See P rob l em 3B.14 and the note following it.) In Chapter 9, we discuss the "generalized F i t t i n g subgroup" F * ( G ) , which always contains its own centralizer, w i t h no extra assumption on G .

We close this section by showing how the Ha l l -H igman L e m m a 1.2.3 can be used. F i rs t , we need a definition. If G is vr-separable, the rr-length of G is the m i n i m u m possible number of factors that are rr-groups i n any normal series for G i n which each factor is either a vr-group or a vr'-group. For example, G has vr-length 1 precisely when G is not a vr'-group and there exist normal subgroups N C M of G such that N and G / M are vr'-groups and M / N is a vr-group.

3.22. Theorem. L e t G be a I T - s e p a r a b l e g r o u p , a n d suppose t h a t a H a l l I T - s u b g r o u p of G i s a b e l i a n . Then the v r - l e n g t h of G i s a t most 1.

Proof. Consider the vr-separable group G = G / C v ( G ) , and observe that C v ( G ) = 1. Let i f be an abelian H a l l vr-subgroup of G , and note that

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H is a H a l l vr-subgroup of G . _ T h e n H contains every normal vr-subgroup of G , and i n part icular, C v ( G ) C H . B u t # is abelian, and thus H C C c ( C v ( G ) ) C C v ( G ) , where_the last containment follows by the H a l l -H i g m a n L e m m a 1.2.3. Thus H = 0 * ( G ) , and we have H < G. Thus 1 C N C T V i i C G is a normal series for G w i t h just one rr-factor, and the proof is complete. •

P r o b l e m s 3 D

3D.1 . Let G be a p-solvable group, and let P G S y l p ( G ) .

(a) If C y (G) = 1, show that Z ( P ) C O p ( G ) .

(b) Show that the p-length of G is at most the nilpotence class of P .

3D.2. Let Z C Z ( G ) and write G = G / Z . Show that C v ( G ) = O ^ G ) for every set rr of primes.

3D.3. Let G be rr-separable, and assume that C v ( G ) C v ( G ) C Z ( G ) . Show that G is abelian.

3 D . 4 . Let H act by automorphisms on G , and assume that \ H \ and | $ ( G ) | are coprime. Suppose that the induced act ion of # on G / $ ( G ) is t r i v i a l , which means that each coset of $ ( G ) i n G is setwise invariant under H . Show that the action of H on G is t r i v i a l .

Hint . It is no loss to assume that H is a g-group for some prime q. Consider the decomposit ion of each coset of $ ( G ) into P - o r b i t s .

3D.5. Let G be p-solvable, and let P e S y l p ( G ) . Suppose that P C N G ( K ) , where K C G is a subgroup of order not divisible by p. Show that i f C Op / (G).

Hint . F i r s t , consider the case where C y ( G ) = I . To do the general case, consider the group G = G / C y (G) .

3 E

Let A be a group that acts v i a automorphisms on some group G . B y definit ion, A permutes the elements of G , but because we have an action by automorphisms, A also permutes many other kinds of group-theoretic objects associated w i t h G . For example, A acts on the set of a l l Sylow p-subgroups of G for any given prime p , and A acts on the set of a l l conjugacy classes of G . It is of interest to find, or at least to prove the existence of, ^- invar iant objects of various kinds. Once one has found an A-invar iant object, there are usually further actions that can be studied. For example, i f H is an


A-invar iant subgroup of G, then A acts on the set of right cosets or of left cosets of H i n G, and on the set of conjugacy classes of H .

O f course, the most basic type of A-invariant object i n G is an A¬invariant element. These A-f ixed points i n G clearly form a subgroup, which is denoted C G { A ) . To see why this centralizer notat ion is really quite natural , embed A and G in their semidirect product V = G x A . We know that the original action of A on G is realized by conjugation wi th in T, and thus g a = g i f and only i f g and a commute i n V. Thus the A-fixed-point subgroup of G is exactly the centralizer of A in G, and this explains the notat ion C G ( A ) , which is used to denote the fixed-point subgroup even if the semidirect product has not been mentioned. Similar ly , working i n the semidirect product, we see that C A ( G ) is the set of elements of A that act t r iv i a l ly on G, or i n other words, it is the kernel of the action of A on G. T h i s notat ion also can be used without mentioning the semidirect product.

It is especially easy to find A-invar iant objects i n G i f A and G have co-prime orders, and this is the si tuat ion we study i n this section. For example, we w i l l prove the following.

3.23. Theorem. L e t A a c t v i a a u t o m o r p h i s m s o n G, w h e r e A a n d G a r e finite g r o u p s , a n d suppose t h a t ( |A | , \ G \ ) = 1. Assume a l s o t h a t a t l e a s t one of A o r G i s s o l v a b l e . Then f o r each p r i m e p , t h e f o l l o w i n g h o l d .

(a) T h e r e exists a n A - i n v a r i a n t Sylow p - s u b g r o u p of G.

(b) / / S a n d T a r e A - i n v a r i a n t Sylow p - s u b g r o u p s of G, then S° = T f o r some element c £• C G { A ) .

Note that Theorem 3.23(a) is a generalization of the Sylow E-theorem, to which it reduces when A is the t r i v i a l group, and similarly, 3.23(b) is a generalization of the Sylow C-theorem. Another way to th ink about this result is to imagine that we have a pair of magic eyeglasses through which only A-invar iant objects can be seen. Theorem 3.23(a) asserts that even when we look at G through these A-glasses, the Sylow E-theorem can be seen to hold: there exists a v i s i b l e Sylow p-subgroup for each prime p .

The Sylow C-theorem also remains true when G is viewed through the magic A-glasses. W h a t the usual C-theorem says, of course, is that given two Sylow p-subgroups of G, say S and T , we can find an element g G G such that S9 = T. N o w suppose that S and T are Sylow p-subgroups that we can see after donning our glasses. (In other words, they are A¬invariant.) The glasses version of the Sylow C-theorem says that we can see an element g of G such that S9 = T. The assertion, in other words, is that S and T are conjugate by some A-invariant element: an element of the fixed-point subgroup C G ( A ) . (This, of course, is exactly the assertion

3 E 97

of Theorem 3.23(b).) We w i l l also prove the glasses version of the Sylow D -theorem: i f P is a visible (A-invariant) p-subgroup of G , then P is contained i n some visible (A-invariant) Sylow p-subgroup.

H o w can the glasses version of the Sylow E-theorem and similar results be proved? In the case where A is a g-group for some prime q, an easy counting argument suffices, as follows. F i r s t , observe that we can assume that A is nontr ivial , and thus q divides | A | , and so q does not divide \ G \ . Since | S y l p ( G ) | divides \ G \ , it follows that q does not divide | S y l p ( G ) | . Now A acts on S y l p ( G ) , and hence this set decomposes into A-orbi t s , at least one of which must have size not divisible by q. B u t we are assuming that A is a g-group, and so a l l A-orbi ts have g-power size. There must be an orbit of size 1, therefore, which means that the set S y l p ( G ) contains an A-invar iant member, as required.

A s imilar (but sl ightly more technical) argument can be used to prove 3.23(b) i n the case where A is a g-group, but it appears to be ut ter ly i m possible to make a counting argument like this work if | A | is not a pr ime power. T h e following result of G . G laube rman provides an effective substitute, however. Unfortunately, the proof of Glauberman 's l emma relies on the conjugacy part of the Schur-Zassenhaus theorem, and thus we must either appeal to the Fei t -Thompson theorem, or else assume that something is solvable. (The solvabil i ty assumptions that appear i n the statements of Theorem 3.23 and other results about coprime actions are not really necessary, of course, since at least one of | A | or \ G \ is odd, and hence by the Fe i t -Thompson theorem, at least one of A or G must be solvable.)

3.24. L e m m a (Glauberman) . L e t A a c t v i a a u t o m o r p h i s m s o n G, w h e r e A a n d G a r e finite g r o u p s , a n d suppose t h a t ( | A | , | G | ) = 1. Assume a l s o t h a t a t l e a s t one of A o r G i s s o l v a b l e . Suppose t h a t A a n d G each a c t o n some n o n e m p t y set Vl, w h e r e t h e a c t i o n of G i s t r a n s i t i v e . F i n a l l y , assume t h e c o m p a t i b i l i t y c o n d i t i o n :

( * ) ( a - g ) - a = ( a - a ) - g a

f o r a l l a < E V l , a < E A a n d g < E G. T h e f o l l o w i n g t h e n h o l d .

(a) T h e r e exists a n A - i n v a r i a n t element a e f i .

(b) Ifa,P G f l a r e A - i n v a r i a n t , t h e n t h e r e exists c G C G ( A ) such t h a t a-c = (3.

Before proceeding w i t h the proof of Glauberman 's lemma, we show how it can be used to prove Theorem 3.23.

P r o o f of T h e o r e m 3.23. The set S y l p ( G ) is nonempty by the Sylow E -theorem. E a c h of G and A act on S y l p ( G ) , where G acts by conjugation and the action of A is induced by the given act ion of A on G . (Since each

element of A induces an automorphism of G, it carries Sylow p-subgroups of G to Sylow p-subgroups.) A l so , by the Sylow C-theorem, the action of G on S y l p ( G ) is transitive. Once we check the compat ib i l i ty condit ion of Glauberman 's lemma, it w i l l follow by 3.24(a) that S y l p ( G ) contains an A-invar iant member, thereby establishing (a). We need to check that

[ s 9 ) a = (saya

for S G S y l p ( G ) , a G A and g G G. (Recall that here, exponentiation by g means conjugation by g i n G, but exponentiation by a denotes the given action of a . ) Since A acts by automorphisms, however, we have

( S 9 ) a = ( g ^ S g ) * = { g a ) - 1 S a g a = (S") 9 " ,

as required. Asser t ion (b) is immediate from 3.24(b). •

In the previous proof, i t was t r i v i a l to check the compat ib i l i ty condit ion of Glauberman 's lemma. Indeed, it seems to be equally t r i v i a l to check it i n v i r tua l ly every applicat ion of 3.24.

P r o o f of L e m m a 3.24. To prove (a), let T = G x > A be the semidirect product w i t h respect to the given action of A on G, and as usual, identify A and G w i t h appropriate subgroups of T. Every element of T is uniquely of the form a g , where a G A and g G G, and so we can unambiguously define an act ion of r on ft by setting a - ( a g ) = { a - a ) - g . To verify that this really is an action, let a , b G A and g , h G G. T h e n for a G ft, we have

( a - { a g ) ) i b h ) = (((a-a)- 5)-6)-/z = { { ( a - a ) - b ) - g b ) - h

= a - { ( a b ) ( g b h ) )

= a - ( ( a g ) ( b h ) ) ,

as wanted, where the second equality holds because of the compat ib i l i ty condit ion, which we are assuming.

Now fix a G ft and let U = Ta, the stabilizer of a i n T. If x G T, then since by assumption, G is transitive on ft, we can write a-x = a - g for some element g G G, and thus x g ~ l G U. It follows that x G U G , and so U G = T. N o w G < T , and hence U n G < \ U . A l so ,

\ U : U n G \ = \ U G : G \ = \ T : G \ = \ A \

is coprime to \ U D G\, and thus the Schur-Zassenhaus theorem applies in U w i t h respect to the normal subgroup U n G. Let H be a complement for U n G i n U, and note that \ H \ = \ U : U n G\ = \ A \ , and thus H is a complement for G in T. Since A is also a complement for G in T, the conjugacy part of the Schur-Zassenhaus theorem allows us to write A = H x

for some element x G T. B u t H C U, and so H stabilizes a . T h e n A = H x

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stabilizes a-x, and thus a-x is an A-invar iant member of ft. Th i s completes the proof of (a).

To prove (b), let a and P be A-invar iant members of ft. Since G is transit ive on ft, the set X = { g G G \ a - g = ¡3} is nonempty, and we need to find an. A-invar iant element in X . For this purpose, we propose to use part (a) of the lemma, w i t h the set X in place of ft. If x G X , then a-x = /3, and so for a € A , we have

a-xa = ( a - a ) - x a = ( a - x ) - a = (3-a = ¡3 ,

where the first and last equalities hold because a and ¡3 are A-invariant , and the second is a consequence of the compat ib i l i ty condi t ion. Thus x a G X , and so the act ion of A on G maps X into (and hence onto) itself, and we have an action of A on X .

N o w let H = G p , the stabilizer of ¡3 i n G. If we apply the above argument w i t h 0 i n place of a and H i n place of X , we conclude that the action of A on G maps H to itself, and this defines an action v i a automorphisms of A on H . A l so , since H is a subgroup of G, we see that ( |A | , \ H \ ) = 1 and that at least one of A or H is solvable.

Now A acts on H and on X , and we show next that H acts t ransi t ively on X by right mul t ip l ica t ion. If x G X and h G H , then

a - ( x h ) = ( a - x ) - h = p - h = P ,

and thus x h G X and H acts on X by right mul t ip l ica t ion , as wanted. In fact, this act ion is transitive, since if x,y G X , then a r 1 ^ fixes P, and so x - 1 ? / G i f , and right mul t ip l ica t ion by x~ly carries x to y .

A t this point we know that A acts v i a automorphisms on the group H , that ( |A | , \ H \ ) = 1, and that at least one of A or H is solvable. Also , each of A and H acts on the nonempty set X , and the action of H on X is transit ive, and thus we have almost everything we need to apply part (a) of the lemma to conclude that X contains an A-invar iant element. W h a t remains is to check the compat ib i l i ty condit ion. W h a t we need, i n other words, is ( x h ) a = x a h a for x G X , h G H and a G A . Th i s is obvious, however, since A acts on G v i a automorphisms, and thus X contains an A-invar iant element, as required. •

T h e magic-eyeglasses version of the Sylow D-theorem follows easily from Theorem 3.23(a); its proof does not require a direct appeal to Glauberman 's lemma.

3.25. Corol lary. L e t A a c t v i a a u t o m o r p h i s m s o n G, w h e r e A a n d G a r e finite g r o u p s , a n d suppose t h a t ( \ A \ , \ G \ ) = 1. Assume a l s o t h a t a t l e a s t one of A o r G i s s o l v a b l e . L e t P C G be a n A - i n v a r i a n t p - s u b g r o u p f o r some p r i m e p . T h e n P i s c o n t a i n e d i n some A - i n v a r i a n t S y l o w p - s u b g r o u p o f G .

Proof. Replacing P by a subgroup containing it i f necessary, we can assume that P is not contained i n any str ic t ly larger A-invar iant p-subgroup of G, and we show that in this case, P £ S y l p ( G ) . Let N = N G ( P ) , and observe that N is A-invar iant since it is uniquely determined by the A-invar iant subgroup P . B y Theorem 3.23(a), we can choose an A-invar iant member S € S y l p ( / V ) , and we observe that since P is a normal p-subgroup of N , it must be contained i n the Sylow p-subgroup S. B y our assumption, however, P cannot be properly contained i n S since S is A-invar iant , and thus P = S. Thus P is a Sylow p-subgroup of its normalizer N , and it follows that P is a Sylow p-subgroup of G. (To see this, let P C Q G S y l p ( G ) , and suppose that P < Q. T h e n since normalizers grow i n p-groups, Q n N is a p-subgroup of N s t r ic t ly larger than the Sylow p-subgroup P , and this is impossible.) •

Next , we consider conjugacy classes. In general, if i f is a class of G and H C G is a subgroup, then i f n f f may be empty, and even i f i t is nonempty, a l l we can say is that i f n H is a union of classes of i f . Usually, i f n H is not a single class of H . (Dist inct classes of a subgroup H that are contained in the same class of G are said to be fused i n G.) The si tuat ion is under much tighter control if the subgroup is the set of fixed points of a coprime action. In part icular, there is no fusion i n that case.

3.26. Theorem. L e t A a c t v i a a u t o m o r p h i s m s o n G, w h e r e A a n d G a r e finite g r o u p s , a n d w r i t e C = C G ( A ) . Suppose t h a t ( \ A \ , \ G \ ) = I , a n d assume t h a t a t l e a s t one of A o r G i s s o l v a b l e . T h e n t h e m a p i f ^ i f n C d e f i n e s a b i j e c t i o n f r o m t h e set of A - i n v a r i a n t c o n j u g a c y classes of G o n t o t h e set of a l l c o n j u g a c y classes of C.

Proof. Let i f be an A-invar iant class of G. T h e n A acts on i f , and G acts t ransi t ively on i f by conjugation. The compat ib i l i ty condi t ion of Glauber-man's lemma is that ( x » ) a = ( x a ) 3 a for elements x G i f , a G A and g € G, and as usual, i t is t r i v i a l to see that this holds. B y 3.24(a), therefore, K contains an A-invar iant element, and thus i f n C is nonempty. A l so , by 3.24(b), a l l elements of i f n C are conjugate i n C, and thus i f n C is a class of C.

We now have the map i f ^ i f D C from the set of A-invar iant classes of G to the set of classes of C; we must show that it is a bijection. If i f and L are A-invar iant classes of G, w i t h i f D C = L C ) C , then since i f D C is nonempty and is contained in both i f and L , it follows that i f n L is nonempty, and thus i f = L . Our map, therefore, is injective. For surjectivity, it is enough to show that every element of C lies i n i f n C for some A-invar iant class i f of G. O f course, each element c G C lies in a unique class i f of G, and so it suffices to show that i f is A-invar iant . If a G A , then i f a is a class of G, and since ca = c, we see that c G i f n i f " . Thus the classes i f and i f a are

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not disjoint, and we conclude that K = K a , and hence K is A-invariant , as wanted. •

Now we consider cosets. If A acts on G and H C G is A-invar iant , then as we observed previously, A permutes the left cosets of H i n G and also the right cosets of H i n G. In the case of a coprime action, we can determine which left cosets and which right cosets of H are A-invar iant .

3.27. Theorem. L e t A a c t v i a a u t o m o r p h i s m s o n G, w h e r e A a n d G a r e finite g r o u p s , a n d w r i t e C = C G ( A ) . L e t H C G be A - i n v a r i a n t s u b g r o u p , a n d suppose t h a t ( | A | , \ H \ ) = 1 a n d t h a t one of A o r H i s s o l v a b l e . Then t h e A - i n v a r i a n t left cosets of H i n G a n d t h e A - i n v a r i a n t r i g h t cosets of H i n G a r e e x a c t l y those cosets of H t h a t c o n t a i n elements of C.

Proof. If a G A and c G C, then ( H c ) a = H a c a = H e , and similarly, ( c H ) a = c H , and so cosets of H that contain elements of C are A-invar iant . Now suppose that X is an A-invar iant left coset of H . T h e n H acts transit ively by right mul t ip l ica t ion on X and, of course, A acts on X and A acts v i a automorphisms on H . The compat ib i l i ty condi t ion for Glauberman 's lemma i n this s i tuat ion is [ x h ) a = x a h a for x G X , h G H and a G A , and this holds since the given action of A on G is an action v i a automorphisms. B y 3.24(a), therefore, X contains an element of C, as wanted.

To handle the right cosets, it suffices to observe that every right coset of H in G has the form X " 1 = { x ~ l \ x G X } for some left coset X of H i n G. Since X is A-invar iant i f and only if X " 1 is, the desired result for right cosets is a consequence of that for left cosets. (Alternat ively, i f Y is a right coset of H i n G, then y h = h~ly defines a transit ive action of H on Y, and we can apply Glauberman 's lemma directly.) •

Ac t ions on cosets are especially interesting for normal subgroups. Le t A act on G v i a automorphisms, and suppose that N < G is A-invar iant . O f course, the cosets of N i n G are the elements of the factor group G/N, and since these are permuted by A , we see that A acts on G/N. In fact, this is an action v i a automorphisms since

{ N x N y ) a = { N x y ) a = Nxaya = N x a N y a = { N x ) a ( N y ) a .

The A-invar iant cosets of N i n G, therefore, are exactly the elements of the fixed-point subgroup C G / N ( A ) . We can use the bar convention to give a clean restatement of Theorem 3.27 i n this si tuation, a statement that can be paraphrased as "fixed points come from fixed points (in coprime actions)".

3.28. Corol lary. L e t A a c t v i a a u t o m o r p h i s m s o n G, w h e r e A a n d G a r e finite g r o u p s , a n d l e t N < G be A - i n v a r i a n t . Assume t h a t ( |A | , \ N \ ) = 1 a n d

t h a t a t l e a s t one of A o r N i s s o l v a b l e . W r i t i n g G = G/N, w e h a v e

C C ( A ) = C G T A ) .

Proof. The A-f ixed elements of G are exactly the A-invar iant cosets of N , which by 3.27 are exactly the cosets of N that contain elements of C = C G (A) ._But the cosets of TV that contain elements of C form the image C of C in G . •

3.29. Corol lary. L e t A a c t v i a a u t o m o r p h i s m s o n G, w h e r e A a n d G a r e f i n i t e g r o u p s , a n d assume t h a t { \ A \ , \ G \ ) = 1. If t h e i n d u c e d a c t i o n of A o n t h e F r a t t i n i f a c t o r g r o u p G / * ( G ) i s t r i v i a l , t h e n t h e a c t i o n of A o n G i s t r i v i a l .

Proof. We must show that each element a G A lies i n C A { G ) , and so it is no loss to assume that A = (a). S i n c ^ A is cyclic, it is certainly solvable, and thus Corol la ry 3.28 applies w i t h G = G / $ ( G ) . W r i t i n g C = C G { A ) , we have

G = Cc(A) = C = G $ ( G ) ,

where the first equality holds since A acts t r iv ia l ly on G . It follows that G = G * ( G ) , and thus G = C by the usual argument. (If C < G , then C is contained in some max ima l subgroup of G , which necessarily contains $ ( G ) too, and this is a contradiction.) •

3.30. Corol lary. L e t A a c t v i a a u t o m o r p h i s m s o n G, w h e r e A a n d G a r e finite g r o u p s , a n d assume t h a t ( \ A \ , |G | ) = 1 a n d t h a t t h e a c t i o n of A o n G i s f a i t h f u l . T h e n t h e i n d u c e d a c t i o n of A o n G / $ ( G ) i s f a i t h f u l .

Proof. Let B = C A ( G / $ ( G ) ) and apply the previous corollary to the action of B on G to deduce that B = l . U

Given an arbi trary group action on a finite set, there is an associated set of positive integers: the orbit sizes. A l though there is very l i t t le that can be said about the sets of integers that arise i n this way in general, the si tuat ion is quite different for actions v i a automorphisms on groups, where there is considerably more structure. (See, for example, Corol la ry 8.44.) For coprime actions, we can say even more. In that case, it is true that if m and n are coprime orbit sizes, then m n must also be an orbit size. To prove this, we use a remarkable observation of B . Har t ley and A . Turu l l , which allows us to replace the group being acted on by an abelian group. The key step is the next theorem, which follows fairly easily from results about coprime actions that we have already established.

3.31. T h e o r e m (Har t ley-Turul l ) . L e t A a c t v i a a u t o m o r p h i s m s o n G, w h e r e A a n d G a r e finite g r o u p s , a n d suppose t h a t ( \ A \ , |G | ) = 1. Assume a l s o t h a t

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a t l e a s t one of A o r G i s s o l v a b l e . T h e n A acts v i a a u t o m o r p h i s m s o n some a b e l i a n g r o u p H i n such a w a y t h a t every s u b g r o u p B C A has e q u a l n u m b e r s of f i x e d p o i n t s o n G a n d o n H . A l s o , t h e r e i s a s i z e - p r e s e r v i n g b i s e c t i o n f r o m t h e set of A - o r b i t s o n G t o t h e set of A - o r b i t s o n H .

We extract part of the proof as a separate lemma, which has other applications.

3.32. L e m m a . L e t A a c t v i a a u t o m o r p h i s m s o n G, w h e r e A a n d G a r e finite g r o u p s , a n d suppose t h a t ( | A | , \ G \ ) = 1. Assume a l s o t h a t a t l e a s t one of A o r G i s s o l v a b l e . L e t C = C G { A ) , a n d suppose t h a t P £ S y l p ( G ) i s A - i n v a r i a n t . T h e n P n C G S y l p ( C ) .

P r o o f . Le t S G S y l p ( C ) , and note that S is A-invar iant . B y Corol la ry 3.25, some A-invariant Sylow p-subgroup Q of G contains S. B y Theorem 3.23(b), we have Qc = P for some element c G C, and thus

Pnc = Qcnc = ( Q r ] C ) c 5 Sc.

The result now follows because P n C is a p-subgroup of C that contains the Sylow p-subgroup Sc. •

T h e following is a very general result about groups act ing on finite sets.

3.33. L e m m a . L e t A be a g r o u p t h a t acts o n finite sets ft a n d A , a n d assume t h a t each s u b g r o u p B C A has e q u a l n u m b e r s of fixed p o i n t s i n ft a n d i n A . T h e n t h e r e i s a b i j e c t i o n f : ft A such t h a t f ( a - a ) = f ( a ) - a f o r a l i a e V l a n d a G A . I n p a r t i c u l a r , f maps A - o r b i t s o n ft t o A - o r b i t s o n A , a n d so t h e sets of o r b i t sizes i n t h e t w o a c t i o n s a r e i d e n t i c a l .

P r o o f . We prove the existence of / by induct ion on |ft|; the assertions about orbits are then immediate. Let n e U , where U is an A-orb i t of smallest possible size i n ft. T h e n | A : A ^ | = \ U \ , and by the min imal i ty of \ U \ , no subgroup properly containing A ^ can fix a point in ft. B y hypothesis, therefore, no such subgroup can fix a point i n A . A l so by hypothesis, A M

fixes some point v G A , and it follows that A M is the full stabilizer of v in A , and we have A ^ = A U .

We now define / on U by setting / ( u - x ) = vx for x G A . Th i s is well defined since i f = f i - y , then xy~l G A M = A„, and thus vx = vy. Also , / is injective on U since i f vx = vy, then xy~l G A v = A M , and thus H - x = n-y. Furthermore, i i a e U , we can write a = px for some element x G A , and hence if a G A , we have

f ( a - a ) = f { f i ' { x a ) ) = v ( x a ) = ( v x ) - a = f { a ) - a .

N o w let V = f ( U ) , so that V is the orbit of v and | V | = \ U \ . B y tak ing B = 1, we see that |ft| = | A | , and thus if U = ft, we have V = A and there is

nothing further to prove. We can assume, therefore, that U < Cl and V < A , and we observe that A acts on Q - U and on A - V .

We w i l l argue that the hypotheses of the lemma are satisfied for the actions of A on Q - U and A - V . Once this is established, the inductive hypothesis w i l l guarantee the existence of an appropriate bijection from Cl - U to A - V , and this can be used to extend the definition of / to a l l of Cl.

If B C A , we wish to show that B has equal numbers of fixed points i n Q. - U and i n A - V , and for this purpose, it suffices to show that B has equal numbers of fixed points i n U and V . Th i s is clear, however, since for each element a £ A and each point a G U, we have f ( a - a ) = f ( a ) - a , and so a fixes a i f and only i f it fixes f { a ) . U

P r o o f of T h e o r e m 3.31. We proceed in two steps to prove the first part; the final assertion then follows by L e m m a 3.33. F i rs t , we show that G can be replaced by a nilpotent group, and then we complete the proof by showing that i f G is nilpotent, it can be replaced by an abelian group.

For each prime divisor p of \ G \ , choose an A-invariant Sylow p-subgroup of G, and let N be the nilpotent group constructed as the external direct product of the set V of selected Sylow subgroups. Since A acts on P for each member P G V, there is a natural action of A on N , and we argue that \ C N ( B ) \ = \ C G ( B ) \ for each subgroup B C A . Since an element of the direct product N is B- f ixed if and only i f each of its components is fixed, it is clear that

\ C N ( B ) \ = Yl \ C P ( B ) \ , Pev

and thus for each prime p , the p-part of \ C N ( B ) \ is equal to \ C P ( B ) \ = \ P C \ C \ , where C = C G { B ) . B u t P n C G S y l p ( C ) by L e m m a 3.32, and thus \ P n C\ is the p-part of \ C \ . Th i s shows that \ C N ( B ) \ and \ C \ have equal p-parts for a l l primes p, and hence \ C N { B ) \ = | C | , as wanted.

To complete the proof, we can now assume that G is nilpotent, and in part icular it is solvable wi th some derived length d. We proceed by induct ion on d. If d = 1, then G is abelian, and there is nothing to prove. We can thus assume that d > 1, and we define the external direct product K = G' x G / G ' , which has derived length d - 1. Since G' is characteristic i n G, it is A invariant, and thus A acts v i a automorphisms on G' and on G / G ' , and this induces an action v i a automorphisms of A on K . Since |AT| = \ G ' \ \ G / G ' \ = \ G \ is coprime to A , the inductive hypothesis applies, and thus there exists an abelian group H , acted on by A , and such that \ C H ( B ) \ = | C K ( - B ) | for every subgroup B C A .

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It suffices now to show that \ C \ = \ C K ( B ) \ , where as before, C = C G { B ) . Us ing Corol la ry 3.28 to obtain the second equality below, we obtain

\ C K ( B ) \ = \ C G / G , ( B ) \ \ C G , { B ) \ = \ C G ' / G ' \ \ C n G ' \

= \ C : C n G ' \ \ c n G ' \

= \c\, and the proof is complete. •

We can now prove the result about coprime orbit sizes that we mentioned earlier.

3.34. Theorem. L e t A a c t v i a a u t o m o r p h i s m s o n G, w h e r e A a n d G a r e finite g r o u p s , a n d suppose t h a t ( | A | , \ G \ ) = 1. Assume a l s o t h a t a t l e a s t one of A o r G i s s o l v a b l e . Suppose t h a t t h e r e e x i s t A - o r b i t s of size m a n d n i n G, w h e r e m a n d n a r e c o p r i m e . T h e n t h e r e i s a l s o a n o r b i t of size m n .

Proof. B y Theorem 3.31, we can replace G by an abelian group of the same order without affecting orbit sizes, and thus we can assume that G is abelian. Now let X and y be A-orbi ts i n G of sizes m and n , respectively, and let x G X and y G y. T h e n \ A : A x \ = \X\ = m and \ A : A y \ = \y\ = n , and since m and n are coprime, it follows by L e m m a 3.16 that A = A x A y and that \ A : A x n A y \ = m n . Now write z = xy, and observe that A x ( ~ ) A y C A z . We w i l l show that A x n A y = A z , and consequently, the size of the A-orb i t of z is IA : A x D A y | = m n . We work in the next few paragraphs to prove that A z Q Ay. S imi la r reasoning then shows that A z C A x , and this w i l l complete the proof.

Define the map r : G - > G by setting

a e A y

Because G is abelian, the order of the factors in the product is irrelevant, and thus r is well defined. Furthermore, and also because G is abelian, r is a group homomorphism. We argue that r maps each element of the orbit X into C G ( A ) . To see this, recall that A = A x A y , and so A y acts transit ively on the set X . If u G X , therefore, then as a runs over the elements of A y , each element of X occurs equally often i n the form ua. We conclude that T ( U ) is a power of the product of the elements of the orbit X , and thus r (u ) is A- invar iant , as claimed.

Let X = ( X ) and Y = ( y ) , so that X and Y are A-invar iant subgroups of G. A l so , since r is a homomorphism that maps every element of X into C G ( A ) , it follows that T ( X ) consists of A-f ixed elements. Let Z = X n Y, and note that Z is also an A-invariant subgroup of G. Let B C A be the (setwise) stabilizer of the coset Zy, and observe that A y C B since

A y stabilizes bo th Z and y. Since B stabilizes the coset Zy, it follows by Theorem 3.27 that Zy contains a B- invar iant element c, and thus y £ Z c and we can write y = t c for some element i £ Z .

Since c is B- invar iant and A y C B , we see that r (c) = cl A «l, and so r (c) is /^-invariant. Also , since t £ Z C X , we know that r ( i ) is A-invariant , and in part icular, it is B- invar iant . We conclude that y \ A * \ = r { y ) = r ( i ) r ( c ) is ^ - invar ian t . Since \ A y \ is coprime to \ G \ , however, it follows that y is a power of y l ^ l , and hence y is ^ - invar ian t and B C A r

We argue next that A z C £?, and thus 4 C i r To see this, note that z = xye Xy, and thus Xy = Xz, and this coset is stabil ized by A z . Since y £ Y, it is easy to check that n Y = ( X n >")y = Z y , and since A 2

stabilizes both Xy and y , we conclude that A z stabilizes Zy, and so A z C £?, as claimed. We have now shown that A z C A y , and as we remarked earlier, s imilar reasoning yields A z C A x . T h i s completes the proof. •

We mention that in the proof of Theorem 3.34, the orbit of z = xy is exactly the set xy. T h i s is easy to see, but this observation does not yield a t r i v i a l proof of the theorem because it is not obvious that \Xy\ = \X\\y\.

We close this section w i t h one addit ional remark. G iven an arbitrary set TV of integers exceeding 1, one can construct a graph wi th vertex set M by creating an edge jo in ing m and n in TV whenever m and n have a common prime divisor. A n immediate consequence of Theorem 3.34 is that i f M is the set of nontr iv ia l orbit sizes of a coprime finite group action, then the m a x i m u m distance between any two points in this graph is 2. For a not-necessarily coprime group action, it is s t i l l possible to say something about the corresponding graph: it has at most two connected components. (This theorem of the author and C . Praeger is presented in Chapter 8,)

P r o b l e m s 3 E

3E.1 . Le t A act on G v i a automorphisms, and assume that at least one of A or G is solvable, but do not assume that A and G have coprime orders. If G is nontr iv ia l , prove that G contains a nontr iv ia l A-invar iant p-subgroup for some prime p .

H i n t . If P < A is a p-subgroup and C G { P ) = 1, show that p does not divide \ G \ . Show i n this s i tuat ion that for each prime q, there is an A-invar iant Sylow g-subgroup i n G.

3E.2. Let A act v i a automorphisms on G, and assume that ( \ A \ , \ G \ ) = 1 and that at least one of A or G is solvable. Suppose that G = H K , where

3 F 107

H and K are A-invar iant subgroups. Show that C — ( C ( ~ ) H ) ( C r \ K ) , where C = C G { A ) .

N o t e . Wear ing magic glasses, H and K are visible. The fact that G — H K means that every element of G is the product of an element of H w i th an element of K . Th i s problem asserts that this is visible through A-glasses. Every element of G that we can see is the product of a visible element of H w i t h a visible element of K .

3E.3. Let A act on G v i a automorphisms, and let N < G be A-invar iant . Assume that [ \ A \ , \ N \ ) = 1 and that A acts t r iv i a l ly on bo th N and G/N. Show that A acts t r iv ia l ly on G.

H i n t . N o solvabili ty assumption is needed here because it is no loss to assume that A is cyclic .

3E.4. In the usual coprime action si tuat ion wi th A act ing on G, let C = C G { A ) . If H C G is A-invariant , show that \ H : H n G | divides | G : C| and that | G : C n B | divides \ G : H \ .

H i n t . Use L e m m a 3.32.

3E.5. Let G be p-solvable, and assume that C y ( G ) = 1. Let P = O p ( G ) and write F = $ ( P ) , so that F < G. Note that there is a natural action of G on P / F , and observe that P is contained i n the kernel of this action since P / P is abelian. Show that P is the full kernel of this action.

N o t e . In other words, G / P acts faithfully on P / F , and so G / P is isomorphic to a subgroup of A u t ( P / P ) . A l so , P / F is an elementary abelian p-group, of order p n , say, and so it is isomorphic to the additive group of an n-dimensional vector space over a field of order p . It follows that G / P is isomorphic to a group of invertible n x n matrices over this field.

3 F

We close this already long chapter on split extensions w i t h a discussion of certain extensions that are not generally spli t . G i v e n a group N and a positive integer m, we wish to construct a l l (up to isomorphism) groups G , where G has a normal subgroup 7V0 = N such that G/N0 is cyclic of order m. A s was the case when we discussed semidirect products earlier i n this chapter, our "synthesis" w i l l be preceded by some "analysis" that w i l l enable us to understand groups of the type we wish to construct.

Suppose N < G and that G / N is cyclic of order m. Let g N be a coset that generates the cyclic group G/N, and observe that g m G N . It should be clear that the element g m of N and the automorphism of N induced by g

are relevant to this si tuation, and i n fact, these two objects, along wi th the group TV and the integer TO uniquely determine G up to isomorphism. In the following, we w i l l again use the convention that i f N = N 0 , then x £ N and x0 G N 0 are corresponding elements under the given isomorphism.

3 . 3 5 . Theorem. L e t G a n d G 0 be g r o u p s , a n d l e t N < G a n d N 0 < G 0 , w h e r e G / N a n d G Q / N Q a r e c y c l i c of finite o r d e r TO. Suppose t h a t n ^ n 0 i s a n i s o m o r p h i s m f r o m N t o N 0 , a n d l e t g N a n d g 0 N r j be g e n e r a t i n g cosets of G / N a n d G Q / N Q r e s p e c t i v e l y . Assume t h a t

( 9 m ) o = (ft))™ and { x 9 ) 0 = ( x 0 ) 9 0

f o r a l l elements x G N . Then t h e r e i s a u n i q u e i s o m o r p h i s m 9 : G - > G 0

t h a t extends the g i v e n i s o m o r p h i s m N -> 7V0 a n d t h a t c a r r i e s g t o g Q .

Proof. The dist inct cosets of N in G are g l N w i th 0 < i < TO, and so each element u G G is uniquely of the form u = glx for some element x G N and some integer i i n this range. We are therefore forced to define 9 by setting 9 ( u ) = 9 { g i x ) = { g o Y x o , and we must show that 9 is indeed an isomorphism from G onto G 0 . The elements of G 0 are uniquely of the form ( g o ) l x 0 for 0 < i < TO, and it follows easily that 9 is bo th injective and surjective.

It remains to prove that 9 is a homomorphism. We show first that e(gix) = ( g o Y x o for x e N , and without restriction on the integer i. Wr i te 1 = qm + r, where 0 < r < TO. B y hypothesis, g m e N and { g m ) 0 = { g 0 ) m , and thus { { g m ) q x ) Q = { g 0 ) m g x 0 . We have

d t f x ) = 9 { g r { g m Y x ) = ( g 0 ) r ( ( g m y x ) 0 = ( g 0 ) m q + r x 0 = ( g o f x o ,

as wanted. Now g { x g i y = gi+jx°jy for integers i and j and elements x , y G N . Since ( x ^ ) 0 = ( x 0 ) { 9 o ) 3 , it is easy to compute that 9 ( ( g i x ) ( g i y ) ) = 9 { g i x ) 9 { g i y ) . Th i s completes the proof. •

Now suppose that we are given four ingredients: a group TV, a positive integer TO, an element a G N and an automorphism a of N . The previous theorem guarantees that up to isomorphism, there is at most one group G having TV as a normal subgroup w i t h cyclic factor group G / N of order TO, and having generating coset gN, where g m = a and x9 = xa for a l l x€N.

B u t given N and TO, not every choice of a G N and a G A u t ( / V ) correspond to an actual group. If the group G exists, then since conjugation by g effects the automorphism a on N , and a = g m is fixed by g , we must have a a = a . Another condit ion that clearly must be satisfied is that the m t h power of the automorphism a must be the inner automorphism induced by a . A s the following cyclic extension theorem shows, there are no further requirements.

3 F 109

3.36. Theorem. L e t N be a g r o u p a n d TO a p o s i t i v e i n t e g e r , a n d l e t a e N a n d a e A u t ( / V ) . Assume t h a t

a a = a and = x a

f o r a l l x E N . T h e n t h e r e exists a g r o u p G, u n i q u e u p t o i s o m o r p h i s m , a n d h a v i n g N as a n o r m a l s u b g r o u p w i t h t h e f o l l o w i n g p r o p e r t i e s .

(a) G / N = ( g N ) i s c y c l i c of o r d e r TO.

(b) g m = a .

(c) x a = x 3 .

Proof. The uniqueness follows by Theorem 3.35, and so it suffices to construct G, which we do by realizing it w i t h i n the symmetric group on the set ft of ordered pairs ( i , y ) , where 0 < i < TO and y G M . F i r s t , we define an action of N on ft by setting ( i , y ) - x = ( i , y x ) , where 0 7V0 C Sym(ft) , where x 0 is the permutat ion effected by x for x G N . For convenience, we identify N w i t h N 0 v i a this isomorphism, so that N C Sym(ft) .

Next , we define a function g : ft -> ft by setting ( i , = (t + l , y a ) for 0 < i < m - 1 and y e N , and (TO - 1, y ) 5 = (0, a y a ) . Since a a = a and cr is an automorphism, we see that ( i , a y ) ? = (i + 1, a y a ) for 0 < i < m - 1, and it follows easily that

( i , y ) g m = ( i , ay°m) = { i , a y a ) = ( i , y a ) = ( i , y ) a

for a l l ( i , y ) £ Vl. (To check the first equality, observe that each applicat ion of g increases the first component by 1 modulo m. We pick up one extra factor of a when the first component changes from TO - 1 to 0, and that factor is unaffected by addi t ional applications of the function g . ) Thus g m = a , and in particular, g is a permutat ion of Q.

We show next that x g = gxa for a l l x G N . To verify this, it suffices to check that these functions have the same effect on each element ( i , y ) of ft, and so we want (( i , y ) x ) g = (( i , y ) g ) x a . If 0 < i < TO - 1, we have

(( i , y ) x ) g = ( i , y x ) g = ( i + l , [ y x f ) = ( i + l , y a x a ) = { { i , y ) g ) x a ,

and also

((TO - l , y ) x ) g = (TO - l , y x ) g = ( 0 , a { y x ) a ) = ( 0 , a y a x a ) = (TO - l , y ) g x a ,

as wanted. It follows that x 9 = x a , and in part icular, N < G, where G = ( N , g ) . A l so , G / / V is cyclic, w i t h generating coset g N .

A l l that remains to check is that G / N has order TO. Since g m = a € N , it suffices to show that g i 0 N for 0 < j < TO. B u t (0 ,1)^ ' = (j, 1) and

(0, l ) s = (0, x ) for x E N , and thus g* cannot be an element x £ N when 0 < j < m. Th i s completes the proof. •

A s an example of the use of the cyclic extension theorem, we give another construction of the generalized quaternion groups. We saw these previously, i n P rob l em 3A.2 , where they were constructed as index-2 subgroups of semidihedral groups.

G i v e n a positive integer n , divisible by 8, we start w i t h a cyclic group N of order n / 2 . Let a be the unique element of N of order 2 and let a £ Aut(AT) be the map x x ~ \ so that a fixes a . Tak ing m = 2, we see that a m is the t r i v i a l automorphism of N , and so it is equal to the inner automorphism induced by a . B y the cyclic extension theorem, therefore, the cyclic group N can be embedded as a normal subgroup w i t h index m = 2 i n a group Q , and there exists an element g G Q - N such that g 2 = a and x 9 = x ' 1 for a l l x G N . We argue that every element of Q - N has order 4, and thus a is the only element of order 2 i n Q. To see this, observe that a l l elements of Q - TV have the form gx w i t h x G N , and we have ( g x ) 2 = g x g x = g 2 x 9 x = ax~lx = a , which has order 2. Thus gx has order 4, as claimed. T h e unique group constructed here is denoted Q n ; it is the generalized quaternion group of order n . (See the remark concerning nomenclature following P rob lem 3A.2.)

P r o b l e m s 3 F

3F.1 . Le t G be a group of order n divisible by 8, and suppose that N has a cyclic subgroup C of index 2 such that every element of G - C has order 4. Show that G = Q n , the generalized quaternion group.

3F.2. Show that i n principle, we can construct every solvable group by start ing w i t h the t r i v i a l group and repeatedly applying Theorem 3.36.

3F.3 . Let Q = Q 8 . Show that Q has exactly three cyclic subgroups of order 4, and that these are transit ively permuted by A u t ( Q ) .

Hint . If A and B are cyclic subgroups of Q of order 4, use Theorem 3.35 to construct an isomorphism from Q to Q that carries A to B .

3F.4. Let S = S L ( 2 , 3), which is the group of 2 x 2 matrices of determinant 1 over the integers modulo 3. Show that | 5 | = 24 and that it has a normal Sylow 2-subgroup isomorphic to Q8.

3F.5 . Let S = 5 L ( 2 , 3 ) as in the previous problem, and note that S has index 2 i n G = G L ( 2 , 3), the full group of invertible 2 x 2 matrices over the integers modulo 3. Show that G-S contains an element g of order 2, but

P r o b l e m s 3 F 111

that it contains no element of order 4. Prove that there exists a group H containing S as a subgroup of index 2, where H - S contains an element h of order 4 such that the automorphisms of S induced by g i n G and by h in H are identical . Show that H - S contains no element of order 2 and finally, show that both G — S and H - S contain elements of order 8.

C h a p t e r 4

Commutators

4 A

We begin w i t h a bi t of review. The commutator of two elements x and y in a group G is the element x ^ y ^ x y , which is denoted [ x , y \ . Since [ x , y ] = { y x ) ~ l x y , we can th ink of the commutator of x and y as the "difference" between yx and xy, and i n part icular , [ x , y ] = 1 i f and only i f x and y commute. Alternat ively, we can write [ x , y ] = a r V ' , so [ x , y ] can also be viewed as the "difference" between x and its conjugate x».

If H C G and AT C G are subgroups, we write [ H , K ] to denote the subgroup generated by the set { [ h , k] \ h G H , k G AT} of a l l commutators of elements of J f w i t h elements of K . Th i s subgroup is the c o m m u t a t o r of H and AT, and it is t r i v i a l if and only if H and K centralize each other. We stress that, i n general, [ H , K ] is not e q u a l to the set of commutators of elements of H w i th elements of AT; it is the group g e n e r a t e d by them, or equivalents, it is the unique smallest subgroup of G that contains a l l of these commutators. In part icular, the subgroup [ G , G ] , which is generated by a l l commutators i n G, need not consist entirely of commutators. Reca l l that this subgroup is usually denoted G'; it is the derived subgroup (or commutator subgroup) of G. The derived subgroup G' of G is t r i v i a l if and only i f G is abelian, and as we know, it is the unique normal subgroup of G min ima l w i t h the property that the corresponding factor group is abelian.

There are a number of useful identities satisfied by commutators. It is t r i v i a l to check, for example, that [ x , y ] [ y , x ] = 1 for a l l x,y £ G. In other words, [ y , x ] = [ x , y ] ~ \ and so i f H , K C G, then the generating commutators of [AT, H] are exactly the inverses of the generating commutators of [ H , K ] , and it follows that [ K , H ] = [ H , K ] . Another often-used identity,

113

114 4 . C o m m u t a t o r s

which the reader should check, is that [ x y , z] = [ x , z ] y [ y , z] for a l l x , y , z E G. (In the language of Section 3 B , this says that for each element z E G , the map z] from G to G is a crossed homomorphism w i t h respect to the conjugat ion action of G on itself.) Th i s identity is the key, for example, to proving the following fact.

4.1. L e m m a . L e t H a n d K be s u b g r o u p s of a g r o u p G. T h e n H a n d K n o r m a l i z e [ H , K ] , o r e q u i v a l e n t l y , [ H , K } < ( H , K ) , t h e s u b g r o u p g e n e r a t e d by H a n d K .

Proof. Since [ H , K ] = [AT, H ] , it suffices by symmetry to prove that H C N G { [ H , K } ) . Let h,x e H and k G K . T h e n [ h x , k ] = [ h , k ] x [ x , k ] , and so [ h , k ] x = [ h x , k ] [ x , k ] - 1 e [ H , K ] . In other words, conjugation by x e H maps each of the generators [ h , k ] of [ H , K ] back into [ H , K ] , and thus [ H , K ] x C [ H , K ) . It follows that [ H , K ] X = [ H , K ) , as wanted. (As usual, we substitute x ' 1 for x to deduce the reverse containment.) •

Not surprisingly, there is a formula for [ z , x y ] analogous to the identity [ x y , z] = [ x , z ] y [ y , z ] , but it seems pointless to memorize it since i t can easily be derived using the formula for [ x y , z ] . To be specific, we have the following:

[ z , x y ] = [ x y , z ] - 1 = ( [ x , z ] * [ y , z\yl = [y, z \ - \ [ x , z ] ^ ) " 1 = [ z , y ] [ z , x \ y .

We have already said that K centralizes H i f and only i f [ H , K ] = 1. Th i s can be generalized slightly, as follows.

4.2. L e m m a . L e t N be a n o r m a l s u b g r o u p of a g r o u p G, a n d suppose t h a t H , K C G a r e a r b i t r a r y s u b g r o w p s . W r i t e G = G / N a n d f o l l o w t h e standard " b a r c o n v e n t i o n " , so t h a t H a n d K a r e t h e i m a g e s of H _ a n d K i n G u n d e r t h e c a n o n i c a l h o m o m o r p h i s m G ^ G . T h e n [ H , K ] = [ H , K ] , a n d i n p a r t i c u l a r , H a n d K c e n t r a l i z e each o t h e r i n G if a n d o n l y if [ H , K ] C N .

Proof. Since overbar is a homomorphism, we have [ u , v ] = [ u , v ] for arbit rary elements u,v G G, and it follows that [ H , K ] = [ H , K \ . Th i s subgroup is t r i v i a l precisely when [ H , K ] C N . •

We can also express the relation "AT normalizes H " i n the language of commutators.

4.3. L e m m a . L e t H a n d K be s u b g r o u p s of a g r o u p G. T h e n K C N G ( J i ) if a n d o n l y if [ H , K ] C H . I n p a r t i c u l a r , H < G if a n d o n l y if [ H , G] C H .

Proof. Let h € H and k G AT. Since [ h , k ] = h " 1 ^ , we have h k = h [ h , k ] , and thus h k G H if and only i f [ h , k] e H . The result is now immediate. •

4 A 115

If H and K normalize each other, therefore, then [ H , K ] C H n K . If also H n K = 1 , then [ H , K ] — 1, and so H and K centralize each other. (We have seen and used this fact previously.)

We can use Lemmas 4.2 and 4.3 to reexamine the notion of a central series in a group. Reca l l that a series of subgroups

1 = H 0 C H i C • • • C H r = G

is a central series in G i f a l l H i are normal i n G, and also i C Z ( G / H i - i ) for 0 < i < r . B y L e m m a 4.2, the latter condi t ion can be restated as [ H i , G ] C for 0 < i < r . Conversely, suppose we have a series of subgroups H h as above, and assume that [ H U G ] C i i ^ i for 0 < i < r . F i r s t , observe that since H i - i C we have [ i ^ , G ] C H i }

and thus L e m m a 4.3 yields H i < G for a l l i . The subgroups thus form a normal series, and indeed they form a central series by L e m m a 4.2.

To continue this discussion of central series, it is convenient to define commutators of three or more elements or of three or more subgroups. The standard notat ional convention is to use "left association". For three elements, for example, the triple commutator is denned by [ x , y , z ] = [ [ x , y } , z ] , and for three subgroups, the definition is [ X , Y , Z ] = [ [ X , Y ] , Z ] . (Caut ion: i n general i t is not true that [ X , Y, Z] is generated by the elements of the form [ x , y , z] w i th x e X , y e Y and z G Z . ) More generally, for n > 2, we define

[ x i , X 2 , . . . , X n ] = [ [ x 1 , x 2 , . . . , X n - 1 ] , X n ] .

and s imi lar ly for subgroups.

N o w suppose that { H i | 0 2, we have

[ G , G , G ] = [ [G,G] ,G] C [ H r - U G ] C P r _ 2 ,

and i f r > 3,

[G, G , G , G] = [[G, G , G] , G] C [ F r _ 2 , G] C P r _ 3 .

Cont inuing like this, we see that i f k < r + 1, then [G, G , . . . , G] C 7 f r _ f c + 1 , where there are k copies of G i n the commutator on the left.

Clearly, we need some better notat ion here, and so we write G 1 = G , G 2 = [ G , G ] , G 3 = [G, G , G] and in general, Gk = [ G k ~ \ G ] for k > 1. (This notat ion is not completely standard; some authors write 7

f c ( G ) to denote our subgroup Gk, and one can find other notations in the literature too.) Note that G 2 = [G, G] = G', the derived subgroup, but i n general, Gk w i t h k > 2 is not one of the groups G^" 1) of the derived series, and i n part icular,

116 4. C o m m u t a t o r s

G<2) = G" = [ [ G , G ] , [ G , G ] ] is not usually one of our subgroups Gk. (We shall see, however, that G" C G 4 , and in general, G ^ C G 2 \ )

Note that the subgroups Gk are characteristic i n G . In part icular, G k < G , and so by L e m m a 4.3, we have G k + l = [ G k , G ] C Gk. The subgroups G f c , therefore, form a descending series. (Another way to see this is by induct ion on k. Assuming that Gk C Gk~\ we get G k + l = [ G k , G] C [ G f c _ 1 , G] = G f c . )

Re turn ing now to the subgroups E % of onr central series for G , we have Gk C F r _ f c + 1 for 1 < k < r + 1, and i n particular, G r + 1 C H 0 = 1. N o w consider an arbi t rary nilpotent group, which, we recall, s imply means that G has a central series. It follows that Gm = 1 for some positive integer m. M o r e precisely, i f G is nilpotent and has a central series of length r , then Q r + i = 1 ( T h e l e n g t h o f a s e r i e g i g t h e n u m b e r 0 f containments.)

Conversely, suppose that G m = 1 for some positive integer m. Consider the series

1 = G m C G m _ 1 C ••• C G 1 = G .

(Note that the numbering of these subgroups is "backwards", running from high to low.) Since [ G k , G ] = G k + \ it follows that this is a central series, and thus G is nilpotent. Furthermore, we have seen that Gk C J f r _ f c + 1 , where the H i form an arbi trary central series i n G . Since the Gk are contained term-by-term i n the subgroups of an arbi trary central series, the series of Gk is called the lower central series of G . The length of this series is, of course m - 1, where m is the smallest positive integer such that Gm = 1.

Reca l l from Chapter 1 that the nilpotence class of a nilpotent group G is the smallest integer r such that G is equal to the term Z r of the upper central series of G . We saw that it is also the smallest integer r such that G has a central series { H i | 0 < i < r } , w i th H 0 = 1 and H r = G. (In other words, the nilpotence class is the length of the shortest possible central series.) We now know that if G is nilpotent of class r then G r + 1 = 1. A l so , i t is not possible that Gm = 1 w i th m < r + 1 because otherwise, the lower central series of G would be a central series of length m - 1 < r . Th i s shows that if G is nilpotent of class r , then m = r + 1 is the smallest integer such that G m = 1.

O f course, i f G is an arbi t rary (and not necessarily nilpotent) group, one can s t i l l consider subgroups of the form Gm = [G, G , . . . , G ] , where G occurs m times. B y abuse of notat ion, the series { G i \ i > 1} is often called the "lower central series" of G , even though i n the nonnilpotent case, i t is not a true central series because the identity is not one of its terms. If G is finite, then since its lower central series cannot involve infinitely many strict containments, there must be some superscript m such that G m = G m + 1 . It is then immediate that Gm = Gn for a l l integers n > m, and this final te rm of the lower central series of G is denoted G ° ° . It is not hard to see

4 A 1 1 1

that if_/V < G , then G / N is n i l p o t e n U f a n d o n l y i f G°° C TV. (To see this, write G = G / N , and observe that G™ = G m by repeated applicat ion of L e m m a 4.2.) It follows that if G is finite, then G°° is the unique normal subgroup of G m i n i m a l w i t h the property that the corresponding factor group is nilpotent.

A s we mentioned in Chapter 1, the groups of nilpotence class 1 are exactly the nont r iv ia l abelian groups. Somewhat more interesting are groups of nilpotence class 2. For these nonabelian groups, we have 1 = G 3 = [G, G , G] = [ G 1 , G ] , so that the derived subgroup G' C Z ( G ) , or equivalently G / Z ( G ) , is abelian. In this case, a l l commutators are central, and hence [ x y , z ] = [ x , z \ y [ y , z \ = [ x , z ] [ y , z \ , and so the map [ - , z ] defines a homomor-phism from G into G ' . The following easy consequence of this is sometimes useful. (Before we state our result, we recall that the exponent of a group G is the smallest positive integer n such that x n = 1 for a l l x e G . In other words, the exponent of a group is the least common mult iple of the orders of its elements.)

4.4. L e m m a . L e t P be a p - g r o u p of n i l p o t e n c e class 2, a n d assume t h a t P ' has e x p o n e n t p e . T h e n t h e e x p o n e n t of P / Z ( P ) d i v i d e s p e . I n p a r t i c u l a r , if P ' i s e l e m e n t a r y a b e l i a n , t h e n P / Z ( P ) i s e l e m e n t a r y a b e l i a n , a n d thus $ ( P ) C Z ( P ) .

Note that i f P has class 2, then P ' C Z ( P ) , and so both P ' and P / Z ( P ) are abelian. Thus P ' is elementary abelian if and only if e = 1, and to prove that P / Z ( P ) is elementary abelian, i t suffices to show that its exponent divides p , and this is the assertion of the l emma when e = 1. The last conclusion of the lemma, therefore, is an immediate consequence of the characterization of the F ra t t i n i subgroup of a p-group as the unique smallest normal subgroup having an elementary abelian factor group. Th i s fact was mentioned previously, but we never gave a formal proof, and so we digress to do so now.

4.5. L e m m a . L e t N < P , w h e r e P i s a p - g r o u p . T h e n P / N i s e l e m e n t a r y a b e l i a n if a n d o n l y i / $ ( P ) C N .

Proof. Le t M < P be a max ima l subgroup. Since P is nilpotent, M < P , and we see that P / M has no subgroups other than itself and the identity. Thus P / M has prime order, which must be p . N o w P / M is abelian, and so P ' C M , and also, given x e P , we have X P e M . Thus P ' and x? lie in the intersection of a l l max ima l subgroups of G , or in other words, they lie in $ ( P ) . We conclude that P / $ ( P ) is elementary abelian, and thus also P / N is elementary abelian i f $ ( P ) C TV C P .

Conversely, suppose P / N is elementary abelian, and assume that N < P . B y the fundamental theorem of abelian groups, P / N is a direct product of

118 4. C o m m u t a t o r s

nontr iv ia l cyclic subgroups, each of which must have order p . The product of a l l but one of these factors has index p , and so is m a x i m a l i n P / N . The intersection of those max ima l subgroups of P / N obtained by deleting one of its direct factors of order p is t r i v i a l , and thus N is the intersection of some collection of max ima l subgroups of P . We conclude that $ ( P ) C TV, as wanted. •

P r o o f of L e m m a 4.4. To prove that P / Z ( P ) has exponent d iv id ing p e , we must show that x ^ £ Z ( P ) for a l l i e P . We have [ x P e , y ] = [ x , y ] p B = 1, where the first equality holds since [ - , y \ is a homomorphism, and of course, the second equality holds because [ x , y ] G P ' , which has exponent p e . Th is shows that x P e commutes w i t h a l l elements y G P , and so x ? e G Z ( P ) , as required. B y L e m m a 4.5, the proof is now complete. •

Next , we present a result relating centers and commutator subgroups. Its proof is based on an exception to the general principle that elements of the commutator subgroup need not be commutators.

4.6. L e m m a . L e t A be a n a b e l i a n n o r m a l s u b g r o u p of a g r o u p G, a n d suppose t h a t G/A i s c y c l i c . Then G' = [ A , G ] a n d

G' = A / [ A n Z ( G ) ) .

I n p a r t i c u l a r , if A i s finite, t h e n \ A \ = \ G ' \ \ A n Z ( G ) | .

Proof. Choose a generating coset A g for G / A , and let 9 : A - > A be the map defined by 9 ( a ) = [ a , g ] . (Note that [ a , g] G A since A < G.) If a , b G A , then

9 ( a b ) = [ a b , g ] = [ a , g ] b [ b , g ] = [ a , g ] [ b , g ] = 9 ( a ) 9 ( b ) ,

where the th i rd equality holds since [a,g] and b lie in the abelian group A . Thus 9 is a homomorphism, and ker(0) = C A ( g ) . A l so , 9 ( A ) is a subgroup, and 9 ( A ) C [ A , G] C G'.

Since A g is a generating coset for G = G / A , we see that J g ) = (g) = G, and thus G = A ( g ) , and it follows that C A ( g ) C A n Z ( G ) . The reverse inequali ty is clear, and thus ker(0) = C A ( g ) = A n Z ( G ) . Th i s yields 0 (4 ) = A / ( A n Z ( G ) ) , and so to complete the proof, it suffices to show that 9 ( A ) = G'.

We already know that 9 ( A ) is a subgroup of G', and so it suffices to show that 9 ( A ) < G and that G / 9 ( A ) is abelian. F i r s t , since A is abelian and 9 ( A ) C A , it is clear that A C N G (0(A)) . A l so , 9 ( a ) 9 = [ a , g } 3 =

[ a 9 , g [ = 9 ( a 9 ) G 9 ( A ) for a l l a G A , and thus 5 G N G (0(A)) . It follows that G = A ( 5 ) C N G (0(A)) , and so 0(A) < G , as required.

N o w (redefining overbars) write G = G / 9 ( A ) , so that G = A j g ) . Since [ a , g ] = [a, 5 ] = 9 ( a ) is t r i v i a l for a l l a G A , it follows that the elements o for

4 A 119

a G A together w i t h g form a pairwise commut ing set of generators for G, which is therefore abelian. Th i s completes the proof. •

A s an example of how L e m m a 4.6 can be used, we offer the following.

4 .7. Theorem. L e t A < P , w h e r e P i s a p - g r o u p , a n d suppose t h a t A i s a b e l i a n a n d \ A \ = p m . Assume t h a t P / A i s c y c l i c , a n d t h a t \ A n Z ( P ) | = p . T h e n t h e n i l p o t e n c e class of P i s TO.

Proof. Let Z = A n Z ( P ) , and note that P ' C A and \ A : P ' \ = \ Z \ = p by L e m m a 4.6. If TO = 1, then A = Z is central i n P , and since P / A is cyclic, it follows that P is abelian, and thus its nilpotence class is 1 = TO, as wanted. We can thus assume that TO > 1, and we proceed by induct ion on TO.

Since \ A : P ' \ = p and \ A \ > p , we have P ' > 1, and thus P ' meets Z ( P ) nont r iv ia l ly by Theorem 1.19. T h e n 1 < P ' n Z ( P ) C A n Z ( P ) = Z , and since | Z | = p, we have Z = P ' nZ (P ) , _and i n part icular , Z C P ' . N o w wri te P = P/Z,_and_observe that ( P ) ' = P ' has index p i n the abelian normal subgroup A of P . A l so , ~ P / A = P / A is cycl ic , and hence by L e m m a 4.6, we have \ A n Z ( P ) \ = \ A : ( P ) ' | = p, and thus P satisfies the hypotheses of the theorem.

Since \ A \ = p m " \ we conclude from the induct ive hypothes is that P has nilpotence class_m - 1, and thus i n the lower central series of P , we have ( P ) m = 1 and ( P ) 7 " " 1 ^ 1. Since P m = ( P ) m = 1, we have P m C Z , and thus P m + 1 = [ P m , P ] C [ Z , P ] = 1.

To complete the proof, we must show that P m ^ 1. If TO = 2, then pm _ p i > ^ a n ( j g o w e c a n S U p p o s e that TO > 2, and hence P m _ 1 C P 2 =

P ' C A. We have P ^ 3 1 = ( P ) ™ " 1 ^ 1, and thus P ™ " 1 ^ = 4 n Z ( P ) . It follows that P m ~ l % Z ( P ) , and thus P m = [ P m - \ P ] ^ 1, as required. •

Suppose now that P is a nonabelian p-group of order p e . Assume that | Z ( P ) | = p, and that P has an abelian subgroup A of index p (which, of course, is necessarily normal) . Since P is nonabelian, A is nontr ivia l , and thus A n Z ( P ) > 1, and hence Z ( P ) C A . A l s o , P / A is cycl ic , and it follows by Theorem 4.7 that that the nilpotence class of P is e - 1.

To put this into context, let P be an arbi t rary p-group of nilpotence class r , and order p e , where e > 2. Since P has at least one normal subgroup of index p 2 and the corresponding factor group is abelian, it follows that \ P : P ' \ > p 2 . Now consider the lower central series

P = P 1 > P 2 > P 3 > ••• > P r + 1 = 1,

and observe that \ P : P 2 | = | P : P ' | > p 2 , and of course, \ P l : P m | > p for 2 r + 1, and

120 4. C o m m u t a t o r s

thus the nilpotence class r of a p-group P of order p e w i t h e > 2 is at most e - l . If i t happens that r = e - 1, we say that P has m a x i m a l c lass .

If \ P \ = P e > P 2 and | Z ( P ) | = p and P has an abelian subgroup of index p , we have seen that P has nilpotence class e - 1, and thus P has m a x i m a l class. In part icular, it is easy to see that dihedral , generalized quaternion and semidihedral 2-groups satisfy these hypotheses, and so a l l of these groups have max ima l class. In fact, these are the only max ima l class 2-groups, but we w i l l not prove that here. A l so , we mention without proof that for p > 2, there exist max ima l class p-groups that fail to have an abelian subgroup of index p .

We close this section wi th a discussion of the set of elements x (E P such that x p = 1, where P is a p-group. In general, these elements do not form a subgroup. (If P = D 8 , for example, there are exactly six such elements including the identity, and of course a set of six elements i n a group of order 8 cannot be a subgroup.) The group generated by a l l elements x £ P such that x P = 1 is denoted fii(P), and more generally, for integers r > 1, one defines ftr(P) = ( x G P \ x p " = 1). (There does not seem to be any standard name for this characteristic subgroup, which is usually just called "omega-r" of P . )

If p > 2 and P is a p-group of class at most 2, then the elements x G P such that x p = 1 actually do form a subgroup. (Equivalently, S l ^ P ) has exponent d iv id ing p in this case.) A s is shown by D 8 , which has class 2, the condi t ion p > 2 is essential. B u t the assumption that the nilpotence class is at most 2 is unnecessarily strong; in fact, i t suffices to assume that the nilpotence class is less than the prime p. The proof of this result of P h i l i p H a l l uses "commutator collect ion", which is the basic idea that underlies our proof for class 2 groups. The proof of the full result is considerably more complicated, however, and we do not present it here.

G i v e n x , y ( E P w i th X P = 1 = y P , we want to show that ( x y ) p = 1, and so we investigate how to compute ( x y ) n for arbi trary elements x,y G P and positive integers n . The idea here is as follows. We have

( x y ) n = x y x y x y • • • xy

where on the right, there are n copies of x al ternating w i t h n copies of y . If P is abelian, it is t r i v i a l to simplify this. We can move each copy of x as far as necessary to its left, freely passing through the intervening copies of y , thereby collecting the n copies of x at the left, and leaving n copies of y on the right. Th i s , of course, yields the familiar formula ( x y ) n = x n y n for abelian groups.

W h a t if P is not abelian? In general, i n any group, i f we have the string y x and we want to move the x to the left, across the y , we can use the

4 A 121

identi ty yx = x y [ y , x } . In other words, we can s t i l l pass the x through the y , but to do so, we must pay a price. T h e penalty is that we must insert the commutator [y, x] to the right of the y .

For simplici ty, consider the case n = 3. M o v i n g the second x left across the first y (and inserting [ y , x ] a a payment for the privilege of doing this) we have

{ x y f = xyxyxy = x x y [ y , x ] y x y .

Next , we want to move the th i rd x leftward i n order to j o i n the first two, and we do this i n three steps, first moving across y , then across [ y , x ] , and finally across another y . To pay for these transits, we must insert [y, x ] , and then [ y , x , x ] , and then another [ y , x ] . We thus have

[ x y f = xyxyxy = x x y [ y , x \ y x y

= x x y [ y , x } x y [ y , x } y

= x x y x [ y , x } [ y , x , x ] y [ y , x } y

= x x x y [ y , x] [y, x] [y, x , x } y [ y , x ] y ,

and at this point, we have collected a l l three copies of x at the left, and these are followed by a mixture consisting of three copies of y , three copies of [y, x ] , and one copy of [y, x , x } .

We can now continue to "simplify" this, by collecting the three copies of y toward the left, to a posit ion just to the right of the three copies of x . In order to do this, we w i l l need to move the second y across copies of [y, x] and [ y , x , x ] , and so we must pay penalties of [ y , x , y ] and [ y , x , x , y ] . To collect the th i rd y , our penalties w i l l be [ y , x , y ] , [ y , x , x , y ] , and also [ y , x , y , y ] and [ y , x , x , y , y \ . W h e n this process is finished, we w i l l have ( x y f = x 3 y 3 w , where w is some complicated product of one or more copies of each of [ y , x ] , [ y , x , x ] , [ y , x , y ] , [ y , x , x , y ] and [ y , x , x , y , y ] .

We could continue. The simplest commutator appearing as a factor of w is [y, x ] , and there are three of these. Suppose we collect these leftward, to a posi t ion just to the right of the three copies of y . To do this, we w i l l need to pay "transit fees" by inserting commutators like [ [ y , x , x ] , [y,x]] and others. T h i s w i l l result i n a formula of the form ( x y ) 3 = x 3 y 3 [ y , x ] 3 w ' , where w' is a product of commutators of "weight" 3 and higher. T h i s seems like quite a mess, but i t is possible to do a l l of this i n a systematic way, and to keep track of everything, and this is the H a l l "commutator collection process", which we w i l l not pursue further.

If P has class at most 2, everything becomes much easier. In that case, the commutator [y, x] is central, and so every copy of [y, x] can be moved to the far right, where it w i l l not interfere w i t h further collection. We s imply need to count how many copies of [ y , x ] we produce as we collect the n

122 4. C o m m u t a t o r s

copies of x leftward i n ( x y f . The leftmost x starts i n its desired posit ion, and it need not be moved. The second x moves across one y , incurr ing one payment of [ y , x ] ; the th i rd x moves across two copies of y generating two addi t ional copies of [y, x } ; the fourth x generates three more copies of [y, x ] , and so on. In total , therefore, we obtain 1 + 2 + • • • + (n - 1) = (n - l ) n / 2 copies of [ y , x ] , and this establishes the formula

( x y ) n = x n y n \ y , x ] ^ n - ^ 2 ,

i n a nilpotent group w i t h class at most 2.

T w o useful consequences of this formula are given i n the following result.

4.8. Theorem. L e t P be a p - g r o u p w i t h n i l p o t e n c e class a t most 2, a n d assume t h a t p > 2 . The f o l l o w i n g then h o l d .

(a) The set { x e P \ x P = 1} i s a s u b g r o u p of P .

(b) / / [ y , x ] P = 1 f o r a l l x , y E P , then t h e map x ^ x p i s a h o m o m o r -p h i s m f r o m P i n t o i t s e l f

Proof. F i r s t , assume that [y, x } p = 1 for al l x , y E P . Since the nilpotence class of P is at most 2, we can apply our commutator collection formula to conclude that { x y ) p = x p y P [ y , x ] m , where m = p ( p - l ) / 2 . Since p ^ 2, however, m is a mult iple of p , and thus [ y , x } m = 1. It follows that ( x y ) P = x P y P , and thus the map i ^ ^ i s a group homomorphism, proving (b).

N o w (without assuming that p th powers of commutators i n P are t r ivial) observe that since a l l commutators i n P are central, the map [ - , x ] from P to P ' is a group homomorphism for every element x E P . If y E P and y P = 1, therefore, we have [ y , x } p = [ y p , x ] = [ l , x ] = 1 for a l l x E P. In

particular, if bo th ^ = 1 and / = 1, then commutator collection yields [ x y ) P = x p y P [ y , x ] m = 1, where as before, m = p ( p - l ) / 2 is a mult iple of p because p ^ 2. The set { x e P \ x P = 1} is thus closed under mul t ip l ica t ion, and so it is a subgroup. •

Observe that the conclusion of Theorem 4.8(a) is equivalent to saying that X P = 1 for every element x 6 fti(P).

P r o b l e m s 4 A

4 A . 1 . Suppose that G = A B , where A and B are abelian subgroups. Show that G' = [ A , B } .

4A.2 . Let H , K and L be subgroups of order 2 i n some group G, and observe that the set { [ h , k , l ] \ h <E H , k <E K , I <E L } contains at most one nonidenti ty element, and so it generates a cyclic group. Now let G = A 5 , the

P r o b l e m s 4 A 123

alternating group of degree 5. Show that i t is possible to choose subgroups H , i f and L of order 2 such that [ P , i f , L] = G.

H i n t . Let P C G have order 5. Choose H , i f C N G ( P ) w i t h H ^ K , and choose L ^ N G ( P ) . Show that [P , i f ] = P and observe that (P, L ) = G.

Note. T h i s problem shows that [P , i f , L] is not necessarily generated by elements of the form [ h , k, 1} w i t h h e H , k e i f and / e L .

4 A . 3 . Say that a group Q is quasiquaternion i f Q = C U , where C and P are cyclic subgroups, w i t h C < Q and C n U = Z ( Q ) . (Note that generalized quaternion and semidihedral groups are quasiquaternion.) Suppose that G / Z is quasiquaternion, where Z C G' n Z ( G ) . Show that Z = 1.

Hint . Let A and B be the subgroups of G = G / Z corresponding to C and U of the definition of "quasiquaternion". Show that A n P = Z ( G ) and apply L e m m a 4.6.

Note . G i v e n any group Q, consider abelian groups Z that can be isomor-phical ly embedded i n some group G i n such a way that Z C G' D Z ( G ) and G / Z = Q. It is a fact that up to isomorphism, there is a unique largest such group Z , depending on Q; it is called the Schur multiplier of Q. The homomorphic images of the Schur mult ipl ier are a l l of the groups Z that satisfy our conditions. Th i s problem asserts that quasiquaternion groups have t r i v i a l Schur mult ipl iers . ( A l i t t le more information about Schur multipliers appears i n Chapter 5.)

4A.4 . A p-group P is said to be extraspecial i f P ' = Z ( P ) has order p . Show that i f P is extraspecial, then P / Z ( P ) is elementary abelian, or equivalents , Z ( P ) = P ' = $ ( P ) .

Note. Nonabel ian p-groups of order p 3 are extraspecial.

4A.5 . Le t P be an extraspecial p-group and let x , y € P w i t h xy + yx. Let U = ( x , y ) a n d V = C P ( U ) .

(a) Show that \ P : C P ( x ) \ = p .

(b) Show that \ U \ = p 3 .

(c) Show that \ P : V \ = p 2 .

(d) If P > U, show that Z ( V ) = Z ( P ) and that V is extraspecial.

(e) Show that | P : Z ( P ) | = p 2 e for some integer e.

H i n t . Use induct ion on | P | for e.

Note. For every odd integer n > 1 and every pr ime p, there are exactly two isomorphism types of extraspecial p-groups of order p n . If p ^ 2, these groups can be distinguished by the fact that one of them has exponent p and the other has exponent p 2 .

124 4. C o m m u t a t o r s

4A.6 . Let P be a p-group having max ima l class, and let N p 2 . Show that N is one of the terms of the lower central series for P , and i n part icular, N is the only normal subgroup of G having order |W | .

4A.7 . F i x a prime p and let A be the external direct product of p copies of a cyclic group C of order p n w i t h n > 0. V i e w the elements of A as p-tuples of the form ( c 1 ; c 2 , . . . , cp) w i t h a E C. Let U = ( u ) be cyclic of order p , and let U act on A by cycl ing the components. In other words,

( c i , c 2 , . . . , C p - i , c p ) u = ( c p , c i , c 2 , . . . , C p - l ) .

Fina l ly , let P = A x P be the semidirect product, and view A and U as subgroups of P . Show that the m a x i m u m of the orders of the elements of P is exactly p n + 1 .

Note . O f course, P is just the regular wreath product P = C \ U . Tak ing n = 1, we get an example where fti(P) does not have exponent p .

4A.8 . Assume the notat ion of P rob lem 4 A . 7 w i t h n > 0.

(a) Show that C A ( U ) = Z ( P ) is the set of p-tuples i n A such that a l l components are equal.

(b) Show that [ A , U] = P ' is the set of p-tuples i n A such that the product of the components (computed i n C ) is the identity.

(c) Show | Z ( P ' P ) | = p .

(d) If n = 1, show that P has max ima l class, and that i n any case, P ' U has max ima l class.

4 A . 9 . Let G = N A , where N < G and A C G, and let M be the final term i n the series TV D [ N , A ] D [ N , A , A ] D [ N , A , A , A ] D • • •.

(a) Show that A « G i f and only i f M C A .

(b) If M C A , show that M C A°°.

4A.10. Let P be a p-group w i t h | P : Z ( P ) | < p n . Show that

| P ' | < p n ( n - 1 ) / 2 .

H i n t . Induct on n . If P is nonabelian, choose Q so that Z ( P ) C Q C P w i t h \ P : Q \ = p . Get a bound on \ Q ' \ and then apply L e m m a 4.6 to the group P / Q ' . Note that { P / Q J = P ' / Q ' .

Note . If the center of a group has smal l index, then the group is i n some sense "nearly abelian", and so it should be no surprise that the derived subgroup is not too big.

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4A.11 . Le t G = A I H be the regular wreath product of an abelian group A w i t h an arbi t rary finite group H , and view the base subgroup B as the group of functions / : H A w i t h pointwise mul t ip l ica t ion . Let K C H . Show that [ B , K ] is exactly the set of functions f e B w i t h the property that the product of the values of / over each left coset of K i n H is the identity. Deduce that \ [ B , K } \ = \ A \ \ H \ - \ H - K \ .

H i n t . The action of H on B is given by f h ( x ) = f { x h ~ l ) for h € H .

4A.12. G i v e n a finite group H , let T ( H ) denote the set of subgroups T of H max ima l w i t h the property that T is abelian and can be generated by two elements. Let G = A I H as i n P rob lem 4 A . 1 1 , where A is abelian, and let B be the base group.

(a) Show that if x,y e G and [ x , y ] € B , then [ x , y ] € [ B , T ] for some member T E T ( H ) .

(b) Show that i f

then G' D B contains elements that are not commutators i n G.

(c) If H is not an abelian group that can be generated by two elements, show that there exists some number m such that the inequali ty of (b) holds whenever | A | > m. In part icular , show that we can take m = 3 if H = D 8 .

4A.13. Let G be nilpotent group w i t h class m > 1, and let a € G. Show that the nilpotence class of the subgroup H = G ' ( a ) is less than m .

which holds for every three elements of every group. Its proof, which proceeds by expansion and cancellation, is completely elementary and ut ter ly routine. It is a computat ion that is fun to carry out, however, and we urge the reader to do so. One can think of the H a l l - W i t t formula as a k i n d of three-variable version of the much more elementary two-variable identity, [ x , y ] [ y , x ] = 1. T h i s observation hints at the possibi l i ty that a corresponding four-variable formula might exist, but i f there is such a four-variable identity, it has yet to be discovered.

We digress briefly to mention that the H a l l - W i t t identi ty is analogous to the somewhat simpler Jacobi identity for addit ive commutators i n a r ing.

4 B

126 4. C o m m u t a t o r s

(We are assuming associativity of mul t ip l ica t ion, of course.) If x,y £ R, where R is a r ing, the additive commutator of x and y is xy - yx, which is zero precisely when x and y commute. Unfortunately, the standard notat ion for the addit ive commutator of x and y i n a r ing is exactly the same as the notat ion for group commutators: [ x , y ] . B u t we w i l l not refer to additive commutators outside of this paragraph, and so we hope that this ambiguity w i l l not cause any confusion. The Jacobi identity asserts that [ x , y , z\ + [ y , z , x ] + [ z , x , y ] = 0 for a l l x , y , z £ R, where R is a r ing, and where as w i t h groups, the triple commutators are denned by left association. We mention that a (nonassociative) r ing L w i t h mul t ip l ica t ion x-y is a L ie ring i f x -x = 0 for a l l x £ L and the Jacobi identity ( x - y ) - z + ( y - z ) - x + ( z - x ) - y = 0 holds for a l l x , y , z £ L . In part icular, every associative r ing is a L i e r ing w i t h respect to the mul t ip l ica t ion x-y = [ x , y ] . In fact, L i e rings provide a powerful tool for the study of p-groups, but we w i l l say no more about that here.

O u r pr inc ipal applications of the H a l l - W i t t identi ty rely on the following l emma and its corollary, bo th of which are referred to as the "three-subgroups lemma."

4.9. L e m m a . L e t X , Y a n d Z be s u b g r o u p s of a n a r b i t r a r y g r o u p G, a n d suppose t h a t [ X , Y, Z] = 1 a n d [ Y , Z , X ] = 1 . T h e n [ Z , X , Y ] = 1 .

Proof. We want to show that [ [ Z , X ] , Y ] = 1, or equivalently, that every element of the group [ Z , X ] commutes w i t h every element of Y. For this purpose, we show that a l l commutators [ z , x] for z £ Z and x £ X centralize each element y £ Y. Th i s is sufficient because C G ( y ) is a subgroup of G, and so i f it contains a l l of the commutators [ z , x ] , then it contains the subgroup [ Z , X ] generated by these commutators. It is, therefore, enough to show that [ z , x , y ] = 1, for a l l x £ X , y £ Y and z £ Z . (But recall that in general, these commutators do not generate [ Z , X , Y ] . ) Equivalently, it suffices to show that [ z , x ' \ y ] = 1 for a l l x , y and z i n X , Y and Z , respectively.

N o w [ x ^ y - 1 ] £ [ X , Y ] , and so [ x , y - \ z ] £ [ X , Y , Z ] = 1, and we have [ x , y ' \ z } y = 1. Similar ly , [ y , Z - \ x ] z = 1, and thus by the H a l l - W i t t identity, [ z , x - \ y } x = 1. Thus [ z , x - \ y ] = 1, as wanted. •

4.10. Corol lary. L e t N be a n o r m a l s u b g r o u p of a g r o u p G, a n d l e t X , Y , Z C G be a r b i t r a r y s u b g r o u p s . I f [ X , Y , Z ] C N a n d [ Y , Z , X ] C N , t h e n [ Z , X , Y] C N .

Proof. Let G = G / N and follow the standard "bar convention". B y L e m m a 4.2, we have [ H j C \ = [ H , K ] for a l l subgroups H and K of G. T h e n

[ X , Y , Z ] = [ [ X , Y ] , Z ] = [ \ X 7 Y ] , Z \ = [ X , Y , Z ] = 1

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and similarly, [ Y , Z , X ] = 1. B y L e m m a 4.9 then, 1 = [ Z , X , Y] = [ Z , X , Y] and thus [ Z , X , Y ] C N . U

A n appeal to the three-subgroups l emma i n a proof often makes it possible to avoid tedious and messy elementwise commutator calculations. U n fortunately, it is not always easy to see how to apply this very useful tool . We give a number of applications here and i n what follows, and these should provide plenty of practice.

F i r s t , we obtain some addi t ional information about the lower central series.

4.11. Theorem. As u s u a l , w r i t e Gn t o d e n o t e t h e n t h t e r m of t h e l o w e r c e n t r a l series of a g r o u p G. T h e n [ G \ GP] C G i + i f o r i n t e g e r s i , j > 1.

Proof. We proceed by induct ion on j , which is the superscript on the right i n the commutator [ & , & ] . Since Gl = G, we see that if j = 1, the formula we need is [ G \ G ] C G i + 1 , and this is va l id since by definition, G i + l = [ G \ G ] . We can assume, therefore, that j > 1, and so we can write 0 = \ G i - \ G \ . T h e n

[ G \ G j ] = [ G j , G<] = [ G j ~ \ G, G*] ,

and to show that this tr iple commutator is contained i n the normal subgroup G i + i , it suffices by the three-subgroups l emma (Corol lary 4.10) to show that

[G, G \ G j ' 1 } C G i + j and [ G \ Gj~\ G] C G i + j .

We have

[G, G\ G j ~ l ] = [ G \ G , = [ G i + \ G j ~ 1 } C = G i + j ,

where the containment is val id by the induct ive hypothesis. A l so ,

[ G \ G j - \ G ] C [ G i + j - \ G] = G i + j ,

where here too, the containment follows from the induct ive hypothesis. Th is completes the proof. •

Suppose we consider commutators of n copies of G that are not necessari ly left associated. For example, i f n = 8, we might have something like [[[G, G ] , [G, G , G]], [[G, G] , G } } . (We refer to such an object as a commutator of weight n.) T h e following result shows that the left-associated weight n commutator of copies of G contains every other such weight n commutator. (Of course, the left associated weight n commutator is exactly Gn, the n t h term of the lower central series.)

4.12. Corollary. L e t G be a g r o u p . T h e n every w e i g h t n c o m m u t a t o r of c o p i e s o f G i s c o n t a i n e d i n Gn.

128 4. C o m m u t a t o r s

Proof. We proceed by induct ion on n . The only weight 1 commutator is G itself, and so there is nothing to prove i n this case. A n arbi trary weight n commutator w i t h n > 1 has the form [ X , Y ] , where X is a weight i commutator , Y is a weight j commutator and i + j = n . Since i < n and j < n , the inductive hypothesis yields X C Gi and Y C G > , and thus [ X , Y] C [ G \ G ' } C G i + ^ ' = G n by Theorem 4.11. •

Consider the terms G^ of the derived series of G. We see that G' = [ G , G] is a weight 2 commutator, G" = [ G ' , G'] = [ [ G , G ] , [ G , G]] is a weight 4 commutator , and G'" = [ G " , G " } is a weight 8 commutator. In general, G » is a weight 2 r commutator for a l l r > 0, and this yields the following.

4.13. Corol lary. We h a v e G » C G 2 " f o r a l l g r o u p s G a n d i n t e g e r s r > 0. A l s o , if G i s m l p o t e n t of class m, t h e n t h e d e r i v e d l e n g t h of G i s a t most l + l o g 2 ( m ) .

Proof. Since G^ is a weight 2 r commutator, the first assertion is immediate from Coro l la ry 4.12. N o w assume that G is nilpotent of class m , and let d be its derived length, so that G^ = 1 and d is the smallest integer w i t h this property. We have

G2"'1 D G^d~l) > 1 = G m + 1 ,

and so 2d~l < m + 1. T h e n 2d~l < m, and hence d - 1 < l og 2 (m) . •

Next , we explore some connections between the conjugacy class sizes and the nilpotence class of a nilpotent group G. Suppose we know the number m of different sizes of conjugacy classes i n G. If m = 1, for example, then since the size of the class of the identi ty of G is 1, it follows that a l l classes of G have size 1, and so G is abelian. (Of course, we d id not use the assumption that G is nilpotent here.) Th i s suggests the (stil l unresolved) question of how the number m of conjugacy class sizes of G affects the structure of G. In part icular , it would be interesting to know if the derived length of G, or perhaps even the nilpotence class of G, is bounded by some function of m.

After many years without progress on these questions, it was proved by K . Ishikawa that if m = 2, then G has nilpotence class at most 3. (In this case too, it is not really necessary to assume that G is nilpotent; a result of N . Ito shows that a finite group w i t h just two class sizes is necessarily nilpotent.) More recently, A . M a n n proved a stronger result w i t h an easier proof, and we present that next.

G i v e n a finite group G w i t h class sizes 1 = m < n 2 < • • • < n m , we write M { G ) to denote the subgroup of G generated by a l l elements ly ing i n conjugacy classes of size at most n 2 . (We w i l l use this notat ion for the

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remainder of this section.) Note that M ( G ) = G i f m = 2, and thus the following result contains Ishikawa's theorem.

4.14. T h e o r e m (Mann) . If G i s a finite n i l p o t e n t g r o u p , then t h e m l p o t e n c e class o / M ( G ) i s a t most 3.

Ac tua l ly , we can replace the hypothesis that G is nilpotent i n Theorem 4.14 by a much weaker one: that G has a self-centralizing normal subgroup. ( B y this, we mean a subgroup A < i G such that A = C G { A ) . Observe that such a subgroup is automatical ly abelian.)

4.15. Theorem. Suppose t h a t a finite g r o u p G c o n t a i n s a s e l f - c e n t r a l i z i n g n o r m a l s u b g r o u p . Then M ( G ) i s n i l p o t e n t , a n d i t s n i l p o t e n c e class i s a t most 3.

Reca l l that a nilpotent group G always has a self-centralizing normal subgroup, and so Theorem 4.14 is immediate from Theorem 4.15. (The existence of self-centralizing normal subgroups i n nilpotent groups is part of P rob lem ID.10 , but nevertheless, we give the easy proof here.)

4.16. L e m m a . L e t G be a finite n i l p o t e n t g r o u p , a n d l e t A be m a x i m a l a m o n g n o r m a l a b e l i a n s u b g r o u p s o f G . Then A = C G { A ) , a n d hence every n i l p o t e n t g r o u p c o n t a i n s a s e l f - c e n t r a l i z i n g n o r m a l s u b g r o u p .

Proof. Wr i t e G = C G ( A ) . T h e n A C C, and we assume that this containment is proper. Since A o G , we have C < G, and thus we can choose a subgroup B C C such that B / A is m i n i m a l normal i n G / A . Since G/A is nilpotent, it follows that B / A C Z ( G / A ) , and thus B / A has pr ime order. In part icular , B / A is cyclic, and since B C G = C G ( A ) , it follows that A C Z ( B ) . We conclude that B is abelian, and since B < G , this contradicts the choice of A . U

The key ingredient i n the proof of Theorem 4.15 is the following.

4.17. L e m m a . L e t K be a n a b e l i a n n o r m a l s u b g r o u p of a finite g r o u p G, a n d l e t x E G be n o n c e n t r a l . Then \ C G ( x ) \ < \ C G ( y ) \ , w h e r e y = [ k , x ] a n d k E K i s a r b i t r a r y .

Proof. Observe that y <E [AT, G] C AT. If y = 1, then C G { y ) = G > C G ( x ) , and there is nothing further to prove, so we assume that y ^ 1. B o t h x and y lie i n the subgroup H = K C G ( x ) , and of course, C H { x ) = C G ( x ) and C H ( v ) Q C G { y ) . It suffices, therefore, to prove that | C H ( a ; ) | < \ C H ( y ) \ .

N o w let 6 : K AT be the map defined by 6 { t ) = [ t , x ] . For s,t€K, we have

6 ( s t ) = [ s t , x ] = [ s , x ] % x ] = [ s , x ] [ t , x ] = e(s)6(t),

130 4. C o m m u t a t o r s

where the t h i rd equality holds because both t and [ s , x] lie i n the abelian group AT. Thus 6 is a homomorphism, and 6 ( K ) is a subgroup, and we have

\ 9 { K ) \ = \ K : ker(0)| = \ K : K n C H { x ) \ = \ H : C H { x ) \ .

We argue next that each of the subgroups AT and C G ( x ) normalizes B ( K ) , and thus B ( K ) < K C G ( x ) = H . Certainly, AT C N G ( 0 ( A T ) ) since 9 { K ) C K and AT is abelian. Also , since 0(AT) is uniquely determined by the normal subgroup AT and the element x , we see that C G ( x ) C N G ( 9 ( K ) ) . N o w 0(AT) < H , and y is a nonidenti ty element of 6 { K ) , and thus the entire i f -conjugacy class Y of y consists of nonidenti ty elements of 6 { K ) . Then |JJ : C H ( y ) \ = | ^ | < |#(AT)| = \ H : C H { x ) \ , as required. •

4.18. Corol lary. L e t K be a n a b e l i a n n o r m a l s u b g r o u p of a n a r b i t r a r y finite g r o u p G, a n d w r i t e M = M ( G ) . Then [AT, M ] C Z ( G ) .

Proof. Wr i t e G = G / Z ( G ) . It suffices to show that [AT~, M ] = 1, and for this purpose, it is enough to check that x centralizes K for a l l elements x i n some generating set for M . It suffices, therefore, to observe that [ k , x ] e Z ( G ) for a l l k <E K , where x lies i n a conjugacy class of size 1 or n 2 in G. O f course, i f x e Z ( G ) , there is nothing to prove, and if x lies i n a class of size n 2 , then by L e m m a 4.17, the element [ k , x ] lies i n a smaller class, and hence [ k , x ] e Z ( G ) , as required. •

P r o o f of T h e o r e m 4.15. Let K < G be self-centralizing, and write M = M ( G ) . B y Corol la ry 4.18, we have [AT, M ] C Z ( G ) , and thus [AT, M , M ] = 1 and [ M , A T , M ] = 1. B y the three-subgroups lemma, [ M , M , K ] = 1, and thus [ M , M ] C C G ( A T ) = AT. T h e n M 4 = [Af, M , Af, Af] C [ K , M , M ] = 1, and it follows that A f is nilpotent w i t h class at most 3. •

We clearly cannot drop the assumption i n Theorem 4.15 that G contains a self-centralizing normal subgroup since, for example, i f G is nonabelian and simple, then M ( G ) = G is certainly not nilpotent. We do, however, obtain the following very general result.

4.19. Theorem. L e t G be a finite g r o u p , a n d w r i t e F = F(Af), w h e r e M = M ( G ) . Then t h e n i l p o t e n c e class of F i s a t most 4.

Proof. Let n be the nilpotence class of F , so that F n > 1 and F n + 1 = 1. Assuming , as we may, that n > 3, consider the characteristic subgroup F n ~ 2 of G . If this subgroup is abelian, then by Corol la ry 4.18, we have [ F n ~ 2 , M ] C Z ( G ) , and thus F n = [ F n " 2 , F , F ] C [ F n " 2 , M , F ] = 1, which is a contradiction. We conclude that F n ~ 2 is nonabelian, and therefore 1 < [ F n ' 2 , F n ' 2 } C F 2 n - 4 by Theorem 4.11. B u t F n + l = 1, and thus 2n - 4 < n + 1. Th i s yields n < 4, as wanted. •

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P r o b l e m s 4 B

4B.1 . Let G be nilpotent of class exceeding 2. Show that G has a characteristic abelian subgroup that is not central.

4B.2. Let G be nilpotent. Show that G has a characteristic subgroup K such that K D C G ( K ) and the nilpotence class of K is at most 2.

H i n t . Use P rob lem 4 B . 1 .

4B.3. Let { Z j \ j > 1} be the upper central series for an arbi t rary group G. Show that [ G \ Zj] C Z j - i for a l H > 0 and al l j . In part icular , [ G \ Z { \ = 1 for i > 0.

4B.4. Le t X , Y C G be arbi trary subgroups, and suppose that Y centralizes [ X , Y ] .

(a) Prove that Y' centralizes X .

(b) N o w assume that X < G. Show that [ X , Y) is abelian.

4B.5 . Let G be supersolvable. Show that M ( G ) is nilpotent w i t h class at most 3.

4 C

Let A act on G v i a automorphisms. Since A and G can be viewed as subgroups of the semidirect product r = G x A , the commutator [ G , A ] can be computed as a subgroup of T. B u t G < T, and so [G, A] C G , and i n fact i t is possible (and often convenient) to th ink of [G, A ) as being computed entirely w i t h i n G , thereby rendering the semidirect product irrelevant. To do this, observe that i f g e G and a G A , then [ 5 ,a] = ^ a ^ a = $"V, where we can view # a either as the result of conjugating g by a in T, or alternatively, as the result of applying a to g i n the or iginal action. The commutator [G, A ] , therefore, is the subgroup { g ~ l g a \ g <E G , o G /4), and this would make sense even i f we d i d not know about semidirect products. Nevertheless, to prove facts about [G, A ] and related objects, we w i l l appeal to semidirect products when it is convenient to do so, and i n part icular, we w i l l apply the three-subgroups lemma to subgroups of the semidirect product .

We can apply L e m m a 4.1 (in the semidirect product) to deduce that [G, A ] < G. A l so A normalizes [G, A ) in G x A , which means that [G, A ] is invariant under conjugation by elements of A , or equivalently, it is invariant under the original action of A on G . In general, i f H C G is an arbi t rary A-invar iant subgroup, we have a natural action of A on H , and sometimes i n this context, we say that H a d m i t s the act ion of A . In part icular, since

132 4. C o m m u t a t o r s

[ G , A ] admits the action of A , we can compute [ G , A , A ] , which is normal i n [ G , A ] , but not necessarily in G. (It is, of course, s u b n o r m a l i n G.) Also , [ G , A , A ] admits A , and thus we can compute [ G , A , A , A ] , and we can continue like this. A l l of these repeated commutators, of course, admit A and are subnormal i n G.

Reca l l that i f A acts v i a automorphisms on G and H C G is A-invar iant , then A permutes the set of right cosets of H in G and also the set of left cosets of H in G. In part icular, i f H < G, we have an action of A on G / H , and it is easy to see that this is an action v i a automorphisms. (This is sometimes called the i n d u c e d a c t i o n of A_on G / H . ) For g G G and a G A , we have ( g H ) a = g a H , and so if we write G = G / H we have ( g ) a = g 5 , and hence

[9, a] = g ' 1 g a = ( g - 1 ) ^ ) = ( g ) ~ H 9 ) a = foa] • It follows that [ G , A ] = [ G , A \ .

The following characterization of [ G , A ] is sometimes useful.

4.20. L e m m a . L e t A a c t v i a a u t o m o r p h i s m s o n G, w h e r e A a n d G a r e g r o u p s . T h e n [ G , A ] i s t h e u n i q u e s m a l l e s t A - i n v a r i a n t n o r m a l s u b g r o u p of G such t h a t t h e i n d u c e d a c t i o n of A o n t h e f a c t o r g r o u p i s t r i v i a l .

P r o o f . Suppose that N < G, where N is A-invariant , and write G = G/N. T h e n A acts t r iv ia l ly on G / N if and only i f 1 = [ G , A ] , which by the previous computat ion is equivalent to 1 = [ G , A ] . Th i s , of course, is the same as [ G , A ] C N , and the result is now immediate. •

4 .21. C o r o l l a r y . L e t A a c t v i a a u t o m o r p h i s m s o n G, w h e r e A a n d G a r e g r o u p s , a n d l e t H Q G . T h e n t h e f o l l o w i n g a r e e q u i v a l e n t .

(a) E v e r y r i g h t coset of H i n G i s a n A - i n v a r i a n t subset.

(b) E v e r y left coset of H i n G i s a n A - i n v a r i a n t subset.

(c) [ G , A ] C H .

I n p a r t i c u l a r , [ G , A] i s t h e u n i q u e s m a l l e s t s u b g r o u p of G w i t h t h e p r o p e r t y t h a t a l l of i t s r i g h t cosets a r e A - i n v a r i a n t , a n d s i m i l a r l y if " l e f t " r e p l a c e s " r i g h t " .

P r o o f . F i r s t , observe that i f X is a subset of G and we write X ~ l = { x ' 1 \ x G X } , then the map X ^ X ~ l takes the right cosets of H to the left cosets of H and v i c e v e r s a . It follows that an automorphism of G fixes a l l right cosets of H i f and only i f it fixes a l l left cosets of H . We see, therefore, that (a) and (b) are equivalent.

Now assume (b), so that every left coset of H is A-invar iant . T h e n for g G G and a G A , we have g a G ( g H ) a = g H , and so [g, a] = g ' 1 g a G H and (c) follows.

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Fina l ly , assume (c), so that [ G , A ] C H . E a c h right coset of H i n G , therefore, is a union of right cosets of [G, A ] i n G . B u t a l l cosets of [G, A] i n G are A-invar iant by L e m m a 4.20, and so unions of such cosets are also A-invar iant , and (a) follows. •

Reca l l that an action of a group A on a set ft is faithful i f the identi ty of A is the only element that fixes every point . In general, the kernel of an arbi t rary action of a group A is the (necessarily normal) subgroup B of A consisting of the elements of A that act t r iv ia l ly . In this si tuation, a l l elements of each coset B a of B i n A mduce the same permutat ion on ft, and so we have a well denned action o f A = A / B on ft, where a - a = a - a . T h i s action of A / B is clearly faithful.

Now suppose that A acts on G v i a automorphisms. Work ing i n the semidirect product G x > A , we see that the kernel of the action of G on A is C A { G ) . (As we remarked previously, it is common to write C A ( G ) to denote the kernel of the action of A on G even if we do not expl ic i t ly construct the semidirect product.) Th i s kernel is the largest subgroup B C A such that [ G , B ] = 1. If B = C A { G ) , then B < A , and wr i t ing A = A / B as before, we see that the natural action of A on G is given by g u = g a for g <E G and a E A , and it follows that a subgroup H C G is A- invar iant if and only if it is A-invar iant . (In other words H admits A if and only i f it admits A . ) Also , [ g , a \ = [g, a] for g € G and a G A , and so if H admits A and A , we have [ H , A ] = [ H , A \ .

W h a t can we say about the structure of a group A i f we know that it acts v i a automorphisms on some group G in such a way that [ G , A , . . . , A ] = 1, where there are (say) m copies of A i n this commutator? T h e answer is that we can say precisely nothing. T h i s is because an arbi t rary group A can act t r iv ia l ly on G , i n which case [G, A ] = 1. M o r e generally, i f B is the kernel of the action of A on G , then B and G cannot "see" each other, and so the fact that [ G , A , . . . , A ] = 1 tells us nothing about B . To draw a conclusion about the structure of A from some assumption about how G sees the action of A , we must therefore assume that the action is faithful.

To facilitate the discussion, we introduce some (nonstandard) notat ion. We write [ G , A , . . . , A ] m to denote the commutator w i t h exactly m copies of A . T h e n [ G , A , . . . , A ] i is s imply [G, A ] and i n general [[G, A ] , A , . . . , A ] m = [ G , A , . . . , A ] m + 1 .

4.22. Theorem. L e t A a n d G be g r o u p s . Suppose t h a t A acts f a i t h f u l l y o n G v i a a u t o m o r p h i s m s , a n d assume t h a t [ G , A , . . . , A ] m = 1. T h e n A i s s o l v a b l e , a n d i t s d e r i v e d l e n g t h i s a t most m - 1 .

Proof. We begin by reformulating the result i n such a way that we need not assume that the action is faithful. In the general case, we cannot hope

134 4. C o m m u t a t o r s

t o prove that A is solvable, but we show by induct ion on m that A ^ m ~ ^ C C A ( G ) , where we recall that C A ( G ) is the kernel of the action. In the original s i tuation, where A acts faithfully, we have C A { G ) = 1, and it w i l l follow that A ^ - 1 ) = 1, and so A is solvable w i t h derived length at most m - 1, as required.

If m = 1, then [ G , A ] = 1, and thus A = C G ( A ) and there is nothing more to prove. Assume then, that m > 2, and let H = [ G , A ] , so that H admits A and [ H , A , . . . , A ] m _ i = [ G , A , . . . , A ] m = 1. B y the inductive hypothesis applied to the action of A on H , therefore, we have A ^ " 2 ) C C A ( H ) , and thus 1 = [ H , A ^ m - ^ ] = [ G , A , A ^ m ~ \ Since A D A ( m " 2 ) , it follows that [ G , A ^ m ~ 2 \ A ^ m ~ 2 ^ ] = 1, and thus also [ A ^ m ~ 2 \ G, A ( m " 2 ) ] = 1 since interchanging the first two entries in a mult iple commutator of subgroups has no effect. The three-subgroups lemma now yields

[ A ^ m - 2 \ A ^ m - 2 \ G } = l ,

and since by definition, j ^ m - l ) _ ^(m-2) ^(m-2)j

we have [ A ^ m ~ l \ G \ = 1. Thus A ^ m ' 1 ^ C C A ( G ) , as wanted. •

W h e n m > 2, the step i n the proof of Theorem 4.22 where A is replaced by A ( m " 2 ) throws away information, and this suggests that perhaps more is true than we proved. In fact more is true: P . H a l l showed that under the hypotheses of our theorem, A is actually nilpotent, and not merely solvable. A l so , the nilpotence class of A is at most m { m - l ) / 2 . Reca l l that by Coro l la ry 4.13, the derived length of a nilpotent group is logari thmical ly bounded i n terms of its nilpotence class, and therefore it follows by Hal l ' s theorem that the linear bound on the derived length of A that we established i n Theorem 4.22 is too weak; it is of the "wrong" order of magnitude.

If m = 2, then A ^ m ~ 2 ^ = A ^ = A , and we lose nothing i n the replacement step of the previous proof. In this case, Ha l l ' s result coincides w i t h Theorem 4.22 since a nontr iv ia l group is solvable w i t h derived length 1 if and only i f it is abelian, which is exactly when it is nilpotent w i t h class 1. We list this case as a separate corollary.

4.23. Corol lary. L e t A a n d G be g r o u p s . Suppose t h a t A acts f a i t h f u l l y o n G v i a a u t o m o r p h i s m s a n d t h a t [ G , A , A] = 1. T h e n A i s a b e l i a n .

Proof. T h i s is just the case m = 2 of Theorem 4.22. •

W i t h o u t using addi t ional machinery, it seems to be difficult to establish Hal l ' s upper bound on the nilpotence class of A . B u t it is not too hard to show that A is nilpotent i n the si tuat ion we have been discussing, and the proof offers more opportunities for using the three-subgroups lemma.

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4.24. Theorem. L e t A a n d G be finite g r o u p s . Suppose t h a t A acts f a i t h f u l l y v i a a u t o m o r p h i s m s o n G, a n d assume t h a t [ G , A , . . . , A ] — 1. T h e n A i s n i l p o t e n t .

Proof. A s before, we drop the assumption that A acts faithfully. We prove by induct ion on \ G \ that i n general, the last term A°° of the lower central series of A acts t r iv ia l ly on G. Once this is established, we see that if the action of A is faithful, then A°° = 1, and hence A is nilpotent.

Since we can assume that G > 1 and we know that [ G , A , . . . , A ] = 1, it follows that [ G , A ] < G. A l so , [ G , A ] admits the action of A and [ [ G , A ] , A , . . . , A ] = 1. B y the inductive hypothesis, therefore, A°° acts t r iv ia l ly on [ G , A ] , and hence [ G , A , A°°] = 1, or equivalently, [ A , G, A 0 0 } = 1. O u r goal w i l l be to show that [ G , A°°,A] = 1 since, i f we can accomplish that, the three-subgroups lemma w i l l y ie ld [A°°,A,G] = 1. B u t [A°°,A] = A°° because A°° is the last term of the lower central series of A , and it w i l l follow that [A°°, G] = 1. The proof w i l l then be complete.

Our strategy for showing that [G,A°°,A] = 1 is to find a nontr iv ia l normal subgroup C of G on which A acts t r iv ia l ly . Once we have found C , we can apply the inductive hypothesis to the induced action of A on G = G / C . (This is va l id since we have [ G , A , . . . , A ] = [ G , A , . . . , A ] = 1.) B y the inductive hypothesis, 1 = [G,A°°] = [G,A°°], and so [G,A°°] C C. Since we are assuming that A acts t r iv ia l ly on C, this yields [ G , A°°,A} = 1, which is what we want.

We can assume that [ G , > 1, or else there is nothing to prove. Take C = C l G i A o o ] ( A ) , so that A certainly acts t r iv i a l ly on C, as required. To see that C > 1, observe that [ [ G , A°°],A, . . . , A ] = 1 , and thus if we start w i t h the nontr iv ia l group [ G , A 0 0 } and repeatedly compute commutators w i t h A , we get a sequence of subgroups of [ G , A°°] te rminat ing at the identity. The last nonidenti ty term i n this sequence, therefore, is centralized by A , and this shows that C > 1, as required.

To complete the proof now, it suffices to show that C < G, and our argument for this involves two further applications of the three-subgroups lemma. F i r s t , we argue that [G,A°°] centralizes [ G , A } . We have already mentioned that A°° acts t r iv ia l ly on [ G , A ] , so that [[G,A\,A°°} = 1, and thus we have [[G,A],A°°,G] = 1. A l s o , [ G , A ] < G, and thus [ G , [ G , A } } C [G, A ] , and we have [G, [G,A],A°°] C [[G, A ] , A°°] = 1. N o w we apply the three-subgroups lemma to deduce that [A°°, G , [G, A}] = 1, and thus [G, A] centralizes [A°°, G] = [G, A 0 0 ) , as claimed. In part icular , since C C [G, A 0 0 } , we have [ G , A , C ] = 1. A l so , of course, [ A , C , G ] = 1 since A centralizes C. B y the three-subgroups lemma, once again, we have [G, G , A ] = 1, and so A centralizes [G, G] .

136 4. C o m m u t a t o r s

Reca l l now that C C [G,A°°] < G, and thus [ C , G ] C [G,A°°]. We just showed, however, that A centralizes [ C , G ] , and thus [ C , G ] C C [ G i X = o ] ( A ) = C , where the equality holds by the definition of C. We have now shown that [ C , G] C C , or equivalently, that C < G, as wanted. The proof is now complete. •

Let us continue to assume that [ G , A , . . . , A ] = 1 for some unspecified number of copies of A . We have seen that i f the action is faithful, then A must be nilpotent. It is impossible, however, to draw any conclusions about the structure of G since, for example, it could be that A is t r iv ia l . B u t al though we can say nothing about G itself, we shall see that the subgroup [ G , A ] is under control. We begin w i t h a special case.

4.25. L e m m a . L e t A a n d G be g r o u p s . Suppose t h a t A acts o n G v i a a u t o m o r p h i s m s , a n d assume t h a t [ G , A , A ] = 1. T h e n [ G , A ] i s a b e l i a n .

Proof. Since [ [ G , A } , A ] = 1, we certainly have [ [ G , A ] , A , G ] = 1. Also , [ G , A] < G, and thus [ G , [ G , A ] , A ] C [[G, A ] , A] = 1. B y the three-subgroups lemma, we have

1 = [ A , G , [ G , A \ ] = [ [ G , A ] , [ G , A ] ] = [ G , A } ' ,

and so [ G , A ] is abelian. •

4.26. Theorem. L e t A be a p - g r o u p t h a t acts v i a a u t o m o r p h i s m s o n a finite g r o u p G, a n d suppose t h a t [ G , A , . . . , A ] = 1. T h e n [ G , A ) i s a p - g r o u p .

Proof. We proceed by induct ion on \ G \ . We can certainly assume that G > 1, and thus since [ G , A , . . . , A ] = 1, it follows that [ G , A ] < G. For convenience, write N = [ G , A ] , and observe that N < G and that N admits the action of A . Since N < G , the inductive hypothesis applies in N , and we deduce that [ N , A ] is a p-group. Also , since [TV, A ] < N , we have [AT, A ] C U, where U = O p ( N ) is characteristic in N < G, and thus U < G .

N o w consider the induced action of A on G = G / U . Since [TV, A ] C U, we see that A acts t r iv ia l ly on N . A l so , [G, A ] = [ G , A \ = N , and thus A acts t r iv i a l ly on G/N. We Jiave J G , A , A] = [ N , A ] = 1, and thus by L e m m a 4.25, we conclude that N = [ G , A ] is abelian. B u t O p ( N ) is t r iv ia l , and since N is abelian, it follows that N is a p'-group.

Since A is a p-group and N is a normal p'-subgroup of G, we can apply Corol la ry 3.28 to the action of A on G. Tha t corollary tells us that "fixed points come from fixed points" ,_and_ since N is contained in the fixed-point subgroup and every element of G / N is A-f ixed, we conclude that the whole group G consists of fixed points. T h e n 1 = [ G , A ] = [ G , A ] , and hence [ G , A ] C U. Since U is a p-group, we are done. •

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In the si tuat ion of Theorem 4.26, we can conclude that [G, A] is nilpotent even without assuming that A is a p-group.

4.27. Theorem. L e t A a n d G be finite g r o u p s . Suppose t h a t A acts v i a a u t o m o r p h i s m s o n G, a n d assume t h a t [ G , A , . . . , A ] = 1. T h e n [ G , A ] i s n i l p o t e n t .

Proof. We proceed by induct ion on \ A \ . (Of course, i f A is t r iv i a l then [ G , A ] = 1 and there is nothing to prove.) Suppose that B is an arbi t rary proper subgroup of A . T h e n [ G , B , . . . , B ] C [ G , A , . . . , A ] = 1, and so by the inductive hypothesis, [ G , B ] is nilpotent. Since [G, B ] < G , we have [G, B ) C F ( G ) , the F i t t i n g subgroup, and thus B acts t r iv i a l ly on G / F ( G ) .

We now know that every proper subgroup of A is contained i n the kernel of the induced action of A on G / F ( G ) , and so i f A is generated by its proper subgroups, then this kernel must be a l l of A . In this case, [ G , A ] C F ( G ) , and hence [G, A ] is nilpotent, as wanted. We can assume, therefore, that A is not generated by its proper subgroups. Eve ry finite group, however, is generated by its Sylow subgroups, and thus some Sylow subgroup of A is not proper. In other words, A is a p-group for some prime p , and hence [G, A ] is a p-group by Theorem 4.26. In part icular, [G, A ] is nilpotent. •

P r o b l e m s 4 C

4 C . 1 . Le t A act on G v i a automorphisms, and suppose that

1 = H 0 C H i C • • • C H m = G

is a chain of A-invar iant subgroups such that each right coset of H ^ i n H i is A-invar iant for a l l i w i t h 0 < i < m. In the older li terature, this s i tuat ion was described by saying that A stabilizes the chain { H i } . If A stabilizes a chain of subgroups i n G , show that i t stabilizes a chain of subnormal subgroups.

4C.2 . Assume that A acts faithfully on G v i a automorphisms, and assume that A stabilizes a chain { H i | 0 < i < m } of n o r m a l subgroups of G . Show that A is nilpotent w i t h class at most m - 1.

4C.3 . Le t A act v i a automorphisms on G , and suppose that N < G and that A acts t r iv ia l ly on N . Show that N centralizes [G, A ] ,

4 D

We continue our study of commutators that arise when A acts v i a automorphisms on G , but now we focus on coprime actions, where ( | G | , \ A \ ) = 1.

138 4. C o m m u t a t o r s

The following ul t imate ly relies on Glauberman's lemma, and that is the reason for the solvabil i ty assumption, which, as usual, is not really necessary i f one is wi l l ing to appeal to the Fei t -Thompson theorem.

4.28. L e m m a . L e t A a n d G be finite g r o u p s . L e t A a c t v i a a u t o m o r p h i s m s o n G, a n d suppose t h a t ( \ G \ , \ A \ ) = 1 a n d t h a t one of A o r G i s s o l v a b l e . T h e n G = C G ( A ) [ G , A \ .

Proof. Wr i t e G = G / [ G , A ] . Since fixed points come from fixed points in coprime actions (Corol lary 3.28) we have C U ( A ) = C G ( A ) . B u t A acts t r iv ia l ly on G / [ G , A ] , and so the left side of this equation is the whole group G. Thus

C G ( A ) [ G , A ] = C G ( A ) = G,

and it follows that [ G , A ] C G ( A ) = G by the correspondence theorem. •

To demystify this proof somewhat, we observe that what is going on here is the following. E a c h coset of [ G , A] i n G is A-invariant , and so by a fairly standard Glauberman's l emma argument, each such coset contains an A-f ixed element.

The following result is used frequently.

4.29. L e m m a . L e t A a c t v i a a u t o m o r p h i s m s o n G, w h e r e A a n d G a r e finite g r o u p s , a n d suppose t h a t ( \ G \ , \ A \ ) = 1. T h e n [ G , A , A ] = [ G , A \ .

Proof. O f course [ G , A , A] C [ G , A ] , and so it suffices to prove the reverse containment. For this purpose, we show that [g, a] G [ G , A , A ] for a l l g G G and a G A . The result w i l l then follow since [ G , A ] is generated by these commutators [ g , a ] .

Suppose first that A is solvable. L e m m a 4.28 yields G = C G ( A ) [ G , A ] , and so i f g G G, we can write g = cx, where c G C G ( A ) and x G [ G , A } . Now let a G A , and observe that [c, a] = 1 . T h e n

[g, a] = [ c x , a] = [c, a ] x [ x , a] = [ x , a] G [ G , A , A] ,

as wanted. Th i s proves that the result is true i f A is solvable.

Now i n the general case, again let g G G and a G A . T h e n

[g, a] G [ G , (a)} = [ G , ( a ) , (a)} C [ G , A , A]

since the cyclic group ( a ) is certainly solvable. Th i s completes the proof. •

We can apply this to the si tuation that we studied i n the previous section.

4.30. Corol lary. L e t A a c t f a i t h f u l l y v i a a u t o m o r p h i s m s o n G, w h e r e A a n d G a r e finite g r o u p s , a n d assume t h a t [ G , A , . . . , A ] = 1. T h e n every p r i m e d i v i s o r of \ A \ a l s o d i v i d e s \ G \ .

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Proof. Let P € S y l p ( A ) where p is a pr ime that does n o t divide | G | . T h e n [ G , P] = [ G , P , . . . , P ] = 1, where the first equality holds by repeated appl i cat ion of L e m m a 4.29. Since the action of A is faithful and P acts t r ivia l ly , we deduce that P = 1, and thus p does not divide \ A \ . •

Next , we prove the so called P x Q l emma of Thompson , and we show how it is relevant to the study of local subgroups.

4.31. T h e o r e m (Thompson) . L e t A be a finite g r o u p t h a t acts v i a a u t o m o r p h i s m s o n a p - g r o u p G, a n d suppose t h a t A = P x Q i s a n i n t e r n a l d i r e c t p r o d u c t of a p - g r o u p P a n d a p 1 - g r o u p Q. Suppose t h a t Q fixes every element of G t h a t P fixes. Then Q acts t r i v i a l l y o n G.

Before we begin the proof of Thompson 's theorem, we establish some easy facts about p-groups acting on p-groups.

4.32. L e m m a . L e t P be a p - g r o u p t h a t acts v i a a u t o m o r p h i s m s o n a n o n -t r i v i a l p - g r o u p G. Then [ G , P ] < G a n d C G ( P ) > 1.

Proof. The semidirect product r = G x P is a p-group, and hence it is nilpo¬tent. It follows that some repeated commutator of the form [T, T , . . . , T] is t r iv ia l , and thus we have [ G , P , . . . , P ] = 1. Successive commutat ion w i t h P , therefore, yields a descending sequence of subgroups start ing w i t h G > 1 and ending wi th 1. Thus [ G , P ] < G, as wanted. Also , i f H is the last nonidenti ty group i n this sequence, then [ H , P ] = 1, and so 1 < H C C G { P ) . •

Assuming that the p-group G is nont r iv ia l i n the s i tuat ion of the P x Q lemma, it follows by L e m m a 4.32 that C G ( P ) > 1. B y hypothesis, however, C G ( P ) C C G ( Q ) , and therefore Q has at least some nontr iv ia l fixed points i n G. W h a t we need, however, is that Q fixes every element of G, and the proof of this, while not hard, is a l i t t le more subtle; i t gives us another opportuni ty to use the three-subgroups lemma.

P r o o f of T h e o r e m 4.31. We can assume that G > 1, and we proceed by induct ion on | G | . We have [G, P] < G by L e m m a 4.32. A l so , [ G , P ] is A¬invariant since bo th G and P are normalized by A i n the semidirect product G x > A . The hypothesis that Q centralizes every element of G centralized by P is inherited by the action of A on [G, P ] , and so the inductive hypothesis applies to this action. We conclude that Q acts t r iv i a l ly on [G, P ] , and hence [ G , P , Q ] = 1. A l so , [ P , Q , G ] = 1 since P and Q centralize each other in A . It follows by the three-subgroups lemma that [ Q , G , P ] = 1, and thus P acts t r iv ia l ly on [ Q , G ] = [ G , Q ] . B y hypothesis, Q acts t r iv ia l ly on this subgroup too, and so [G, Q , Q ] = 1. F ina l ly , since Q is a p'-group and G is

140 4. C o m m u t a t o r s

a p-group, we have [G, Q, Q] = [ G , Q] by L e m m a 4.29, and so [G, Q] = 1, as required. •

A s an applicat ion of the P x Q lemma, we prove the following "local to global" result for p-solvable groups. (Recall from Section 2C that i f p is a prime, a subgroup H C G is said to be p-local in G i f H = N G ( P ) , for some nont r iv ia l p-subgroup P of G.)

4.33. Theorem. L e t G be a finite p - s o l v a b l e g r o u p f o r some p r i m e p . T h e n O p , ( H ) C C y (G) / o r every p - l o c a l s u b g r o u p H C G .

Proof. Le t Q = C y ( P ) , where i f is p-local in G. F i r s t , we consider the case where C y (G) = 1, and where our goal, therefore, is to show that Q = l . Since H is p-local , we can write H = N G ( P ) for some p-subgroup P C G , and we observe that both P and Q are normal i n P . Since P is a p-group and Q is a p'-group, we have P n Q = 1, and thus P Q is the direct product of P and Q.

N o w let U = O p ( G ) , so that U is a p-group, and observe that P Q acts on U by conjugation since U < G. We argue that Q fixes every element of U that is fixed by P , or equivalently, that Q centralizes Q / ( P ) . To see this, observe that C V { P ) C P n N G ( P ) = P n P . B u t P n P is a normal p-subgroup of H , and Q is a normal p'-subgroup of P , and so Q centralizes U n P , and hence it centralizes C [ / ( P ) , as wanted. We conclude by the P x Q l emma that Q C C G ( P ) . B u t G is p-solvable and we are assuming that C y ( G ) = 1, and thus U contains its own centralizer i n G . (This is L e m m a 1.2.3 of H a l l and Higman , which is our L e m m a 3.21.) We now have Q Q C G ( P ) C U, and since Q is a p'-group and U is a p-group, we deduce that Q = 1, as required i n this case.

In the general si tuation, let TV = C y ( G ) and write G = G/N, so that C y ( G ) = 1. If H C G is p-local , then since \ N \ is not divisible by p, L e m m a 2.17 guarantees that H is p-local in G , and thus by the first part of the proof, Op/(IF) = 1. Now Q is a normal p'-subgroup of P , and so Q Q O p / ( P ) = 1. F ina l ly , since Q = 1, we have Q C N , and the proof is complete. •

We return now to the si tuat ion considered at the beginning of this sect ion. We had A acting coprimely on G v i a automorphisms, and either G or A is solvable, and we showed i n L e m m a 4.28 that G = C G ( A ) [ G , A ] . A s the following result of H . F i t t i n g shows, more is true i f G is abelian.

4.34. T h e o r e m (F i t t ing) . L e t A a c t v i a a u t o m o r p h i s m s o n a n a b e l i a n g r o u p G, a n d assume t h a t A a n d G a r e finite a n d t h a t ( | G | , \ A \ ) = 1. T h e n G = C G { A ) x [G, A ] .

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P r o o f . A s G is abelian, bo th factors are normal , and so it suffices to show that G = C G ( A ) [ G , A ] and that C G ( A ) n [ G , A ] = 1. O f course, G is solvable, and so we can appeal to L e m m a 4.28 to deduce the first of these two facts, and so it suffices to prove the second.

Let 9 : G -> G be the map denned by

0 ( g ) = H g b , beA

and note that since G is abelian, the order of the factors i n this product is irrelevant, and hence 9 is well defined. A l so , i f x , y e G , then ( x y ) b = x b y b

for a l l b e A , and so again using the fact that G is abelian, it is easy to see that 9 ( x y ) = 6 ( x ) 9 { y ) , and thus 9 is a homomorphism.

If g € C G ( A ) , then g b = g for a l l b e A , and thus 9 ( g ) = g \ A \ . A l so , since \ A \ is coprime to | G | , we see that g \ A \ ^ 1 i f g + 1, and thus C G ( A ) n k e r ( 9 ) = 1. To complete the proof, therefore, it is enough to show that [G, A] C ker(0), and since [ G , A] is generated by commutators of the form [g, a ] , w i th g e G and a e A , it suffices to prove that each of these commutators lies in ker(0).

If g e G and a e A , then

o(ga) = X [ g a b = e{g), beA

where the second equality holds because a b runs over A as 6 runs over A . It follows that

e([g, a}) = 9 ( g - 1 g a ) = 9 { g - l ) 9 { g a ) = 9 ( g ) - 1 9 ( g ) = 1.

Th i s completes the proof. •

A n easy applicat ion of F i t t ing ' s theorem is the following.

4.35. C o r o l l a r y . L e t A a c t o n G v i a a u t o m o r p h i s m s , w h e r e G i s a n a b e l i a n p - g r o u p , a n d assume t h a t A i s a finite p ' - g r o u p . If A fixes every element of o r d e r p i n G, t h e n A acts t r i v i a l l y o n G.

P r o o f . B y F i t t ing ' s theorem, G = C G { A ) x [ G , A ] , and by hypothesis, C G ( A ) contains every element of order p i n G . Since C G ( A ) and [G, A] have only the identity i n common, it follows that [ G , A ] contains no elements of order p . B u t [ G , A] is a p-group and every nontr iv ia l p-group contains elements of order p. We conclude that [ G , A ] = 1, and the proof is complete. •

Surprisingly, the hypothesis that G is abelian i n Coro l la ry 4.35 is unnecessary if p > 2.

142 4. C o m m u t a t o r s

4.36. Theorem. L e t A a c t o n G v i a a u t o m o r p h i s m s , w h e r e G i s a p - g r o u p w i t h p > 2 , a n d assume t h a t A i s a finite p ' - g r o u p . If A fixes every element of o r d e r p i n G, t h e n A acts t r i v i a l l y o n G.

The idea of the proof of Theorem 4.36 is quite simple. We show that if G is a m i n i m a l counterexample, then there is an abelian counterexample of the same order, and this contradicts Corol la ry 4.35. It is straightforward to show that a m i n i m a l counterexample would have nilpotence class at most 2 (and that does not require that p > 2). B u t getting from class 2 to abelian seems to require a new idea: a remarkable tr ick due to R . Baer.

4.37. L e m m a (Baer t r ick) . L e t G be a finite n i l p o t e n t g r o u p of o d d o r d e r a n d n i l p o t e n c e class a t most 2. T h e n t h e r e exists a n a d d i t i o n o p e r a t i o n x + y d e f i n e d f o r elements x,y G G such t h a t G i s a n a b e l i a n g r o u p w i t h r e s p e c t t o a d d i t i o n . A l s o , t h e f o l l o w i n g h o l d .

(a) If xy = yx f o r x,y < E G , t h e n x + y = - x y .

(b) T h e a d d i t i v e o r d e r of each element o f G i s e q u a l t o i t s m u l t i p l i c a t i v e o r d e r .

(c) E v e r y a u t o m o r p h i s m of G i s a l s o a n a u t o m o r p h i s m of t h e a d d i t i v e g r o u p ( G , + ) .

Before we begin the proof of Baer 's lemma, we introduce a bit of suggestive notat ion. If m is an integer and G is a group of order relatively prime to m, then the map x i-> x m has an inverse, namely the map x ^ x n , where n is chosen so that r a n = 1 m o d \ G \ . G iven x G G, therefore, there is a unique element y G G such that y m = x , and in fact, y is a power of x . In the case where | G | is odd and m = 2, we write y = ^/x~ to denote the unique element w i t h square x , and we observe that if H C G is any subgroup, then V h e H for every element he H . A l so , it is easy to see that V o F 1 = ( v ^ ) " 1 and that if x and y commute, then y / x y = y / x ^ y .

P r o o f of L e m m a 4.37. Let x + y = x y ^ ^ x ] . To show that x + y = y + x for a l l x,y e G, we must check that y x y ^ ~ y ] = x y ^ / \ y ~ r ] . T h i s is equivalent to y - ' x ^ y x = / r a v r a r 1 - B u t (^iSl)"1

= v l ^ / F = y / [ y , x ] , and so the identity we must establish is [ y , x ] = ( \ / { y , x \ ) 2 , which is clearly true.

Observe that if x and y commute, then [ y , x ] = 1, and since = 1, it follows that x + y = xy, and this establishes (a). In part icular, since 1 and x commute, we have 1 + x = l x = x for a l l x G G, and thus 1 is an additive identity. A l so , x and x ~ l commute, and hence x + x ~ l = xx~l = 1, and thus x ~ l is the additive inverse of x .

4 D 143

We have

x + (y + z ) = x + y z y / ^ y ] = x y z ^ y \ s ) [ y z ^ y \ , x \ .

Since we are assuming that G has class at most 2, a l l commutators in G are central, and thus the map [ - , x ] is a homomorphism. It follows that

[ y z ^ [ z ~ y ] , x] = [y, x] [ z , x] [ y / & y ] , x] ,

and we observe that the th i rd factor is t r i v i a l since [ z , y] € Z ( G ) , and hence y / [ z ~ y ] € Z ( G ) . We now have

x + (y + z) = x y z y f [ z ~ y ~ W { y , x } [ z , x ] = x y z y / [ z , y ] [ y , x ] [ z , x ] ,

where the second equality holds because [ z , y ] , commutes w i t h [ y , x ] [ z , x ] . O n the other hand,

( x + y ) + z = x y ^ x ] + z = x y ^ x \ z y J [ z , x y ^ / [ y T x \ } ,

and this can be simplified in a similar manner. (Here, we use the fact that [ z , •} is a homomorphism.) We obtain

( x + y ) + z = x y z y / [ y , x ] [ z , x ] [ z , y ] ,

which is equal to the result of our previous computat ion since the commutators commute. Th i s shows that addi t ion on G is associative, and hence ( G , +) is a group.

We have already proved (a). For (b), we use the customary notat ion for addi t ively wr i t ten groups, and we write nx to denote the sum of n copies of x , where n is a positive integer and x <E G. We argue by induct ion on n that nx = x n , which is certainly val id for n = 1. For n > 1, we have

nx = x + ( n - l ) x = x + x n ~ l = xxn~l = x n ,

where the second equality holds by the inductive hypothesis and the th i rd holds by (a), since x and x n ~ l commute in G . It follows that nx = 1 if and only if x n = 1, and since 1 is the additive identi ty (as well as the mult ipl icat ive identi ty), it follows that the addit ive and mult ipl icat ive orders of x are equal.

F ina l ly , the addi t ion in G is uniquely determined by the mul t ip l ica t ion , and thus any permutat ion of the elements of G that preserves the mul t ip l i cation w i l l also preserve the addit ion. Asser t ion (c) follows and the proof is complete. •

P r o o f o f T h e o r e m 4 .36 . We can assume that G > 1, and we proceed by induct ion on G . If H < G and H admits the act ion of A , then since every element of order p in H is fixed by A , the inductive hypothesis guarantees that A acts t r iv ia l ly on H , and hence [ H , A] = 1. In part icular, if [G, A ] < G, then since [G, A ] admits A , we have 1 = [G, A , A ] = [G, A ] , and there is

144 4. C o m m u t a t o r s

nothing further to prove i n this case. (The second equality holds, of course, by L e m m a 4.29, which applies since = 1.)

The derived subgroup G' is characteristic i n G, and hence it admits A . Also , the p-group G is certainly solvable, and thus G' < G. We deduce that [ G ' , A] = 1, and so [<?', A , G] = 1. A l so , since G' < G, we have [ G , G'\ C G', and thus [ G , G ' , A ] C [ G ' , A ] = 1. The three-subgroups lemma now yields [ A , G, G'j = 1. B u t [ A , G ] = [ G , A ] = G, and so [ G , G'] = 1 and G' C Z ( G ) . In other words, the nilpotence class of G is at most 2, and we can apply the Baer tr ick to construct the abelian group structure ( G , +) on the underlying set of G.

N o w an element a G A induces an automorphism of G, and we conclude by L e m m a 4.37(c) that this permutat ion of the elements of G is also an automorphism of ( G , +). In other words, A acts v i a automorphisms on the abelian group ( G , +) . B y L e m m a 4.37(b), the elements of order p in ( G , + ) are exactly the elements of order p i n G, and by hypothesis, they are a l l fixed by A . B y Coro l la ry 4.35, therefore, A acts t r iv i a l ly on ( G , + ) , and hence its action on G is t r iv i a l . •

A s imilar argument allows us to prove a stronger version of the T h o m p son P x Q l emma i n the case where p is odd. Reca l l that in Theorem 4.31 we had a group A = P Q acting on a p-group G, where P is a p-subgroup and Q is a p'-subgroup of A , and both are normal i n A . ( A n d we assumed that C G ( P ) C C G ( Q ) . ) We now drop the assumption that P < A .

4 .38 . T h e o r e m . L e t A be a finite g r o u p t h a t acts v i a a u t o m o r p h i s m s o n a p - g r o u p G, w h e r e p > 2 , a n d l e t P a n d Q be s u b g r o u p s of A . Suppose t h a t P i s a p - g r o u p a n d t h a t Q < A a n d Q i s a p ' - g r o u p . Assume a l s o t h a t Q fixes every element of G fixed by P . T h e n Q acts t r i v i a l l y o n G.

P r o o f . F i r s t , observe that both G and Q are normalized by A i n the semidi-rect product G x > A . Therefore [ G , Q] is A-invariant , and it admits the action of A .

N o w assume that G is abelian, so that by F i t t ing ' s theorem, G = C G { Q ) x [ G , Q ] . If [ G , Q ] > 1, then by L e m m a 4.32 applied to the act ion of P on [ G , Q ] , we have C [ G i 0 ] ( P ) > I . Thus P has nontr iv ia l fixed points in [ G , Q ] , and by hypothesis, these P-f ixed points are also Q-fixed points. Th i s is a contradict ion, however, since [ G , Q ] n C G ( Q ) = 1, and it follows that [ G , Q] = 1 in this case, as wanted.

Now for the general case, we can assume that G > 1. We proceed by induct ion on G, observing that if P < G and H admits the action of A , then the inductive hypothesis yields that [ H , Q] = 1. In particular, since [ G , Q] admits the action of A , we see that if [ G , Q] < G, then 1 =

P r o b l e m s 4 D 145

[G, Q, Q] = [ G , Q ] , and we are done. We can assume, therefore, that [ G , Q] = G. N o w G' < G admits the action of A , and hence [ G ' , Q] = 1, and we have [ G ' , Q , G ] = 1 and [ G , G ' , Q ] C [ G ' , Q ] = 1. Thus [ Q , G , G ' \ = 1, and since [ Q , G ] = [ G , Q ] = G, we conclude that G' C Z ( G ) . Thus G has nilpotence class at most 2, and we can apply the Baer t r ick.

Now A acts on the abelian group ( G , +) and every element fixed by P is fixed by Q. B u t we have already proved the theorem for abelian groups, and so we deduce that Q acts t r iv ia l ly on ( G , +), and hence also on G . •

P r o b l e m s 4 D

4D.1 . Let A act v i a automorphisms on G , and assume that N < G admits A and that N D C G ( 7 V ) . Assume that (|AT|, = 1.

(a) If A acts t r iv ia l ly on N , show that its act ion on G is t r iv i a l .

(b) If A acts faithfully on G , show that its act ion on N is faithful.

H i n t . For (a) show that [G, A ] C N . Consider C r ( N ) , where T = G x A

N o t e . If G is solvable and AT = F ( G ) , then N D C G ( N ) . A l so , if G is is p-solvable w i t h C y ( G ) = 1 and N = O p ( G ) , then TV D C G ( 7 V ) . Thus in either of these cases, if A acts faithfully and coprimely on G , then A acts faithfully on N .

4D.2. Let G be nilpotent w i t h class at most 2, and assume | G | is odd. Fol lowing the proof of L e m m a 4.36, construct an additive structure on G , and (as usual for addit ively wri t ten groups) write x - y to denote the sum of x and the additive inverse of y . Show that xy - yx = [ x , y } .

4D.3. Let A act v i a automorphisms on G , where ( | G | , \ A \ ) = 1, and assume that one of A or G is solvable. Suppose that A acts t r iv i a l ly on every A¬invariant proper subgroup of G , but that the action of A on G is nontr iv ia l . Prove a l l of the following.

(a) G is a p-group.

(b) G' C Z ( G ) .

(c) N o A-invariant subgroup H exists w i t h G' < H < G.

(d) G I G ' is elementary abelian.

(e) G' is elementary abelian (and possibly t r iv ia l ) .

(f) If p > 2, then = 1 for a l l x G G .

(g) If p = 2, then x 4 = 1 for al l x e G .

H i n t . For (c), apply F i t t ing ' s theorem to the action of A on G/G'.

146 4. C o m m u t a t o r s

4D.4. Let A act v i a automorphisms on G , where G is a 2-group and A has odd order. Show that i f A fixes every element x G G such that x 4 = 1, then the action of A on G is t r iv ia l .

4D.5 . Let A be solvable w i t h derived length n , and suppose that A acts faithfully v i a automorphisms on an abelian group B , where ( \ B \ , \ A ^ n ~ ^ \ ) = 1. Show that G = B x > A has derived length n + 1. Deduce that there exist solvable groups w i t h arbi t rar i ly large derived lengths.

H i n t . For the second part, suppose A has derived length n and let C be cyclic of prime order p not d iv id ing \ A < - n - % Show that if G = G U is the regular wreath product, then G has derived length n + 1.

4D.6. Let # : G -> G be the map that appeared in our proof of F i t ting's theorem (Theorem 4.34.) Show that (in the notat ion of that theorem) 0 ( G ) = C G ( A ) and ker(0) = [G, A ] .

4D.7. Let G be p-solvable, and suppose that a Sylow p-subgroup of G is cyclic . Let K C G be a p'-subgroup, and suppose that p divides | N G ( A " ) | . Show that K C C y ( G ) .

H i n t . Induct on | G | . If C y ( G ) > 1, apply the inductive hypothesis to G = G / C y (G) . Otherwise, show that G has a normal Sylow p-subgroup.

C h a p t e r 5

Transfer

5 A

We seek results that can be used to prove that a group is not simple. In other words, we want to find techniques that can produce normal subgroups of a group G , or equivalently, homomorphisms 9 from G into some group H , and we want to find hypotheses sufficient to guarantee that k e r ( 9 ) is neither t r iv i a l nor is the whole group G. A s we shall see, the "transfer" homomorphism, which we discuss i n this chapter, is a powerful tool for proving such "good" theorems.

W h a t groups H are available to serve as targets for homomorphisms from G ? One possibility, of course, is to take H to be a symmetric group Sn. (Recal l that this was how we proved Theorem 1.1 and its corollaries.) Another useful and well studied si tuat ion is where H is the group of in -vertible n x n matrices over a field F . (Recall that this group is the general linear group, denoted G L ( n , F ) . ) A homomorphism from G into G L ( n , F ) is called an F-representation of G , and representation theory has proven to be a fertile source of good theorems, especially when the field F is taken to be the complex numbers C . We digress to mention that an F-representation 9 of G determines a function x • G - > F , called the character associated w i t h 9, which is defined by setting x { s ) to be the trace of the mat r ix 9 ( g ) . Perhaps surprisingly, it turns out that in the case where F = C , it suffices to study these characters i n order to determine most of the interesting properties of the corresponding representations. In part icular , it is not terr ibly difficult to prove that ker(0) = { g e G | x { g ) = x ( l ) } , and so characters can be used as proxies for C-representations i n order to prove good theorems. Indeed, it is character theory that provided W . Burnside w i t h the key to his

147

148 5. T r a n s f e r

pY-theorem, which asserts that a group whose order has exactly two prime divisors cannot be simple.

We can also choose subgroups of G to serve as targets for homomor-phisms denned on G. Tha t is approximately what the transfer is: a certain homomorphism from G into an arbitrary subgroup H . A s we shall see, however, technical considerations necessitate that the target group for the transfer homomorphism should be abelian, and so instead of a subgroup H , we use its commutator factor group H / H ' as the target.

Let G be an arbi t rary (not necessarily finite) group, and let H C G be a subgroup of finite index. In this section, we define the transfer map v : G -»• H / H ' , and we prove that it is a homomorphism. We start by fixing a set of representatives for the right cosets of H i n G. Such a set, called a r i g h t t r a n s v e r s a l for H in G, is s imply a set of elements of G chosen so that each right coset of H contains exactly one member of the set. If T is a right transversal for H in G, therefore, the right cosets H t for t G T are distinct, and they are a l l of the right cosets of H i n G. We can thus use the elements of T as labels for the right cosets of H , and in particular, \T\ = \ G : H \ .

Reca l l that G acts by right mul t ip l ica t ion on the set of right cosets of H , and so we can write ( H x ) - g = H x g for each right coset H x of H in G and each element g G G. Since the elements of the right transversal T can be viewed as labels for the right cosets of H , the action of G on the set { H t | t G T } of right cosets uniquely determines an action of G on T, and we define t - g accordingly. In part icular, ( H t ) - g = H ( t - g ) , or in other words, if t G T and g G G, then t - g is the unique element of T that lies in the coset ( H t ) - g = H t g . Since our "dot" action of G on T is really just the action of G on the right cosets of H , we see, for example, that the stabilizer i n G of an element t G T is exactly the stabilizer of the coset H t , which we know is H l . Mos t importantly, since we know that right mul t ip l ica t ion on right cosets really is an action, it follows that the dot act ion of G on T is a true action, and thus, for example, ( t - x ) - y = t - ( x y ) for t G T and x,y G G.

N o w suppose that t G T and g G G. T h e n t g G H t g = H ( t - g ) , and hence there is a unique element he H such that t g = h ( t - g ) . Since t g { t - g ) ~ l = h , it follows that for a l l elements t G T and g G G, we have t g { t - g ) ~ l G H . G i v e n g G G, we would like to mul t ip ly a l l of the elements t g i t - g ) - 1 for i e T t o obtain an element of H depending on g . There is an obvious difficulty here, however, since i f H is nonabelian, the order in which this mul t ip l ica t ion is carried out may matter. In order to resolve fhis issue, we suppose that some specific (but arbitrary) ordering of the elements of T has

5 A 149

been declared, and we let V : G - > H be the function defined by the formula

t e T

where i n comput ing the product, the elements t E T are taken in the specified order. A l t h o u g h this nomenclature is not standard, we refer to any map V : G —> H constructed in this way as a pretransfer map. There are usually many different pretransfer maps from G to each subgroup H because, in general, the map depends bo th on the choice of the right transversal T and on the ordering of T.

G i v e n a pretransfer map V : G - > H , where H C G is an arbi t rary subgroup, we can define a new map v : G -> H / H " by composing V w i t h the canonical homomorphism h -> 7i from H t o H = H / H ' . In other words

ufo) = V 7 ^ ,

and we observe that since H / H ' is abelian and each factor t g i t - g ) - 1 in the product defining V { g ) lies i n If , it follows that the map v is independent of the specific ordering on T that we used to define V. A s we shall see presently, v is also independent of the choice of the right transversal T, and so for each subgroup H C G, the map v is unambiguously defined. It is the transfer map from G to H / H ' , which is often inaccurately referred to as the transfer from G to H . (Perhaps we should mention that it is customary to use the letter "v" for the transfer because i n German , "transfer" is "Verlagerung").

Before we present any theorems or proofs, we introduce some convenient notat ion. G iven two elements x and y in our subgroup H , we can assert that x and y have the same image in H / H ' by wr i t ing H ' x = H ' y or x = y . Unfortunately, if either x or y is some complicated expression such as the product defining V { g ) above, the coset notat ion is somewhat unwieldy, and overbars are even worse. For that reason, we introduce the alternative notat ion x = y to mean the same thing: that x = y . (It would be more correct to write x = y m o d H ' , but we w i l l usually suppress explicit mention of the modulus when no confusion is l ikely to arise.)

5.1. Theorem. L e t G be a g r o u p , a n d suppose t h a t H i s a s u b g r o u p of finite i n d e x . T h e n t h e t r a n s f e r m a p v . G - > H / H ' i s i n d e p e n d e n t of t h e c h o i c e of t h e r i g h t t r a n s v e r s a l used t o d e f i n e i t .

Proof. Let S and T be two right transversals for H in G, and observe that for each element t E T, there is a unique element st E S that lies in the same right coset of H . We can thus write st = h t t for some uniquely determined element h t E H , and we have

S = {St I t E T } = { h t t \ t E T } .

150 5. T r a n s f e r

Now let s G S be arbitrary, and let g G G. T h e n s = st for some element i 6 T and s - g is the unique element of S in the coset H s g = H t g = H { t - g ) . It follows that s - g = st.g = h t . g ( t - g ) . We thus have

s g ( s - g ) - 1 = h t t g { h t . g { t - g ) ) - 1 = h t { t g { t - g ) - l ) K ; } g .

W r i t i n g Vs and VT to denote pretransfers constructed from S and T , we have

V s ( g ) = J] ^ ( s - g ) - 1 = [J h t i t g i t - g ) - 1 ) ^ . ]

Here, the first congruence is not necessarily an equality because we made no assumption about how the arbi trary orderings on S and T are related. The second congruence is va l id because we are working modulo H 1 , and so we are free to rearrange factors that lie in H . F ina l ly , t - g runs over a l l of T as t does, and thus the first and th i rd factors on the right cancel (modulo H ' ) , and we conclude that V s ( g ) = V T { g ) , as required. •

A l t h o u g h it may be reassuring to know that the transfer map from G to H is unambiguously defined, as we have just shown, the really v i t a l fact about the transfer is that it is a homomorphism, and the proof of that fact does not rely on Theorem 5.1.

5.2. Theorem. L e t G be a g r o u p , a n d suppose t h a t H i s a s u b g r o u p of finite i n d e x . T h e n t h e t r a n s f e r m a p v : G - > H / H ' i s a h o m o m o r p h i s m .

Proof. Let T be a right transversal for H i n G, and let V be a pretransfer associated w i t h T. For t G T , we have t - { x y ) = ( t - x ) - y , and thus

t i x y ^ H x y ) ) - 1 = { t x { t - x ) - l ) { { t - x ) y { { t - x ) - y y l ) .

Since bo th factors on the right lie in H , we have

V ( x y ) = l [ t ( x y ) ( t - ( x y ) ) - 1 = JJ t ^ i - x ) " 1 n ^xM( i -x ) -y ) - 1

= V { x ) V { y ) ,

where the last congruence holds because t-x runs over a l l of T as i does. It follows that v ( x y ) = v ( x ) v ( y ) , and the proof is complete. •

A l t h o u g h Theorem 5.2 is not very deep, it enables us to prove the following, which is our first appl icat ion of transfer theory.

5.3. Theorem. Suppose t h a t G i s a finite g r o u p , a n d l e t p be a p r i m e d i v i s o r of \ G ' n Z ( G ) | . T h e n a S y l o w p - s u b g r o u p of G i s n o n a b e l i a n .

5 A 151

Proof. Let P G S y l p ( G ) , and assume that P is abelian. Let T be a right transversal for P i n G , and consider the transfer homomorphism v : G -> P . Since P is abelian, we have u = V, where V is the pretransfer computed using T.

N o w G' n Z ( G ) is a normal subgroup of G w i t h order divisible by p, and so its intersection w i t h the Sylow subgroup P is nontr iv ia l . We can thus choose a nonidenti ty element z G P D G' n Z ( G ) , and we compute v { z ) . If t G T , then since z G Z ( G ) , we have P t z = P z t = P t , where the last equality holds because z G P . Thus i is the element of T i n the coset P i s , and we conclude that t-z = t. T h e n t z { t - z ) ~ l = tzt~l = z, where the last equality follows because z is central. We conclude that v { z ) = V { z ) = z\T\ = z\G:P\.

O n the other hand, v is a homomorphism from G into the abelian group P , and hence the derived subgroup G' is contained i n ker(w). Since z G G ' , we have 1 = v ( z ) = z\G:P\, and thus z has order d iv id ing the p'-number | G : P | . B u t z G P , and so z has p-power order, and we conclude that z = l . Th i s contradict ion shows that P must be nonabelian. •

5.4. Corol lary. L e t Z C Z ( r ) n T ' , w/iere r i s a finite g r o u p . T h e n a S y l o w p - s u b g r o u p of T / Z i s n o n c y c l i c f o r every p r i m e d i v i s o r p of \ Z \ .

Proof. Let P G S y l p ( r ) . We are assuming that p divides \ Z \ and that Z C Z ( r ) n r ' , so it follows by Theorem 5.3 that P is not abelian. Since P n Z C Z ( P ) and P is nonabelian, we conclude that P / ( P n Z ) cannot be cyclic . B u t P / ( P n Z ) = P Z / Z G S y l p ( r / Z ) , and thus T / Z has a noncyclic Sylow p-subgroup. •

We digress to provide some context for Coro l la ry 5.4. G i v e n a finite group G , consider pairs of groups ( T , Z ) , where Z C Z (T) and T / Z = G . (We say that T is a central extension of G i n this situation.) O f course, it is t r iv i a l to find central extensions of G since we could s imply take T = G x Z , where Z is any abelian group. To prevent this, and to make the si tuat ion more interesting, we insist that Z should be contained i n the derived subgroup T' as well as in the center Z ( r ) . W i t h this addi t ional requirement, the problem is much more restrictive. Indeed, one can prove that among al l pairs ( T , Z ) , where Z C Z (T) n T' and T / Z = G , there is a unique (up to isomorphism) largest group Z that can occur as a second component, and that every other possible second component is a homomorphic image of this largest one. T h e largest possible second component of a pair (T, Z ) associated w i t h a given group G is called the Schur multiplier of G , and it is denoted M ( G ) .

If (T, Z ) is one of our pairs in which Z = M ( G ) , then, of course, |T | = \ G \ \ M ( G ) \ , and i n this case, T is said to be a Schur representation group for G . If the group G is perfect, which, we recall , means that G' = G ,

152 5. T r a n s f e r

then it has a unique (up to isomorphism) Schur representation group, but i n general, a given group G can have more than one isomorphism type of Schur representation group. For example, if G is noncyclic of order 4, it is not very hard to show that \ M ( G ) \ = 2, and thus bo th the dihedral group D g and the quaternion group Q8 are Schur representation groups for G.

Corol la ry 5.4 can be restated this way: If p is a prime divisor of \ M ( G ) \ , then a Sylow p-subgroup of G must be noncyclic, and in particular, it is nontr iv ia l . For example, the only prime p for which a Sylow p-subgroup of the alternating group A h is noncyclic is p = 2, and it follows that M ( A 5 ) is a 2-group. In fact, the Schur mult ipl ier of A b has order 2. (We cannot resist mentioning that \ M ( A n ) \ = 2 for a l l integers n > 4 except n = 6 and n = 7; i n those two cases, the Schur mult ipl ier has order 6. Th i s is another example of what seems to be common phenomenon in finite group theory: general patterns w i t h finite lists of exceptions.)

P r o b l e m s 5 A

5 A . 1 . Suppose that G is abelian and H C G has index n . Show that the transfer homomorphism from G to H is the map g g n .

5A.2 . Show that the transfer homomorphism v : G ->• G / G ' is just the canonical homomorphism from G to G / G ' .

5A.3 . Let H C K C G, where \ G : H \ < oo, and suppose that S is a right transversal for H in K and that T is a right transversal for K i n G. Wr i t e ST — { s t \ s e S, t e T } .

(a) Show that each element of ST is uniquely of the form st w i t h s G S and i G T.

Show that S T is a right transversal for H in G. Deduce that | G : H | = \ G : K \ \ K : H \ .

Let s G -S and t G T, and suppose that g E G. Show that

(si).gr = S'(i3(i«5)_1)(i-g). (d) Let V r : G —> i f be a pretransfer computed w i t h respect to T , and

let v : A" -»• Tf/Tf ' be the transfer homomorphism. Show that the composite map w : G TJ/TJ ' denned by w ^ ) = v { V T ( g ) ) is the transfer homomorphism.

5A.4 . Let H C Z ( G ) , and assume that | G : H \ = n < oo. Let v : G -> H be the transfer homomorphism.

(a) I f h e H , show that = h n .

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(b) N o w assume that H is finite and that \ H \ is coprime to n . Show that G = H x ker(v).

N o t e . T h e fact that H is a direct factor of G i n (b) can also be proved by an appeal to the Schur-Zassenhaus theorem.

5A.5 . Let G be finite, and suppose C is a cycl ic normal subgroup such that G / C is cyclic. Show that \ M ( G ) \ divides \ C : G'\.

H i n t . A p p l y L e m m a 4.6 to a group Y containing a subgroup Z such that Y/Z = G and Z C T ' n Z ( T ) .

5A.6 . Let D be dihedral of order 2 n where n > 2. Show that the Schur mult ipl ier M ( D ) has order 2.

5A.7 . Let B and C be cyclic subgroups of a finite group G, and suppose that B C = G and B n C > 1. A s in P rob l em 5 A . 5 , assume that C « G and that | C : G'\ = n . Show that the order of M ( G ) is s t r ic t ly less than n .

N o t e . In part icular, semidihedral and generalized quaternion groups have t r i v i a l Schur multipliers.

5A.8 . Let A and B be arbi trary finite groups.

(a) Show that \ M ( A x £ ) | > \ M ( A ) \ \ M ( B ) \ .

(b) Assuming that | A | and | B | are coprime, show that M ( A x B ) = M ( A ) x M ( £ ) .

5 B

It is often difficult to obtain useful information about the transfer map i f we work direct ly from the definition. B u t if we appeal to the transfer-evaluation lemma, which we present i n this section, computations become easier, and transfer becomes more useful as a tool for proving theorems.

We continue to allow our groups to be infinite, but of course, in order for the transfer from a group G to a subgroup H to be defined, we must assume that the index \ G : H \ is finite.

5.5. L e m m a (Transfer Evalua t ion) . L e t G be a g r o u p , a n d l e t H be a s u b g r o u p w i t h finite i n d e x n . L e t V : G - > H be a p r e t r a n s f e r m a p c o n s t r u c t e d u s i n g a r i g h t t r a n s v e r s a l T f o r H i n G, a n d fix a n element g G G. T h e n t h e r e e x i s t a subset T0 C T a n d p o s i t i v e i n t e g e r s n t f o r t e T0 ( d e p e n d i n g o n g ) such t h a t t h e f o l l o w i n g h o l d .

(a) t g ^ t - 1 6 H f o r a l l t G T0.

(b) V { g ) =. n t g n n ~ l . t e n

154 5. T r a n s f e r

(c) £ TH = n . ier 0

(d) If g has f i n i t e o r d e r m, t h e n n t i s a d i v i s o r of m f o r a l l i G T 0 .

O f course, the product i n assertion (b) is not well defined unless some order i n which to take the factors is established. Th i s is really irrelevant, however, since we are working modulo H ' and by (a), each factor t g n n ~ x

lies i n H . If H happens to be abelian, so that H ' = 1, then of course, assertion (b) says that v ( g ) = n*0nt*_1> w h e r e v i s t h e transfer map.

P r o o f of L e m m a 5.5. The cyclic subgroup (g) C G acts on T v i a our "dot" action, and thus T decomposes into (#)-orbits. Let T 0 C T be a set of representatives of these orbits (so that each orbit contains exactly one member of T 0 ) and let n t be the size of the orbit containing t € T0. Of course, the sum of the orbit sizes is | T | = n , and this establishes (c). Also , (d) is immediate since if o ( g ) = m < oo, then \ ( g ) \ = m, and so the size of each (#)-orbit on T divides m.

Now let (!) be a (#)-orbit on T, and let t G T 0 lie i n O, so that \ 0 \ = n t . Since

O = { t - x \ x £ ( g ) } = { t - g * \ i E Z } ,

we observe that the elements t - g l must be distinct for 0 < i < n t , and that these are exactly the elements of O. A l so , it should be clear that t - g n t = t.

Reca l l that for every element s € T , we have s g ( s - g ) - 1 € H , and the product of these elements (in some order) is equal to V ( g ) . Tak ing s = t - g \ we have

{ t - g ^ g i t - g ^ 1 ) - 1 - s g ( s - g ) - 1 € H for a l l integers i. Since t - g n t = t, we see that the product of the elements of this form w i t h 0 < i < n t - 1 (taken i n order of increasing i ) is precisely t g n t t - \ which therefore lies i n H . Th is proves (a), and it shows that except possibly for the order of the factors, the element t g n t t ~ l is the contr ibut ion to the product denning V ( g ) coming from the elements of T that lie i n O. Since T is the union of the various orbits and T0 is a set of representatives of these orbits, assertion (b) follows. •

A s our first appl icat ion of the transfer-evaluation lemma, we compute the transfer into a central subgroup.

5.6. Theorem. L e t G be a g r o u p , a n d suppose t h a t Z i s a c e n t r a l s u b g r o u p of f i n i t e i n d e x n . T h e n t h e t r a n s f e r f r o m G t o Z i s t h e m a p g ^ g n . T h i s m a p i s t h e r e f o r e a h o m o m o r p h i s m f r o m G i n t o Z .

Note that since Z is abelian, the transfer to Z really does map to Z ; there is no need to work modulo anything. A l so , observe that Theorem 5.6

5 B 155

generalizes both P rob lem 5A.1 and P rob lem 5A.4(a) . F ina l ly , we mention that since, in the si tuat ion of the theorem, G / Z is a group of order n , it is obvious that g n G Z . W h a t is not at a l l obvious, however, is that the map g i-> g n is a homomorphism, so that { x y ) n = x n y n for a l l x,y£G.

P r o o f of T h e o r e m 5.6. Choose a right transversal T for Z i n G and let g G G . Choose a subset T 0 C T and integers n 4 for i G To, as i n the transfer-evaluation lemma (5.5). For t G T0, therefore, we have t g ^ t ' 1 G Z , and since Z C Z ( G ) , we have

t ^ ' f - 1 = ( i ^ T 1 ) ' = t ^ i t g ^ t - ^ t = g n t .

B y L e m m a 5.5(b), the product over i G To of these elements is v ( g ) , where v : G Z is the transfer. Since £ n t = n , however, this product is g n . •

We digress from our study of the transfer to discuss an applicat ion of Theorem 5.6.

5.7. T h e o r e m (Schur). Suppose t h a t Z ( G ) has finite i n d e x i n a g r o u p G. T h e n G' i s a finite s u b g r o u p .

We need a few prel iminary results. F i r s t , we show that i n the s i tuat ion of Schur's theorem, there are only finitely many commutators in G . O f course, this falls far short of the assertion of the theorem since in general, not every element of G ' is a commutator.

5.8. L e m m a . L e t T be a r i g h t t r a n s v e r s a l f o r Z ( G ) i n a g r o u p G. T h e n every c o m m u t a t o r i n G has t h e f o r m [ s , t ] , w h e r e s,t G T . I n p a r t i c u l a r , if | G : Z ( G ) | < oo, t h e n t h e r e a r e o n l y finitely many d i f f e r e n t c o m m u t a t o r s .

Proof. T h e second statement follows from the first since \T\ = \ G : Z ( G ) | . To prove the first assertion, observe that every element of G can be wri t ten i n the form xs, where x G Z ( G ) and s G T. To complete the proof, therefore, it suffices to show that [ x s , yt] = [ s , t ] , where x , y G Z ( G ) and s,teT. We have [ x s , y t ] = [ x , y t } s [ s , y t ] = [ s , y t ] , where the second equality holds because [ x , y t ] = 1 since x is central. Also , [ s , y t ] = [ i / M ] - 1 = ( [ y , s ] % s ] ) " 1 = [i, s } ' 1 = [ s , t ] , where the th i rd equality holds since y is central. •

We also need the following easy consequence of Theorem 5.6.

5.9. Corol lary. Suppose t h a t G i s a g r o u p a n d t h a t \ G : Z ( G ) | = n . T h e n t h e n t h p o w e r of every c o m m u t a t o r i n G i s t h e i d e n t i t y .

Proof. T h e map g ^ g n is a homomorphism from G into the abelian group Z ( G ) , and thus the commutator subgroup G' is contained i n its kernel. •

The final ingredient in the proof of Theorem 5.7 is the following somewhat more general fact due to A . P . Die tzmann .

156 5. T r a n s f e r

5.10 . T h e o r e m (Dietzmann) . L e t G be a g r o u p , a n d l e t X C G be a finite subset t h a t i s c l o s e d u n d e r c o n j u g a t i o n . Assume t h a t t h e r e i s a p o s i t i v e i n t e g e r n such t h a t x n = 1 f o r a l l x e X . T h e n ( X ) i s a finite s u b g r o u p of G.

P r o o f . Let S be the subset of G consisting of a l l products of finitely many members of X , al lowing repeats, and allowing the "empty" product, which is equal to 1. T h e n S is closed under mul t ip l ica t ion , and also, since x n = 1 for each element x G X , we see that x ~ l = x 7 1 ' 1 lies i n S. It follows that S is closed under inverses, and hence S = ( X ) . Every element of ( X ) , therefore, is a product of members of X , and we w i l l show that it is never necessary to use more than (n - 1 ) \ X \ factors. Since X is finite, it w i l l follow that ( X ) is finite, as wanted.

Consider an arbi t rary product g = x i x 2 - - - x m of TO not necessarily dist inct elements of X , and suppose that some part icular element x € X occurs at least k times among the factors x u where k > 1. We argue that g can be rewrit ten as a product of m elements of X i n such a way that the first factor is x , and there are s t i l l a to ta l of at least k factors equal to x . To see why this is true, suppose that t is the smallest subscript such that x t = x . We can assume that i > 1, and we observe that

X \ X 2 • • • X t - \ X t = X \ X 2 • • • X t - \ X = X ( X \ X 2 • • • X t - l ) X

= X ( X 1 ) X { X 2 ) X • • • { x t - l ) X .

The expression on the right is a product of t elements of X because X is closed under conjugation by x , and of course, its first factor is x . A l so , there are at least k - 1 copies of x among the elements x { w i t h t < i < TO, and so i f we mul t ip ly the above expression on the right by x t + i • • • x m , we get a new representation of g as a product of TO elements of X , and these include a to ta l of at least k copies of x . The first factor i n this product is x , as desired.

Assuming now (as we may) that x x = x , suppose that k > 2. We can repeat the above argument for the product x 2 x 3 • • • x m , and so we can assume that x 2 = x . Cont inuing like this, we see that we can assume that g is a product of m elements of X such that the first k factors are a l l equal to x .

Now fix an arbi trary element g € ( X ) and write g = x x x 2 x % • • • x m , where the factors X i 6 X are not necessarily distinct, and where m is as smal l as possible. Supposing that m > ( n - l ) \ X \ , we derive a contradict ion. B y the "pigeon-hole" principle, at least one element x € X must occur at least n times among the factors x % for 1 < i < TO, and by the above reasoning, it is no loss to assume that the first n factors are a l l copies of x . B u t x n = 1,

5 C 157

and thus g = x n + l x n + 2 • • • x m is a product o f m - n elements of X. Th i s is the desired contradict ion by the assumed min ima l i ty of m . •

F ina l ly , we assemble the pieces to obta in a proof of Schur's theorem.

P r o o f of T h e o r e m 5.7. Let X be the set of a l l commutators i n G, and observe that \X\ is finite by L e m m a 5.8. A l so , since [x,y]*> = [x°,y9} for x, y and g in G, we see that X is closed under conjugation. Now write n = \ G : Z ( G ) | , so that by Corol la ry 5.9, we have xn = 1 for a l l xeX. The result now follows by Theorem 5.10. •

P r o b l e m s 5 B

5B.1. Let G be finite, and suppose that P G S y l p ( G ) and that g G P has order p . If g G G', but g £ P ' , show that ^ G P ' for some element i G G w i t h i 0 P .

H i n t . Let u be the transfer homomorphism from G to P / P ' and observe that g G ker(u), and thus V ( g ) G P ' , where V is a corresponding pretransfer map. A p p l y the transfer-evaluation lemma, and note that each integer n t is either 1 or p .

5B.2. In the s i tuat ion of Theorem 5.10, suppose | A | = m. Show that \ ( X ) \ < n m .

H i n t . Wr i t e X = { x u x 2 , . . . , x m } and show that every element of ( X ) has the form ( x i ) e * ( x 2 ) e 2 • • • ( x m ) e ™ , where the e% are integers satisfying 0 < e* < n .

5B.3. Let G be an arbi t rary group. Show that an element x G G lies i n some finite normal subgroup of G i f and only if x has finite order and its conjugacy class in G has finite size.

5 C

We return now to finite groups. If we want to use transfer theory to prove that some group G is not simple, we need to choose a subgroup P C G, and we must show that the kernel of the transfer homomorphism v : G H / H ' is neither t r i v i a l nor the whole group G. If P is proper i n G, then automatically, k e r ( v ) > 1 since i n that case, v cannot be injective. (Recal l that we are assuming that G is finite.) To prove "good" nonsimpl ic i ty theorems, therefore, it is sufficient to find conditions that w i l l guarantee that ker(?j) < G, or equivalently that there exists at least one element x G G such that v ( x ) ^ 1.

158 5. T r a n s f e r

G i v e n an element x G G , it is sometimes easier to compute v ( x ) if x happens to lie i n the transfer target H . O f course, i f we l imi t our attention to elements of H and we can show that not every element of H lies i n ker(u), this is good enough. (Obviously, i f H % ker(u), then ker(?j) < G, and i n that case, G is not simple.) In fact, i f H is a Sylow subgroup of G, or more generally, if H is a H a l l subgroup, then the converse of this assertion is true too: if k e r ( v ) < G, then H % ker(v). In the s i tuat ion where H is a H a l l subgroup, therefore, we lose nothing by computing the transfer only on elements of H . The following easy lemma proves somewhat more than this.

5.11. L e m m a . L e t H be a H a l l s u b g r o u p of a f i n i t e g r o u p G, a n d l e t v : G - > H / H ' be t h e t r a n s f e r h o m o m o r p h i s m . T h e n v { H ) = v { G ) , a n d so \ H : H D k e r { v ) \ = | G : k e r ( u ) | .

Proof. We know that \ G : ker(7j)| = \ V ( G ) \ , and since v { G ) C H / H ' , we deduce that \ G : ker(v)| divides \ H \ . T h e n \ G : ker(v)| is coprime to \ G : H \ , and so k e r ( v ) H = G by L e m m a 3.16. The result now follows. •

A g a i n let H C G, and consider the transfer homomorphism v : G - > H / H ' . W h e n we apply the transfer-evaluation lemma to compute v ( x ) for some element x G G, we are led to consider elements of H of the form t x n t t ~ 1

for certain elements i e G and positive exponents n t . U x e H , which as we now know is often the case of interest, then also x n t G H . In this si tuation, the elements x n t and t x n t t ~ l bo th lie i n H , and they are clearly conjugate i n G (via the element i ) , but of course, they need not be conjugate i n H .

Mot iva ted by this, we consider the intersection of a conjugacy class X of G w i t h a subgroup H , where we assume that X n H is nonempty. Clearly, if x G A n J f and y lies i n the conjugacy class of H containing x , then y £ X D H , and this shows that I n i i i s a union of conjugacy classes of H . We say that classes Y\ and F 2 of H are fused i n G i f Yx and F 2 are contained i n the same class of G . A l so , if H C AT C G , we say that AT controls G-fusion in H if every two classes of H that are fused in G are already fused i n K . Equivalently, AT controls G-fusion i n H if and only i f every pair of G-conjugate elements of H are ^-conjugate . In particular, G always controls G-fusion i n H , and H controls G-fusion i n itself precisely when no two distinct classes of H are fused in G . A s we shall see, the notion of "control of fusion" is v i t a l for understanding transfer, especially transfer to Sylow subgroups, and more generally, to H a l l subgroups.

We begin our study of fusion w i t h the following general result.

5.12. L e m m a . L e t P G S y l p ( G ) , w h e r e G i s a f i n i t e g r o u p . T h e n N G ( P ) c o n t r o l s G - f u s i o n i n C G ( P ) .

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P r o o f . Assuming that x , y G C G { P ) are conjugate i n G , we must show that i n fact, these elements are conjugate in N G ( P ) . Wr i t e y = x 9 w i t h g G G, and observe that since x G C G ( P ) , we have y = x 9 G C G { P ) 9 = C G { P 9 ) . T h e n P 9 C C G ( y ) , and hence P 9 is a Sylow p-subgroup of C G ( y ) . A l so y G C G ( P ) , and so P C C G ( y ) , and hence P is also a Sylow p-subgroup of C G ( y ) . W e can now apply the Sylow C-theorem i n the group C G { y ) to the two Sylow p-subgroups P and P 9 , and we deduce that ( P 9 ) c = P , for some element c € C G ( y ) . T h e n #c G N G ( P ) and x ^ c = y c = y , and so x and y are conjugate i n N G ( P ) , as required. •

We can now prove Burnside 's normal p-complement theorem. (Recal l that i f p is a prime, then a normal p-complement i n a finite group G is a normal subgroup wi th index equal to the order of a Sylow p-subgroup. E q u i v a l e n t s , it is a subgroup N < G such that \ N \ is not divisible by p and \ G : N \ is a power of p. In s t i l l other words, a normal p-complement is a normal H a l l p'-subgroup. A group that has a normal p-complement is said to be p - n i l p o t e n t . )

5 .13 . T h e o r e m (Burnside) . L e t P G S y l p ( G ) , w h e r e G i s a finite g r o u p , a n d suppose P C Z ( N G ( P ) ) . Then G has a n o r m a l p - c o m p l e m e n t .

P r o o f . In this si tuation, P is abelian, so P C C G ( P ) . B y L e m m a 5.12, therefore, every two elements of P that are conjugate in G are conjugate in N G ( P ) . B u t P is central i n N G ( P ) , so no two distinct elements of P can be conjugate i n N G ( P ) . It follows that no two distinct elements of P can be conjugate i n G .

Now consider the transfer homomorphism v : G -> P , and recall that in this case, where P is abelian, there is no dis t inct ion between the transfer and pretransfer. Let x G P , and apply the transfer-evaluation lemma to compute that

v ( x ) = J] t x n H ~ l , teT0

where T 0 is some subset of a right transversal T for P i n G . A l so , £ n t = \ G : P | , and we know that each of the factors t x ^ t ' 1 lies i n P . Since t x n n ~ l and x n t are G-conjugate elements of P , the first paragraph of the proof guarantees that they are equal, and thus x n t = t x n n ~ l for a l l t G T0. T h e n

v ( x ) = Y [ x n t = x | G : F | , t e n

and so if v ( x ) = 1, then x has order d iv id ing \ G : P | . B u t x G P , and so the order of x also divides | P | . It follows that x = 1, and hence P n k e v ( v ) = 1. B y L e m m a 5.11, we have | G : ker(u)| = | P | , and so ker(?j) is the desired normal p-complement in G . •

160 5. T r a n s f e r

5.14. Corol lary. L e t P e S y l p ( G ) , w h e r e G i s a finite g r o u p a n d p i s t h e s m a l l e s t p r i m e d i v i s o r of \ G \ , a n d assume t h a t P i s c y c l i c . T h e n G has a n o r m a l p - c o m p l e m e n t .

Proof. We know that N G ( P ) / C G ( P ) can be isomorphically embedded i n A u t ( P ) , which, because P is cycl ic , has order 1, and so we can write \ P \ = p" , w i t h n > 0. T h e n < p ( \ P \ ) = p n _ 1 ( p - 1), and since this number has no prime divisor larger than p, it follows that | N G ( P ) : C G ( P ) \ has no prime divisor larger than p . Th i s index also has no prime divisor smaller than p because no such prime divides \ G \ . F ina l ly , the prime p itself does not divide | N G ( P ) : C G ( P ) | since P is abelian, and thus the full Sylow p-subgroup P is contained in C G ( P ) . We conclude that | N G ( P ) : C G ( P ) | has no prime divisors at a l l , and so this index is 1, and N G ( P ) = C G ( P ) . In other words, P C Z ( N G ( P ) ) , and hence by Burnside 's theorem, G has a normal p-complement. •

We mention that if some group G has a cyclic Sylow p-subgroup, then the same is true for every subgroup of G and every homomorphic image of G. (For subgroups, this follows by the Sylow D-theorem since a Sylow p-subgroup of a subgroup of G is contained i n a Sylow p-subgroup of G. For factor groups, our assertion is a consequence of P rob lem I B . 5 , which implies that a surjective homomorphism carries Sylow p-subgroups to Sylow p-subgroups.) In part icular, this shows that the hypotheses of the following corollary are inherited by subgroups and factor groups.

5.15. Corol lary. L e t G be a finite g r o u p , a n d suppose t h a t a l l S y l o w s u b g r o u p s of G a r e c y c l i c . T h e n G i s s o l v a b l e .

Proof. We can assume that G > 1, and we proceed by induct ion on \ G \ . If p is the smallest prime divisor of \ G \ , then G has a normal p-complement N by Coro l la ry 5.14. A l so , N < G and a l l Sylow subgroups of TV are cyclic, and thus N is solvable by the inductive hypothesis. Since G / N is a p-group, it is solvable too, and thus G is solvable by L e m m a 3.10(e). •

In part icular , it follows that the order of a nonabelian simple group cannot be a product of distinct prime numbers.

Ac tua l ly , much more can be said about groups in which a l l Sylow subgroups are cyclic . Not only is such a group solvable, but it is also meta-cyclic, which means that it has a cyclic normal subgroup w i t h a cyclic factor group. T h e following shows that even more is true.

5.16. Theorem. Suppose t h a t a l l S y l o w s u b g r o u p s of t h e finite g r o u p G a r e c y c l i c . T h e n b o t h G' a n d G / G ' a r e c y c l i c , a n d they h a v e c o p r i m e o r d e r s .

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The key step here is the following.

5.17. Theorem. L e t P be a c y c l i c S y l o w p - s u b g r o u p of some finite g r o u p G. T h e n p d i v i d e s a t most one of t h e n u m b e r s \ G ' \ a n d \ G : G ' \ .

Proof. Let N = N G ( P ) , and let K C N be a complement for P in N . (The existence of K is guaranteed by the Schur-Zassenhaus theorem.) B y F i t t ing ' s theorem (4.34), we have P = [ P , K ] x C P ( K ) . If [ P , K ] = 1, then C P { K ) = P , and i n this case, P is central in N = P K . B y Burnside 's theorem, therefore, G has a normal p-complement M , and since G / M is a p-group, it must be cyclic . T h e n G' C M , and so p does not divide \ G ' \ .

If, on the other hand, [ P , K ] > 1, then [ P , K ] contains the unique subgroup of order p i n the cyclic group P . It follows that C P ( K ) contains no subgroup of order p, and hence C P ( K ) = 1. Therefore P = [ P , K ] C G', and so \ G : G'\ is not divisible by p , as required. •

We need one further observation: an abelian group i n which a l l Sylow subgroups are cyclic must itself be cyclic . In fact, i f we choose a generator x p of the Sylow p-subgroup of G for each prime divisor p of \ G \ , it is easy to see that the element T\ X p is a generator for G.

P r o o f of T h e o r e m 5.16. Tha t no prime can divide bo th \ G ' \ and \ G : G'\ follows by Theorem 5.17. A l so , G / G ' is an abelian group w i t h a l l Sylow subgroups cycl ic , and hence it must be cyclic . To show that G' is cyclic, we proceed by induct ion on \ G \ . We can assume, of course, that G > 1, and thus G' < G because G is solvable by Coro l la ry 5.15. T h e inductive hypothesis applied to G' tells us that G" is cycl ic , and thus A u t ( G " ) is abelian. B u t G / C G ( G " ) is isomorphical ly embedded in A u t ( G " ) , and thus G / C G ( G " ) is abelian, and we have G' C C G ( G " ) . Now G ' / G " is abelian and hence is cycl ic , and since G" C Z ( G ' ) , it follows that G' is abelian, and hence G ' is cycl ic , as required. •

Next , we consider what happens if only part of an abelian Sylow p-subgroup P of G is central in N G ( P ) . (In the case of Burnside 's theorem, where the whole of P is central in N G ( P ) , it is automatic that P is abelian, but in this more general si tuation we must assume that P is abelian.) Since Burnside 's theorem (5.13) is an easy consequence of the following, we could have omit ted the proof of that earlier result, which is hereby rendered redundant. We mention also that the following can also be viewed as a stronger form of Theorem 5.3.

5.18. Theorem. L e t P be a n a b e l i a n S y l o w p - s u b g r o u p of a finite g r o u p G. T h e n

G ' n P n Z ( N G ( P ) ) = 1.

162 5. T r a n s f e r

Proof. Let v : G - > P be the transfer, and l e t i G G ' n F f l Z ( N G ( P ) ) . T h e n

1 = v(a;) = JT ^ i - 1 ,

where the first equality holds since x e G ' C ker(v), and the second follows by the transfer-evaluation lemma, where T 0 and the integers n t are as usual, and where each factor t x ^ t ' 1 lies i n P . Since x £ P , bo th x n t and ta^r1

lie i n P , and hence these G-conjugate elements of P C C G ( P ) are conjugate in N G ( P ) by L e m m a 5.12. B u t x is central in N G ( P ) , and thus x n t is also central in this group, and hence is conjugate only to itself i n N G ( P ) . It follows that x n t = t x ^ t ' 1 for a l l t e T0, and thus v { x ) = x \ G : P \ because Y,nt = \ G : P \ . Thus 1 = a ; l G : P l , and x has p'-order. Since x € P , we conclude that x = 1, as desired. •

A s an applicat ion of Theorem 5.18, we offer the following.

5.19. Corol lary. Suppose t h a t a S y l o w 2 - s u b g r o u p of a n o n a b e l i a n finite g r o u p G i s a d i r e c t p r o d u c t of c y c l i c s u b g r o u p s , one of w h i c h i s s t r i c t l y l a r g e r t h a n a l l of t h e o t h e r s . T h e n G i s n o t s i m p l e .

Proof. Let P e S y l 2 ( G ) . We can write P = A x P , where A is cyclic of order a > 2, and where W 2 = 1 for a l l x e B . Let C = { x a / 2 \ x G P } , and observe that C is a nontr iv ia l characteristic subgroup of P of order 2, and thus the unique nonidenti ty element t of C is central i n N G ( P ) . It follows by Theorem 5.18 that t 0 G', and thus G' < G. If G is simple, then G' = 1, and so G is abelian, and this is a contradiction. •

We mention that it is known from a part of the simple group classification that if a Sylow 2-subgroup of a simple group is abelian, then in fact, it must be elementary abelian. In other words, a l l of its cyclic direct factors have order 2.

P r o b l e m s 5 C

5C.1 . Show that Burnside 's normal p-complement theorem is a consequence of Theorem 5.18.

5C.2 . Suppose that U C V C P C G, where P is an abelian Sylow 2-subgroup of G and U and V are characteristic i n P such that \ V : U\ = 2. If G is simple, show that \ G \ = 2.

5C.3 . Le t G be simple and have an abelian Sylow 2-subgroup P of order 2 5 . Deduce that P is elementary abelian.

P r o b l e m s 5 C 163

5C.4. Suppose that al l Sylow subgroups of G are cyclic . Show that for every divisor m of | G | , there exists a subgroup of order m, and show that every two subgroups of order m a r e conjugate in G.

5C.5 . Le t P € S y l p ( G ) and suppose that A and B are G-conjugate normal subgroups of P . Show that A and B are conjugate i n N G ( P ) . If A is characteristic in P , deduce that A = B .

Note . If A is characteristic i n P e S y l p ( G ) , it is certainly possible for A to be G-conjugate to some subgroup B of P w i t h B ^ A . B u t among a l l G -conjugates 73 of A that are contained i n P , the only one that can be normal i n P is A itself.

5C.6 . Le t W C P C G . We say that VP is weakly closed i n P w i t h respect to G if the only G-conjugate of W contained i n P is W itself. Now suppose that P e S y l p ( G ) and that W C P .

(a) If is weakly closed i n P w i t h respect to G , show that W is weakly closed i n Q for each Sylow p-subgroup Q of G that contains W .

(b) Show that W is weakly closed i n P w i t h respect to G i f and only if W is normal i n N G ( P ) and it is also normal i n a l l Sylow p -subgroups of G that contain i t .

(c) Suppose that W is weakly closed in P w i t h respect to G , and let W C P C G . Show that every G-conjugate of W contained i n P is actually P-conjugate to W.

(d) Suppose that W C Z ( P ) and that V7 is weakly closed i n P w i t h respect to G . Show that N G ( W ) controls G-fusion in P .

5C.7 . Le t \ G \ = 3 ° - 5 - l l . Show that G has a normal Sylow 3-subgroup.

5C.8 . Le t p > 2 be the smallest prime divisor of \ G \ , and suppose that \ G \ is not divisible by p 3 . Show that G has a normal p-complement.

5C.9 . Suppose that G is nonabelian simple and has even order not divisible by 8. Show that \ G \ is divisible by 3.

Note. There do exist nonabelian simple groups w i t h order not divisible by 3, but a l l such groups have order divisible by 64. In fact, the only nonabelian simple 3'-groups are the Suzuki groups S z ( q ) , where q = 2 e and e > 1 is odd. T h e group S z { q ) has order q 2 - ( q - l ) - ( g 2 + 1).

164 5. T r a n s f e r

5C.10. Suppose that G is simple and has an abelian Sylow 2-subgroup of order 8. Show that \ G \ has order divisible by 7.

N o t e . The simple group P S L ( 2 , 8) of order 504 = 2 3 -3 2 -7 has an elementary abelian Sylow 2-subgroup of order 8. The same is true for the smallest Janko sporadic simple group J u which has order 175560 = 2 3 - 3 - 5 - 7 - l M 9 .

5C.11. Let H be a H a l l subgroup of G, and assume that H C Z ( N G ( J i ) ) . Show that G has a normal p-complement for every pr ime divisor p of \ H \ .

H i n t . W o r k by induct ion on \ G \ . Show that if P is a Sylow p-subgroup of H and N G ( P ) < G, then P is central in N G ( P ) . If P < G and Q is a Sylow subgroup of H w i t h N G ( Q ) < G, show that P is central in N G ( Q ) .

5C.12. Suppose that G has a cyclic Sylow p-subgroup, and let N < G have index divisible by p . Prove that N has a normal p-complement.

5C.13. (Navarro) Suppose that P is a Sylow subgroup of G such that P = N G ( P ) . Show that N G ( P ' ) has a normal p-complement.

H i n t . W i t h o u t loss of generality, P ' < G. Show that G is p-solvable, and let X be a H a l l p-subgroup. Show that G = P ' N G ( X ) and apply Dedekind's lemma.

5 D

We have seen that transfer theory can sometimes be used to prove that a group is not simple, and as we saw in the previous section, the transfer into a Sylow p-subgroup may be sufficient to accomplish this, especially when the Sylow subgroup is abelian. We ask now, when does this work? In other words, i f v is the transfer from G to a not necessarily abelian Sylow p-subgroup P of G, we want to know when it is true that k e r { v ) < G. Since G / k e r ( v ) = v { G ) C P / P ' , we see that G / k e r ( v ) is an abelian p-group, and hence transfer to a Sylow p-subgroup can prove nonsimplic i ty only for groups that have factor groups that are nontr iv ia l abelian p-groups. One of the results we prove in this section is that if G actually has such a factor group, then this fact can definitely be established by considering the transfer to a Sylow p-subgroup.

In order to make a more precise statement, we introduce some notat ion . G i v e n a pr ime p and a finite group G, let AP(G) denote the unique smallest normal subgroup of G for which the corresponding factor group is an abelian p-group. (That A*(G) is well defined is an immediate consequence of the fact that X M , N < G and both G / M and G / N are abelian p-groups, then G/(M n N ) is also an abelian p group.) There is, of course,

5 D 165

an obvious analogy between the definitions of A*(G) and O p { G ) , which, we recall, is the unique smallest normal subgroup of G whose factor group is a (not necessarily abelian) p-group. In fact, A P { G ) and O p { G ) are more int imately connected than merely by analogy. We have O p ( G ) C A*(G) and G / A p { G ) is isomorphic to the largest abelian factor of G/OP(G). Thus AP(G)/OP(G) = ( G / C F ( G ) ) ' = G ' O p { G ) / O p { G ) , and therefore, A*(G) = G ' O p { G ) . A l so , it should be clear that A P ( G ) / G ' = O p { G / G ' ) is the normal p-complement of G/G'.

5.20 . T h e o r e m . L e t v : G -+ P / P ' be the t r a n s f e r h o m o m o r p h i s m , w h e r e G i s a finite g r o u p a n d P G S y l p ( G ) . Then k e r ( v ) = A P ( G ) .

In the previous section, we were able to obta in results about the transfer from G into an abelian Sylow subgroup P because i n that case, N G ( P ) controls G-fusion i n P . In order to study the general case, where P is not necessarily abelian, we need other techniques that help in the study of fusion, and so we make the following definition, which is the key to the proofs of Theorem 5.20 and other results in this section. If H C G is an arbi t rary subgroup, we define the f o c a l s u b g r o u p of H w i t h respect to G to be the subgroup generated by a l l elements of the form x~xy, where x and y are G-conjugate elements of P . Th i s subgroup is denoted F o c G ( P ) .

If x , h G H , then of course, x and xh are G-conjugate elements of H (because they are P-conjugate) , and so X - l x h lies i n F o c G ( P ) . B u t x~1xh = [ x , h ] , and this shows that F o c G ( P ) contains a l l commutators of elements of P . It follows that H ' C F o c G ( P ) , and in fact, H ' = ¥ocH(H). Now assume x, y G P are G-conjugate, so that y = x9 for some element g G G . T h e n x~xy = [ x , g ] G G ' , and hence F o c G ( P ) C G'. F ina l ly , we recall that if P C AT C G and AT controls G-fusion i n P , then two elements x, y G P are G-conjugate if and only if they are AT-conjugate. It follows in this case that F o c K { H ) and F o c G ( P ) have exactly the same generating sets, and so F O C K ( P ) = F o c G ( P ) .

Throughout much of the rest of this chapter, our pr imary interest w i l l be the transfer into a Sylow p-subgroup of G . M a n y of our arguments, however, would work as well for an arbi t rary H a l l rr-subgroup, where TT is a set of primes. The following "focal subgroup theorem", for example, is val id for an arbi t rary H a l l yr-subgroup, and the proof of the more general version goes through without change. The only extra work required is to define A n { G ) i n the obvious way: it is the unique smallest normal subgroup of G for which the factor group is an abelian rr-group.

5 .21 . T h e o r e m (Focal Subgroup). Suppose t h a t P i s a Sylow p - s u b g r o u p of a finite g r o u p G, a n d l e t v : G - > P / P ' be the t r a n s f e r h o m o m o r p h i s m .

166 5. T r a n s f e r

T h e n F o c G ( P ) = P n G' = P n A p { G ) = P n ker(u) .

Proof. F i r s t , observe that

F o c G ( P ) C P n G ' C P n A * ( G ) C P n k e r ( « ) .

The first of these containments holds because the focal subgroup F o c G ( P ) is contained in P by definition, and we observed previously that F o c G ( P ) C G'. The second containment holds since G / A P { G ) is abelian, and so G' C A P ( G ) , and finally, the th i rd containment holds because G/ke r (v ) = v ( G ) C P / P ' , which is an abelian p-group, and thus A P ( G ) C ker(v). To complete the proof, therefore, it suffices to show that P n ker(v) C F o c G ( P ) .

Let x e P n k e r ( v ) , and let T be a right transversal for P in G . To apply the transfer-evaluation lemma to compute v ( x ) , we choose an appropriate subset T 0 C T and positive integers n t for t G To. T h e n x n t and t x n t t ~ l are G-conjugate members of P , and thus

{ x - n t ) { t x n t r l ) G F o c G ( P )

for a l l i e Tb. It follows that the product of these elements (in any order) lies i n F o c G ( P ) . Also , because P ' C F o c G ( P ) , we can arbi t rar i ly rearrange elements of P i n this product to deduce that

TT x ~ n t TT t x n t t - 1 e F o c G ( P ) , t%To t J o

or equivalents ,

Y[ t x n n ~ l EE ] T x n t m o d F o c G ( P ) . t e r 0 t e r 0

B y the transfer-evaluation lemma, the product on the left is congruent modulo P ' to V { x ) , where V is a pretransfer map from G to P . Since x G ker(u), we have V { x ) G P ' , and thus this product lies i n P ' C F o c G ( P ) . It follows that the product on the right also lies i n F o c G ( P ) . B u t ^ n t = \ G : P | , and this yields x \ G : P \ G F o c G ( P ) . Since | P | and \ G : P \ are coprime, there exists an integer m such that m | G : P | EE 1 m o d | P | , and therefore,

x = ( x l G : P l ) m G F o c G ( P ) ,

as required. •

Theorem 5.20 is an immediate consequence of the focal subgroup theorem.

P r o o f of T h e o r e m 5 .20. Wri te AT = ker(v) and A = A P { G ) , so that A C K , and our goal is to prove that equality holds here. Since \ G : K \ and

5 D 167

| G : A \ are p-powers, it follows that P K = G = P A , and by the focal subgroup theorem (Theorem 5.21), we have P n K = P n A . T h e n

\ G : K \ = \ P : P D K \ = \ P : P H A \ = \ G : A \ ,

and since A C AT, it follows that A = AT, as wanted. •

Now fix a Sylow p-subgroup P of G. Let P C H C G , and let 1; and w, respectively, be the transfer homomorphisms from G and P to P / P ' . We wish to compare the kernels A P ( G ) and A P ( P ) and the images v { G ) and w ( P ) of these two transfer maps. Since P A P ( G ) = G, we have P A P ( G ) = G , and thus P / ( P n A P ( G ) ) = G / A P ( G ) is an abelian p-group. It follows that A P { H ) C P n A P ( G ) , and thus | v (G) | = | G : A P ( G ) | < | P : A P ( P ) | = \ w { H ) \ . It should be clear that equality holds here i f and only if A P ( P ) = P n A P { G ) , and also if and only if \ G : A P ( G ) \ = \ H : A P ( P ) | . If equality does hold, we say that H controls p-transfer in G . (Occasionally, we w i l l neglect to mention the group G , and s imply say that " P controls p-transfer".)

The point here is this. If we know that P controls p-transfer in G , and we also know that A P ( P ) < H , i t follows that A * ( G ) < G , and so G is nonsimple (unless | G | = p). In other words, a result that guarantees that some subgroup P containing a Sylow p-subgroup of G controls p-transfer allows us to shift the burden of proving the existence of a nontr iv ia l abelian p-factor group from G to the smaller group P .

A s before, suppose that P C H C G , where P e S y l p ( G ) . One way to guarantee that P controls p-transfer i n G is to show that P controls G -fusion in P . (But we stress that P can control transfer wi thout control l ing fusion.)

5.22. Corol lary. L e t P e S y l p ( G ) , w h e r e G i s a finite g r o u p , a n d suppose t h a t P C H C G , w h e r e H i s a s u b g r o u p t h a t c o n t r o l s G - f u s i o n i n P . Then H c o n t r o l s p - t r a n s f e r , a n d hence

A p ( P ) = P n A p ( G ) and G / A P ( G ) = P / A P ( P ) .

Proof. Let v and w, respectively, be the transfer maps from G and from P to P / P ' . Our goal is to show that \ G : k e r ( v ) \ = | P : ker(to)|, or equiva lent^ (by two applications of L e m m a 5.11) that | P : P n k e r ( w ) | = | P : Pnker(ti>)|. B u t P n k e r ( u ) = F o c G ( P ) and P n k e r H = F o C / / ( P ) by the focal subgroup theorem (5.21), and we have F o c G ( P ) = F o c H ( P ) since we are assuming that P controls G-fusion in P . The result now follows. B

5.23. Corol lary. L e t G be a finite g r o u p , a n d suppose t h a t P e S y l p ( G ) i s a b e l i a n . Then N G ( P ) c o n t r o l s p - t r a n s f e r .

168 5. T r a n s f e r

P r o o f . We know by L e m m a 5.12 that N G ( P ) controls G-fusion in P . The result follows by the previous corollary. •

In fact, a condi t ion much weaker than that the Sylow p-subgroup P is abelian is sufficient to establish that N G ( P ) controls p-transfer. A theorem of P . H a l l and H . Wie land t asserts that N G ( P ) controls p-transfer whenever P has nilpotence class less than p. (In other words, it suffices that P should not be too severely nonabelian.) Actua l ly , the Ha l l -Wie land t theorem is stronger than we have just stated, and i n Chapter 10, we present an even stronger result due to T . Yoshida .

It is not hard to deduce the Burnside normal p-complement theorem from Coro l la ry 5.23, and in fact, the more general Theorem 5.18 follows fairly easily too. Another applicat ion of Corol la ry 5.22 is the following.

5 .24. T h e o r e m . L e t G be a finite s i m p l e g r o u p , a n d suppose t h a t H C G i s a m a x i m a l s u b g r o u p . If H i s n i l p o t e n t , t h e n H i s a p - g r o u p f o r some p r i m e V-

P r o o f . We derive a contradiction by assuming that p and q are distinct primes that divide | P | . Let P € S y l p ( P ) , and observe that P < H since P is nilpotent. A l so , P is nontr ivia l , and P is not normal i n G since G is simple. T h e n P C N G ( P ) < G , and we have H = N G ( P ) by the maximal i ty of H . We deduce that P is a full Sylow p-subgroup of G since otherwise, P < S for some p-subgroup S of G, and thus P < N 5 ( P ) = Sn N G ( P ) = S n P . Th i s is impossible, however, since S n H is a p-subgroup of P that properly contains the Sylow p-subgroup P of P .

Similar ly , P = N G ( Q ) , where Q € S y l ? ( P ) , and we have Q G S y l ? ( G ) . It follows by L e m m a 5.12 applied to Q that P = N G ( Q ) controls G-fusion in C G ( Q ) . B u t P C C G ( Q ) since P and Q are normal subgroups of P and P n Q = 1. We conclude that H controls G-fusion i n P . Since P e S y l p ( G ) and H = N G ( P ) , we deduce from Corol lary 5.22 that H controls p-transfer i n G .

N o w P is nilpotent and has order divisible by p, and thus (F (P) < P . It follows that AP(P) < P because the nontr iv ia l p-group P /OP(P) must have a nontr iv ia l abelian homomorphic image. Since H controls p-transfer in G and A P ( P ) < P , we have A?(G) < G , and hence since G is simple, AP(G) = 1. T h e n G is a p-group, and this is the desired contradiction since q divides | G | . •

The obvious question at this point is whether or not a nonabelian simple group actually can have a max ima l subgroup that is a p-group for some prime p. The answer is "yes", but only if p = 2. The Sylow 2-subgroup of

P r o b l e m s 5 D 169

the simple group PSL{2,17) of order 2 4 -3 2 -17 is max imal , and the impossib i l i ty for odd primes is a consequence of Thompson 's normal p-complement theorem, which we prove i n Chapter 7.

We close this section w i t h a brief discussion of Tate's theorem, which can be viewed as a k ind of "supercharger" for transfer theory. Suppose that, as usual, we have P C H C G , where P e S y l P ( G ) . We have seen that under suitable hypotheses, it may be possible to show that H controls p-transfer, which, we recall, means that A P ( H ) = P n A ? ( G ) . (We know, for example, that this happens i f H controls G-fusion i n P , and there are also other situations where control of p-transfer can be established.) Tate's theorem says that i f A P { H ) = H n A p ( G ) , then, in fact, ( F ( P ) = H f ) ( F ( G ) . (The converse of this is easy: if C F ( P ) = P n C F ( G ) , it is routine to show that A P ( H ) = P n A P ( G ) . )

In order to see the significance of Tate's theorem, we consider a special case. Suppose that in some group G, we have G ' f l P = P ' , where P G Sy lp(G). Since | A P ( G ) : G ' | is a p'-number, it follows easily that A P ( G ) n P = G' n P , and thus A P ( G ) n P = P ' = A P { P ) . We can now apply Tate's theorem w i t h H = P to conclude that ( F ( G ) D P = O P ( P ) = 1. It follows that |OP(G)| is not divisible by p, and thus C F ( G ) is a normal p-complement i n G . To summarize, we have shown using Tate's theorem that if G ' n P = P ' , where P G S y l P ( G ) , then G must have a normal p-complement.

Unfortunately, we cannot prove Tate's theorem here. There seems to be no easy proof of this result, or even of the corollary that we discussed in the previous paragraph. Tate's original argument used cohomology theory, but we prefer Thompson's pretty character theoretic proof, which is presented in the author's character theory text.

P r o b l e m s 5 D

5D.1. Let P G Sylp(G), where P is abelian. Let P C H C G , and assume that H controls p-transfer i n G . If H has a normal p-complement, show that G has a normal p-complement.

5D.2. Le t P G S y l P ( G ) and assume that P C Z ( G ) . W i t h o u t using B u r n -side's normal p-complement theorem or any other part of transfer theory, deduce that G has a normal p-complement. Use this to derive Burnside 's theorem from Coro l la ry 5.23.

5D.3. Let G be a nonabelian simple group, and suppose that P C G is a p-subgroup that is a max ima l subgroup of G .

(a) Show that P € S y l P ( G ) .

170 5. T r a n s f e r

(b) If 1 < N < P and P / N is abelian, show that P is the unique Sylow p-subgroup of G that contains N . Deduce that N is weakly closed i n P w i t h respect to G. (See Prob lem 5C.6.)

(c) Show that the nilpotence class of P must be at least 3.

H i n t . If P has class less than 3, apply (b) w i t h N = Z ( P ) , and use P roblem 5C.6(d) .

5D.4. Let P G S y l p ( G ) and suppose that P C K C G. Show that O P { K ) C A" n O P ( G ) and that if equality holds, then A ? ( K ) = AT n A P ( G ) .

5D.5. Suppose that P e S y l p ( G ) , and assume that A l l P = P ' , where A = A P { G ) . If P ' < A , prove (without appealing to Tate's theorem) that G has a normal p-complement.

Hint . Us ing the Schur-Zassenhaus theorem, let i f b e a complement for P ' i n A and show that G = N G ( K ) P ' . Us ing the fact that P ' C deduce that P normalizes AT.

5D.6. A s in the previous problem, suppose that P G S y l p ( G ) , and assume that A n P = P ' , where A = A P ( G ) . If P ' is abelian, prove (without appealing to Tate's theorem) that G has a normal p-complement.

Hint . Le t N = N G ( P ' ) and observe that N satisfies the hypotheses. Use Burnside 's normal p-complement theorem i n the group A .

5 E

We return now to the problem of determining when a finite group G has a normal p-complement. O f course, Burnside 's theorem guarantees the existence of such a complement if P C Z ( N G ( P ) ) , where P is a Sylow p -subgroup of G. B u t since this hypothesis can only hold if P is abelian, Burnside 's result tells us nothing if P is nonabelian. The focal subgroup theorem, however, can be used to establish a necessary and sufficient cond i t ion for G to have a normal p-complement, w i t h no assumption that the Sylow subgroup is abelian.

5.25. Theorem. A finite g r o u p G has a n o r m a l p - c o m p l e m e n t if a n d o n l y if a S y l o w p - s u b g r o u p of G c o n t r o l s i t s o w n f u s i o n i n G.

Proof. F i r s t , assume that N is a normal p-complement in G, and let P G S y l p ( G ) . T h e n N P = G and N n P = 1, and thus the restriction to P of the canonical homomorphism G - > G = G / N is an isomorphism of P onto G/N. Now suppose that x,y G P are conjugate in G. T h e n their images x and y are conjugate in G, and it follows v i a the isomorphism u h-> u from P

5 E 171

to G that x and y are conjugate in P . Thus P controls its own fusion i n G, as required.

Conversely, assume that P controls its own fusion i n G. Let N = O p ( G ) , and write Q = N n P , so that Q G S y l p ( / V ) . We w i l l show that TV is a normal p-complement i n G by proving that Q = 1. We argue first that A P ( N ) = N . To see why this is so, observe that A P ( N ) is characteristic in N , and hence is normal in G . Furthermore,

\ G : A P { N ) \ = \ G : i V | | / V : A P ( N ) \ ,

which is a power of p , and thus N = O p ( G ) C A * (AT). It follows that iV = A p { N ) , as claimed. B y the focal subgroup theorem, therefore, Focyv(G) = Q n A p { N ) = Q n N = Q.

N o w let x,y G Q be TV-conjugate, so that x " 1 ? / is a typica l generator for FOCJV(Q). T h e n x and y are G-conjugate elements of P , and so by hypothesis, they are conjugate i n P , and we can write y = x u w i t h u € P . We have x~ly = x ~ 1 x u = [ x , u ] G [ Q , P ] , and it follows that F o c i v ( g ) C [ Q , P ] . Since F o c N { Q ) = Q, we have Q C [ Q , P ] , and thus Q C [Q, P , P , . . . , P ] , where there are arbi t rar i ly many commutations w i t h P . B u t P is nilpotent and Q C P , and thus w i t h sufficiently many commutations, we have [Q, P , P , . . . , P ] = 1. It follows that Q = 1, and so iV is a normal p-complement i n G , as wanted. •

A l t h o u g h it is pleasant that Theorem 5.25 gives a necessary and sufficient condi t ion for G to have a normal p-complement, this result is of l imi ted appl icabi l i ty because i n general, it is difficult to determine whether or not a Sylow p-subgroup of G controls its own fusion. A deeper and much more useful cri ter ion for G to have a normal p-complement is the following result of Frobenius, which shows that the existence of a normal p-complement i n G is determined by the p-local subgroups of G . (Recal l that a subgroup N C G is said to be p-local i n G if N = N G ( A ) , where X is some nonidenti ty p -subgroup of G.)

5 .26 . T h e o r e m (Frobenius). L e t G be a finite g r o u p , a n d suppose p i s a p r i m e . T h e n t h e f o l l o w i n g a r e e q u i v a l e n t .

(1) G has a n o r m a l p - c o m p l e m e n t .

(2) N G p O has a n o r m a l p - c o m p l e m e n t f o r every n o n i d e n t i t y p - s u b g r o u p I C G .

(3) N G ( X ) / C G ( X ) i s a p - g r o u p f o r every p - s u b g r o u p I C G .

Observe that i f we were to drop the word "nonidentity" from assertion (2), the theorem would s t i l l be true, but it would be far less interesting. Th i s is so because if X = 1, then N G ( A ) = G , and the impl ica t ion (2) (1)

172 5. T r a n s f e r

would have no content. O n the other hand, assertion (3) is automatical ly true when X = 1, and hence it is irrelevant whether or not we restrict (3) to nonidenti ty subgroups X . (We have chosen not to include such a restriction.) A l so , we stress that i n general, it is not true that if X C G is a p-subgroup and N G ( X ) / C G ( X ) is a p-group, then N G ( A ) has a normal p-complement. The theorem tells us that if for every p-subgroup X Q G , \ t is true that N G { X ) / C G ( X ) is a p-group, then it is also true that N G ( A ) has a normal p-complement for every p-subgroup I C G . The impl ica t ion (3) => (2) is not val id for ind iv idua l p-subgroups X , however.

The implicat ions (1) (2) => (3) of Frobenius ' theorem are nearly t r iv i a l , and we dispose of them first.

5 .27 . L e m m a . I n t h e s i t u a t i o n of T h e o r e m 5.26, if { I ) h o l d s , t h e n so does (2), a n d if ( 2 ) h o l d s , t h e n so does (3).

P r o o f . To prove that (1) => (2), we show more: i f G has a normal p-complement, then every subgroup H C G has a normal p-complement. To see this, suppose that N is a normal p-complement in G. T h e n J V f l / i i s a normal subgroup of H having p'-order, and since \ H : H n N \ = \ N H : N \ divides \ G : N \ , which is a p-power, it follows that N n H is actually a normal p-complement i n H , as wanted.

Now assume (2), and let X C G be a p-subgroup. If X = 1, then N G ( A ) / C G ( A ) is t r iv ia l , and so is a p-group, as required. Otherwise, we know that N G ( A ) has a normal p-complement AT, and we see that K centralizes X since X n AT = 1 and both X and K are normal subgroups of N G ( A ) . T h e n AT C C G ( A ) , and so | N G ( A ) : C G ( A ) | divides | N G ( A ) : AT|, which is a power of p. Thus N G ( A ) / C G ( A ) is a p-group, as required. •

The key to the proof of Frobenius ' theorem is the following.

5 .28 . L e m m a . Assume t h a t N G { X ) / C G { X ) i s a p - g r o u p f o r every p - s u b g r o u p X of a f i n i t e g r o u p G, a n d l e t P , Q € S y l p ( G ) . T h e n Q = P c , f o r some e l e ment c e C G ( P n Q ) .

N

D

5 E 173

Proof. Let V be the set of pairs ( P , Q ) , where P , Q € S y l p ( G ) are not necessarily distinct, and such that Q = P c for some element c G C G ( P n Q ) . Our goal, of course, is to show that the set V consists of a l l pairs of members of Sy lp (G) . If this is false, choose P , Q G S y l p ( G ) w i t h P n Q as large as possible such that ( P , Q ) 0 V. Wr i te D = P n Q, and observe that D ) . Since F n N and Q n N are p-subgroups of TV, they are contained in Sylow p-subgroups S and T of N , respectively, and S is contained i n some Sylow p-subgroup R , of G as shown i n the diagram. Let C = C G ( D ) , so that i V / C is a p-group by hypothesis. T h e n N = SC since S is a Sylow p-subgroup of N . B y the Sylow C-theorem, T = Sn for some element n € N , and we can write n = sy, where s G 5 and y € C. T h e n T = Sn = Ssy = Sy, and since 5 C H , it follows that T = Sy C Ry.

Now P (~) N C S C R, and s o P f l J V C P n £ B u t P H N = N P ( £ > ) > D since D D = P H Q , and so by the choice of F and Q, we have (F , F ) e F . We can thus write R = P x for some element a; e C G ( F n F ) , and i n part icular, we observe that x centralizes D . A l s o , Q n TV C T C R y , and thus Q n i V C F n g . B u t Q n J V = NQ(D) > F> since D < Q, and thus R y n Q > D = P n Q . A g a i n by the choice of F and Q, we have ( F ^ , Q ) € F , and we can write Q = ( R y ) z for some element z G C G ( F ^ n Q ) , and so i n part icular, z centralizes D . Now P x y z = R y z = Q, and since each of x, y and z centralizes D , it follows that x y ^ centralizes D , and thus (F , Q ) lies i n F , which is a contradiction. •

P r o o f of T h e o r e m 5.26. B y L e m m a 5.27, it suffices to prove that (3) (1), and so we assume (3), and we show that G has a normal p-complement. B y Theorem 5.25, it is enough to prove that a Sylow p-subgroup F of G controls its own fusion i n G . We therefore assume that x,y G P are conjugate in G , and we work to show that x and y are actually conjugate i n F .

Wr i t e y = x° w i t h g G G , and observe that y G P n P 9 . B y L e m m a 5.28, there is an element c G C G ( F n P 9 ) such that ( P 9 ) c = P , and we see that c centralizes y . B y hypothesis, N G ( F ) / C G ( F ) is a p-group, and since F

174 5. T r a n s f e r

is a Sylow p-subgroup of N G ( P ) , we can write N G ( P ) = C G ( P ) P . We have #c G N G ( P ) , and so we can write gc = t u , where t G C G ( P ) and u G P . In part icular, since x G P , we see that t centralizes x , and thus y = y c = x

g c = x t u = x u . Since u G P , it follows that x and y are conjugate in P , as required. Th i s completes the proof. •

How can we check that N G ( X ) / C G { X ) is ap-group for some p-group X I This is equivalent to saying that C G ( X ) contains a full Sylow g-subgroup of N G ( X ) for every prime q different from p. In other words, a group G satisfies (3) of Frobenius ' theorem if whenever a g-subgroup Q of G acts on (normalizes) a p-subgroup X of G, the action is t r iv i a l . Th i s yields the following corollary.

5.29. Corol lary. Suppose G i s a finite g r o u p a n d \ G \ = p a m , w h e r e p does n o t d i v i d e m. Assume t h a t n o p r i m e d i v i s o r q o f m d i v i d e s a n i n t e g e r of t h e f o r m p e - 1 , w h e r e 1 < e < a . T h e n G has a n o r m a l p - c o m p l e m e n t .

It follows, for example, that every nonabelian simple group of even order either has order divisible by 32, or by one of 3, 5 or 7. To see why this is so, write \ G \ = 2 a m , where TO is odd, and suppose that a < 5. Since G is nonabelian and simple and has even order, it does not have a normal 2-complement. B y Corol la ry 5.29, therefore, m cannot be coprime to al l of the numbers 2 e - 1, where 1 < e < 4. Since 2 1 - 1 = 1, 2 2 - 1 = 3, 2 3 - 1 = 7 and 2 4 - 1 = 15, at least one of the primes 3, 5 or 7 must divide m.

O f course, by the Fe i t -Thompson odd-order theorem, every nonabelian simple group has even order. A l so , it follows from the classification of simple groups that the order of every such group is divisible by 3 or 5.

P r o o f of Corol lary 5.29. If G does not have a normal p-complement, then (3) of Frobenius ' theorem must fail , and thus G contains a g-subgroup Q that acts nontr ivia l ly on some p-subgroup X , where q is some prime d i visor of TO. N o w C X ( Q ) < X , so we can write \ X : C X ( Q ) \ = p e , where 1 < e < a . A l l of the elements of X - C X ( Q ) lie i n Q-orbits of size divisible by q, and it follows that q divides \X\ - \ C X { Q ) \ = \ C x ( Q ) \ { p e - 1). B u t | C x ( Q ) | is a power of p and hence is coprime to q, and thus q divides p e - 1. The result now follows. •

A somewhat less t r i v i a l appl icat ion is the following.

5.30. Corollary. L e t p be a n o d d p r i m e , a n d suppose t h a t every element of o r d e r p i n t h e finite g r o u p G i s c e n t r a l . T h e n G has a n o r m a l p - c o m p l e m e n t .

Proof. Otherwise, G has a p-subgroup X acted on nontr ivia l ly by a q-subgroup Q, where q is some prime different from p. Since the elements of

P r o b l e m s 5 E 175

order p i n A are central i n G, a l l of them are fixed by Q, and it follows by Theorem 4.36 that Q acts t r iv ia l ly on X . Th i s is a contradiction. •

A l t h o u g h Corol la ry 5.30 fails for p = 2, it is true that G has a normal 2-complement if every element x e G such that x A = 1 is central in G. The proof of this is essentially the same as that of Coro l la ry 5.30, except that one must appeal to P rob lem 4D.4 i n place of Theorem 4.36.

We mention that if p > 2, there is a much stronger version of the impl icat ion (2) (1) in Frobenius ' theorem. In order to show that G has a normal p-complement, it is not necessary to check that N G ( A ) has a normal p-complement for every nonidentity p-subgroup X of G. In fact, it suffices to consider just t w o such subgroups: the center Z ( P ) , and a certain characteristic subgroup J ( P ) , where P e S y l p ( G ) . Th i s theorem of Thompson is proved in Chapter 7.

P r o b l e m s 5 E

5E.1 . Let G be finite, and suppose that every proper subgroup of G has a normal p-complement but that G itself does not. Show that G has a normal Sylow p-subgroup and that \ G \ has exactly one pr ime divisor other than p .

5E.2. Le t G be a finite group in which every proper subgroup is supersolv-able. Show that G is solvable.

H i n t . If G is a counterexample of m i n i m a l order, show that G is simple. T h e n use P rob lem 3B.10 to show that G must have a normal p-complement, where p is the smallest prime divisor of \ G \ .

5E.3. Suppose that every two-generator subgroup of a finite group G has a normal p-complement. Show that G has a normal p-complement.

H i n t . Consider ( x , y ) , where x lies i n some p-subgroup P and y e N G ( P ) has order not divisible by p .

C h a p t e r 6

Frobenius Actions

6 A

Let A and N be finite groups, and suppose that A acts on N v i a automorphisms. The action of A on N is said to be Frobenius i f n a ^ n whenever n G N and a G A are nonidenti ty elements. Equivalent ly, the action of A on N is Frobenius if and only if C N ( a ) = 1 for a l l nonidenti ty elements a G A , and also if and only if C A ( n ) = 1 for a l l nonidenti ty elements n G N . Yet another point of view is to consider the orbits of the act ion of A on the elements of N . If 1 + n G N , then by the fundamental counting principle, the size of the A-orb i t of n is \ A : C A { n ) \ , and so we see that the act ion is Frobenius if and only i f a l l A-orbi ts on N other than the t r i v i a l orbit {1} have size equal to |A| . (Recall that an orbit of an action of a finite group A on some set is said to be regular if the orbit size is equal to |A | . If a l l A-orbi t s are regular, the action is semiregular.) Thus an action v i a automorphisms of A on N is Frobenius precisely when A acts semiregularly on the nonidenti ty elements of N . F ina l ly , to conclude this in t roduct ion to Frobenius actions, we observe that if A acts on N v i a automorphisms and the action is Frobenius, then the natural act ion of every subgroup of A on N is also Frobenius, and so is the action of A on every subgroup of N that admits A .

B y the following easy observation, Frobenius actions are automatical ly coprime actions. Every th ing we know about coprime actions from Chapter 3, therefore, applies to Frobenius actions.

6.1. L e m m a . L e t A a n d N be finite g r o u p s , a n d suppose t h e r e i s a F r o b e n i u s a c t i o n of A o n N . Then \ N \ = 1 m o d \ A \ , a n d hence \ N \ a n d \ A \ a r e c o p r i m e .

177

178 6. F r o b e n i u s A c t i o n s

Proof. The set N is decomposed into A-orbi ts . One of these is {1} and al l of the others are regular orbits, which have size The result follows. •

6 . 2 . Corollary. L e t A a n d N be finite g r o u p s , a n d suppose t h e r e i s a F r o b e n i u s a c t i o n of A o n N . L e t M be a n A - i n v a r i a n t n o r m a l s u b g r o u p of N . T h e n t h e i n d u c e d a c t i o n of A o n N / M i s F r o b e n i u s .

Proof. Let 1 7 a G A . T h e n ( a ) acts coprimely on N , and we recall from Coro l la ry 3.28 and the discussion preceding it that i n coprime actions, "fixed points come from fixed points" i n the induced action on the factor group N / M = N . (This assumes that one of (a) or M is solvable, but since (a) is cycl ic , this hypothesis is satisfied.) It follows that C F ( ( a » = C N { ( a ) ) , and this is t r i v i a l since C N ( ( a ) ) = 1. Thus C F ( a ) = 1, and so the action of A on N is Frobenius, as required. •

Before we proceed to develop the theory of Frobenius actions, it seems appropriate to discuss some examples. O f course, if either A or N is the t r i v i a l group, then the unique (but not very interesting) action of A on N is Frobenius, but it is almost as easy to construct some nontr iv ia l Frobenius actions. For example, if A has order 2 w i t h nonidenti ty element a , and N is abelian of odd order, we can define a Frobenius action of A on N by setting n a = n _ 1 for a l l n G N .

Next , consider a finite field F of order q. Let N be the additive group of F , and let A be a subgroup of the mult ipl icat ive group F x = F - {0} of F . Thus N is an elementary abelian p-group, where p is the unique pr ime divisor of q, and since F x is cyclic of order q - 1, we can take A to be cyclic of order r for an arbi trary divisor r of q - 1. If n G N and a G A , we define an action by setting n a = n a , where the product n a is denned by viewing bo th n and a as elements of F and using the field mul t ip l ica t ion . Th i s really is an action v i a automorphisms since if x , y G N , then ( x + y ) a = { x + y ) a = x a + y a = x a + y a by the distr ibutive law. To see that this action is Frobenius, suppose that n a = n . T h e n n a = n , and so n ( a - 1) = 0. It follows that either n = 0, which is the identity in N , or a = 1, which is the identity i n A .

G i v e n primes p and r, let N and A , respectively, have orders p and r . B y the construction of the previous paragraph, we see that there is a Frobenius action of A on N provided that r divides p - 1. (Of course, this d iv is ib i l i ty condi t ion is necessary by L e m m a 6.1.) In fact, an arbi trary nontr iv ia l action of a group A of prime order r on a group N of prime order p is necessarily Frobenius since if n G N and a G A are nonidentity elements, then a cannot centralize n or else A = ( a ) centralizes N = ( n ) , and the action is t r iv i a l . In part icular , if the action of A on N is nontr ivia l , then r must divide p - I by L e m m a 6.1.

6 A 179

Not every group A can have a Frobenius act ion on a nontr iv ia l group N , and not every group N can admit a Frobenius act ion of a nontr iv ia l group A . A n easy result of this type is the following, but as we shall see after we develop the necessary machinery, much more is true.

6.3. Theorem. L e t A a n d N be finite g r o u p s , a n d suppose t h e r e i s a F r o b e n i u s a c t i o n of A o n N . Assume t h a t \ A \ i s even a n d t h a t N i s n o n t r i v i a l . T h e n A c o n t a i n s a u n i q u e i n v o l u t i o n , a n d N i s a b e l i a n .

Proof. Since |A | is even by assumption, we can fix some involut ion t G A , and we consider the map x i-> x " V from N to itself. If x , y G N and x ~ l

x t = y - i y t t then y x ~ l = ^ ( x 4 ) " 1 = ( y x " 1 ) ' - Since the action of A is Frobenius and t ^ 1, it follows that y x ' 1 = 1, and thus x = y . Th i s shows that our map x H-» x ~ l x l is injective, and hence it is also surjective because N is finite. G iven an arbi trary element y G N , therefore, we can write y = x " V for some element x£N. Thus

y l = ( x - V ) ' = ( x - ' Y x = ( x ' y ' x = ( x - V ) " 1 = y - 1 ,

where the second equality holds since t2 = 1. If s G A is an arbi trary involut ion, a s imilar calculat ion yields y s = y ~ \ and thus y s = y l . T h e n y

s t ~ x = y for a l l y £ N , and since N is nont r iv ia l by assumption, we can assume that y ^ 1, and thus st'1 = 1 because the action of A is Frobenius. Thus s = t, and hence t is the unique involut ion i n A , as required.

We have seen that t inverts every element of TV, and thus if x , y G N , then ( x y Y = ( x y ) ' 1 = y ^ x ' 1 = ? r V = ( y x ) K It follows that x y = y x , and thus N is abelian. •

A finite group A is said to be a Frobenius complement if it has a Frobenius action on some nonidenti ty group N , and similarly, a finite group N is a Frobenius kernel if it admits a Frobenius action by some nonidenti ty finite group A . (Our definition of "Frobenius complement" is not quite standard, but as we explain later, it is equivalent to the usual definition.) B y Theorem 6.3, we know that a Frobenius complement of even order contains a unique involut ion, and we w i l l prove that a Frobenius complement of odd order is even more t ight ly constrained. A n odd order Frobenius complement, for example, contains a unique subgroup of order p for every prime p d iv id ing its order.

There are also strong structural constraints on Frobenius kernels, as is suggested by the fact that the Frobenius kernel i n Theorem 6.3 is abelian. In general, Frobenius kernels need not be abelian, but they are always nilpo¬tent. The nilpotence of Frobenius kernels had been a long open problem, which was finally settled by J . Thompson in his P h . D . thesis in 1959. We start work toward a proof of Thompson's theorem i n this chapter, and we


complete the proof in Chapter 7, where we present one of the key ingredients: Thompson 's deep normal p-complement theorem.

A l l of the Frobenius complements and Frobenius kernels that we have seen so far are abelian, and i n fact, a l l of our Frobenius complements have been cyclic . (We shall see that an abelian Frobenius complement is necessari ly cyclic.) B u t nonabelian examples of Frobenius complements and kernels exist too. F i rs t , to show that a Frobenius kernel can be nonabelian, let F be a finite field of order q, and let N be the group of "unitary" upper tr iangular 3 x 3 matrices over F . (The elements of TV are a l l of the 3 x 3 matrices over F w i t h ones on the diagonal and zeros below the diagonal.) It should be clear that \ N \ = q3, and it is easy to check that N is not abelian. If q - 1 is not a power of 2, we show that TV is a Frobenius kernel by constructing a nonidenti ty subgroup A of the group G L ( 3 , F ) of al l invertible 3 x 3 matrices over F such that A acts by conjugation on N , and we w i l l show that this action is Frobenius. (Note that al though N is not abelian, it is nilpotent, as is required by Thompson's theorem.)

Let C be a (necessarily cyclic) odd-order subgroup of the mult ipl icat ive group F x of F , and let A be the group of 3 x 3 diagonal matrices of the form a = d i a g ( l , c, c 2 ) , where c runs over C . (Note that we can choose C so that \ C \ is an arbi trary odd divisor of | F X | = q - 1, and so, in particular, i f q - 1 is not a power of 2, we can suppose that A is nontrivial . )

It is easy to check that A acts by conjugation on N , and that in fact,

1 X z a "1 cx c2z~

0 1 y = 0 1 cy 0 0 i _ 0 0 1

where a = d i a g ( l , c, c 2 ) . If a is a nonidentity element of A , then c + 1, and thus also c 2 ^ 1 because we are assuming that C has odd order. If n a = n i n this si tuation, we see that each of the above-diagonal entries x , y and z of n must be zero, and thus n = 1. It follows that the action is Frobenius.

Next , we mention some examples of nonabelian Frobenius complements. F i rs t , consider the group S = 5 L ( 2 , 3 ) of 2 x 2 matrices w i t h determinant 1 over the field of order 3. It is easy to see that \S\ = 24, and it is not hard to check that S has a unique Sylow 2-subgroup, which we w i l l call Q. O f course, \ Q \ = 8, and i n fact, Q is isomorphic to the quaternion group Q8. In part icular , Q is noncyclic, and its unique element of order 2 is the negative of the identity matr ix . Since Q is a subgroup of 5 L ( 2 , 3 ) , it has a natural action on the vector space V of dimension 2 over the field of order 3, and since the unique element of order 2 in Q fixes no nonzero vector, it follows that the action of Q on the additive group of V is Frobenius. (Note that

6 A 181

since \V\ = 1 + \ Q \ , the act ion of Q on the nonidenti ty elements of V is regular, and not just semiregular.)

Next , we appeal to the not completely obvious fact that S L ( 2 , 5) contains an isomorphic copy S of 5 L ( 2 , 3 ) . (In fact, S is the normalizer i n 5 L ( 2 , 5 ) of a Sylow 2-subgroup.) Since S C 5 L ( 2 , 5 ) , there is an act ion of S on a 2-dimensional vector space W over a field of order 5, and it easy to see that no element of order 2 or order 3 i n S L { 2 , 5) can have an eigenvalue equal to 1. N o such element, therefore, can have a nont r iv ia l fixed point on W, and it follows that the action of S on W is Frobenius. (Note that i n this case too, the action on nonidenti ty elements is regular and not just semiregular since \W\ = 25 = 1 + 24 = 1 + \S\.) Somewhat similarly, the group 5 L ( 2 , 5 ) can be embedded i n SL(2,11), and this yields a Frobenius action of 5 L ( 2 , 5 ) on a vector space of order l l 2 = 121. (Since | 5 L ( 2 , 5 ) | = 120, we again have a regular action on nonidenti ty elements.) B u t SL(2,11) is not a Frobenius complement, and so the story ends here. (In fact, SL(2,11) contains a nonabelian subgroup of order 55, but we w i l l prove that no Frobenius complement can have a noncyclic subgroup whose order is a product of two primes.)

Not ice that the Frobenius complement 5 L ( 2 , 5 ) is not solvable. In fact, SL(2,5) is perfect, which, we recall, means that it is its own derived subgroup. A theorem of H . Zassenhaus asserts that S L { 2 , 5) is the only perfect Frobenius complement. (We mention that one of the few serious errors i n W . Burnside 's classic group theory book is his assertion that Frobenius complements are always nilpotent. Th i s false statement is his Theorem V on Page 336 of the 1911 edition.)

Next , we consider the semidirect product G = N x A , where A acts on N , and we ask what can be said about G i f the given act ion is Frobenius. A s usual when we consider semidirect products, we view N and A as subgroups of G, where N is normal and A is a complement for N i n G. (Recal l that this means that N A = G and N n A = 1.) In this s i tuat ion, of course, the original act ion of A on N is exactly the conjugation act ion of A on N i n G.

6.4. Theorem. L e t N be a n o r m a l s u b g r o u p of a finite g r o u p G, a n d suppose t h a t A i s a c o m p l e m e n t f o r N i n G . T h e f o l l o w i n g a r e t h e n e q u i v a l e n t .

(1) T h e c o n j u g a t i o n a c t i o n of A o n N i s F r o b e n i u s .

(2) A C \ A 9 = l f o r a l l elements g € G - A .

(3) C G ( a ) C A f o r a l l n o n i d e n t i t y elements a e A .

(4) C G ( n ) C N f o r a l l n o n i d e n t i t y elements n e N .

If bo th N and A are nontr iv ia l i n the si tuat ion of Theorem 6.4, we say that G is a F r o b e n i u s g r o u p and that A and N are respectively the Frobenius complement and Frobenius kernel of G.

We need the following easy computat ion.

6.5. L e m m a . L e t A be a s u b g r o u p of a finite g r o u p G, a n d suppose t h a t A n A 9 = 1 f o r a l l elements g G G - A . L e t X be t h e subset of G c o n s i s t i n g of those elements t h a t a r e n o t c o n j u g a t e i n G t o any n o n i d e n t i t y element of A . T h e n \X\ = \ G \ / \ A \ .

P r o o f . If A = 1, then A has no nonidentity elements, and so X = G and there is nothing further to prove. Assuming that A > 1, we see that A = N G ( A ) since if A x = A , then A n A x = A > 1, and thus x G A . It follows that A has exactly \ G : A \ distinct conjugates in G. Furthermore, since each of these conjugates satisfies the condi t ion we assumed about A , no two of them can have a nontr iv ia l intersection. The conjugates of A , therefore, account for a total of \ G : A \ { \ A \ - 1) nonidenti ty elements of G, and these are exactly the elements of G that are conjugate to nonidenti ty elements of A . We conclude that \X\ = \ G \ - \ G : A \ ( \ A \ - 1) = \ G : A \ , as required. •

6.6. C o r o l l a r y . L e t N be a n o r m a l s u b g r o u p of a finite g r o u p G, a n d suppose t h a t A i s a c o m p l e m e n t f o r N i n G such t h a t A n A 9 = 1 f o r a l l elements g e G - A . T h e n N i s e x a c t l y t h e set X of L e m m a 6.5. I t i s , i n o t h e r w o r d s , t h e set of elements of G t h a t a r e n o t c o n j u g a t e t o any n o n i d e n t i t y element of A .

P r o o f . Since N n A = 1, it is also true that N n A 9 = 1 for a l l g G G, and thus no element of N can be conjugate to a nonidenti ty element of A . It follows that J V C ! B y L e m m a 6.5, however, \X\ = \ G \ / \ A \ = \ N \ , and the result follows. •

In fact, the set X of L e m m a 6.5 is a l w a y s a subgroup. T h i s remarkable theorem of Frobenius was proved using character theory, and no one has ever found a character-free proof. (We mention that it was Frobenius who invented character theory, and his proof that the set AT is a subgroup is one of its earliest applications.) Observe that once we know that A is a subgroup, it is immediate that it is a normal subgroup. Also , by the definition of X , we know that X n A = 1, and so given that A is a subgroup, it follows by L e m m a 6.5 that \ X A \ = \ X \ \ A \ = \ G \ , and thus A complements the normal subgroup X i n G. In other words, we are i n the si tuat ion of Theorem 6.4, and condi t ion (2) of that theorem holds. It follows that (1) holds, and so the conjugation act ion of A on X is Frobenius. In part icular , if A < G, then A is a Frobenius complement i n G. (Of course, the assumption that A is

6 A 183

proper is needed to guarantee that the group X is nontrivial .) Assuming Frobenius ' theorem, therefore, we have shown that a subgroup A < G is a Frobenius complement in G if and only if A D A 9 = 1 for a l l elements g G G - A . In fact, many authors use this is as the definition of a Frobenius complement. ( A n d indeed, we presented this definition i n the note following problem 1D.3.)

Another way to th ink about Frobenius ' theorem is this. Suppose that G acts t ransi t ively on some set ft, and assume that no nonidenti ty element of G fixes as many as two members of ft. Let a G ft, and let A = Ga, the stabilizer of a . Now suppose that g e G — A , and write ¡3 = a-g, so that p + a . T h e n A 9 = G p , and so every element of A n A 9 fixes bo th a and p . B y assumption, therefore, A n A 9 = 1, and thus the point stabilizer A = Ga satisfies the hypothesis of L e m m a 6.5. The elements of G conjugate to nonidenti ty elements of A are exactly the nonidenti ty elements of G that fix some point of ft, and so i n this context, the set X of L e m m a 6.5 consists exactly of the identi ty together w i t h those elements of G that fix no point of ft. Frobenius ' theorem tells us, therefore, that i f G acts transit ively on a set ft and no nonidenti ty element of G fixes as many as two points, then the set consisting of the identity and the elements of G that fix no points forms a subgroup. In fact, it is not hard to see that this assertion about group actions is equivalent to Frobenius ' theorem.

P r o o f o f T h e o r e m 6.4. Assume (1). To prove (2), suppose that A C \ A X > 1 for some element x € G. We work to show that x G A . Since G = A N , we can write x = a n w i t h a G A and n G N , and thus A x = A a n = A n . T h e n A n n A > 1, and so we can choose a nonidenti ty element bn of this intersection, where bn G A and b G A . We thus have [ b , n ] = b ~ x b n G A , and since N < G, we also have [b, n] G N . T h e n [b, n] G A n /V = 1, and hence 6 centralizes n . B u t the action of A on N is Frobenius by (1) and 6 ^ 1 , and it follows that n = l . Thus x = a n = a lies i n A , as required.

Now assume (2), and let 1 ^ a G A . If x G C G ( A ) , then a G A n A x , and since this intersection is nontr ivia l , it follows by (2) that x G A Thus C G ( o ) C A , proving (3).

Next , we show that (3) (1). If 1 ^ a G A , then by (3), we have CJV(O) = TV n C G ( o ) C /V n A = 1, and thus the action of A on TV is Frobenius, which is assertion (1). We have now established that (1), (2) and (3) are equivalent.

Assuming (1) and (2) now, we prove (4) by showing that if n G N and C G ( n ) 2 N , then n = 1. B y (2) and Coro l la ry 6.6, we know that every element of G outside of N lies i n some conjugate of A , and it follows that some nonidenti ty element of C G ( n ) lies i n a conjugate of A . For an

appropriate conjugate m of n , therefore, some nonidenti ty element of C G ( m ) lies in A . Since m G N and the action of A on N is assumed to be Frobenius, it follows that m = 1, and thus n = 1, proving (4).

F ina l ly , assuming (4), let 1 + n G N . T h e n C A ( n ) = A D C G ( n ) C A i l J V = l , and thus the act ion of A on N is Frobenius, proving (1). •

In fact, the overall assumption i n Theorem 6.4 that G has a normal subgroup N complemented by a subgroup A is not entirely necessary. B y Frobenius ' theorem, condi t ion (2), which concerns the embedding of A i n G, guarantees the existence of the normal subgroup N for which A is a complement. Somewhat analogously, condit ion (4), which concerns the embedding of N i n G, guarantees the existence of the complement A for N . Th i s result, however, is far more elementary than Frobenius ' theorem.

6.7. Theorem. L e t N < G, w h e r e G i s a finite g r o u p , a n d suppose C G ( n ) C N f o r every n o n i d e n t i t y element n G N . T h e n N i s c o m p l e m e n t e d i n G, a n d i f K N < G, t h e n G i s a F r o b e n i u s g r o u p w i t h k e r n e l N .

Proof. Let p be a prime divisor of \ N \ . Choose P G S y \ p ( N ) and S G Sylp(G) , w i t h P C S . T h e n Z ( S ) C C G ( P ) , and since P is nontr iv ia l and is contained i n N , it follows by hypothesis that C G ( P ) C N , and thus Z ( S ) C N . Now S is nontr iv ia l and hence Z ( S ) is a nontr iv ia l subgroup of N . B y hypothesis, therefore, S C C G ( Z ( 5 ) ) C N . Thus N contains a full Sylow p-subgroup of G for every prime p d iv id ing \ N \ , and it follows that no such prime divides \ G : N \ . Th i s shows that TV is a normal H a l l subgroup of G, and hence it has a complement A i n G by the Schur-Zassenhaus theorem. If 1 < N < G then both N and A are nontr ivial , and so G is a Frobenius group by Theorem 6.4. •

P r o b l e m s 6 A

6 A . 1 . Let A be a subgroup of S L ( 2 , p ) of order not divisible by the prime p . Show that the action of A on the vector space of order p 2 is Frobenius.

6A.2 . In the mult ipl icat ive group F x of a field F of order 43, let a have order 7 and let e have order 3. Work ing in the group G L ( 3 , 4 3 ) , let

a = 0

a 4

a 0 0 0

0 0

a 2

and 0 1 0 0 0 1 e 0 0

Show that A = ( a , b) is noncyclic of order 63, and that its natural action on the vector space V of order 4 3 3 is Frobenius.

P r o b l e m s 6 A 185

Hint . Check that (b) has order 9 and normalizes (a), which has order 7. A l so , observe that to show that the act ion is Frobenius, it suffices to check that C v { t ) = 1 for elements i e A o f prime order.

Note. Th i s problem shows that a nonabelian group of odd order can be a Frobenius complement.

6A.3. Show that there is a noncyclic group of order 5 2 - l l that has a Frobenius act ion on some nontr iv ia l group.

H i n t . W o r k i n G L { 5 , p ) for some appropriate pr ime p .

6A.4. Let G be a Frobenius group w i t h Frobenius kernel N . Show that each coset of N i n G other than N itself is contained i n a single conjugacy class of G.

6A.5. Let G be a nonabelian solvable group i n which the centralizer of every nonidenti ty element is abelian. Show that G is a Frobenius group, where F ( G ) is the Frobenius kernel.

6A.6. Let A > 1 and B > 1 satisfy the hypothesis of L e m m a 6.5 i n G. Show that A n B 9 > 1 for some element g G G.

Hint . Consider X and Y corresponding to A and B as i n L e m m a 6.5.

Note . Since we have not proved Frobenius ' theorem that the set appearing i n L e m m a 6.5 is actually a subgroup, you should not appeal to that fact i n these problems.

6A.7. Let A satisfy the hypotheses of L e m m a 6.5 i n G, and assume A C H C G.

(a) G i v e n g G G, show that if A 9 n H > 1, then g G H .

(b) If H < G, show that H = G.

Hint . For (a), check that A 9 n H satisfies the hypothesis of L e m m a 6.5 i n H and use the previous problem.

6A.8. Le t G, A and X be as i n L e m m a 6.5. If M < G, show that either M C X or X C M .

Hint . If M n A > 1, use the previous problem to show that A M = G.

6A.9. Le t G, A and A be as i n L e m m a 6.5, and suppose that t G A is an involut ion.

(a) Show that there are at least \ G : A \ - 1 elements i e G - i such that re* = x - K


(b) Show that t inverts every element of the set X .

(c) Show that t is the unique involut ion in A , and i n part icular, t is central i n A .

(d) Show that every element of X has odd order.

(e) Suppose that x and y lie in X and that xy~l € A . Show that x 2 = y 2 and deduce that x = y .

(f) Show that X is a subgroup.

H i n t . L e m m a 2.14(b) is relevant for (a).

N o t e . Th i s problem provides a proof of Frobenius ' theorem i n the case that the subgroup A has even order.

6 A . 1 0 . Let G , A and X be as in L e m m a 6.5.

(a) Show for each prime divisor p of \ A \ that A contains a full Sylow p-subgroup P of G and that A controls G-fusion i n P .

(b) If A > 1, show that G ' A = G and G ' n A = A ' .

(c) If A is solvable, show that the subset X is a subgroup.

N o t e . B y this problem and the previous one, we know that the conclusion of Frobenius ' theorem holds i f either \ A \ is even or A is solvable. B y the Feit-Thompson odd-order theorem, one of these possibilities must necessarily occur, and this shows that Frobenius ' theorem is a consequence of the odd-order theorem. B u t since the Fei t -Thompson proof uses character theory heavily, this certainly does not y ie ld a character-free proof of Frobenius ' theorem.

6 A . 1 1 . Le t A C G . Show that A satisfies the hypothesis of L e m m a 6.5 i n G if and only if N G ( T ) C A for every nonidentity subgroup T of A .

6 B

In this section we present necessary conditions for a group A to be a Frobenius complement. (Recal l that we have defined this to mean that there exists some nonidenti ty group N and a Frobenius action of A on N . ) The key idea is to consider "parti t ioned" groups. A p a r t i t i o n of G is a collection II of nonidenti ty proper subgroups of G such that | J II = G and X n Y = 1 for every two dist inct members X and Y of II. For example, if G = D 2 n , the dihedral group of order 2n w i t h n > 1, then G is part i t ioned by the set II consisting of the cyclic subgroup C of order n and the n subgroups of order 2 that are not contained i n C.

A n elementary abelian p-group G of order p e w i t h e > 1 provides another example of a part i t ioned group. In this si tuation, the set II of a l l

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subgroups of order p in G is a par t i t ion. It is clear that every two dist inct members of H" intersect t r iv ia l ly , and also, since every nonidenti ty element of G has order p , each such element lies i n some subgroup of order p , and so U n = G, as required. In what follows, the cardinal i ty of a par t i t ion w i l l be relevant, and i n this case, it is especially easy to compute. E a c h member of n contains exactly p - 1 nonidenti ty elements, and since G has a to ta l of p e - 1 nonidenti ty elements, we see that II consists of exactly ( p e - 1)/(p - 1) subgroups. In part icular, if e = 2, then |II| = p + 1.

We mention one more family of examples. If G is a Frobenius group w i t h kernel N and complement A , let II be the collection of subgroups consisting of N and a l l conjugates of A . Since N n A = 1, it is clear that N intersects t r iv ia l ly w i t h every conjugate of A . A l so , every two dist inct conjugates of A intersect t r iv ia l ly because A (and each of its conjugates) satisfies statement (2) of Theorem 6.4. Furthermore, by Coro l la ry 6.6, we know that every element of G not i n N lies in some conjugate of A , and thus [j U = G. In this si tuation, A = N G ( A ) , and so the number of conjugates of A i n G is \ G : A \ = \ N \ , and it follows that |II| = 1 + \ N \ .

6.8. L e m m a . L e t U be a p a r t i t i o n of a finite g r o u p A , a n d suppose t h a t A acts v i a a u t o m o r p h i s m s o n a n a b e l i a n g r o u p U . I f U has a n element w i t h o r d e r n o t d i v i d i n g |XT| - I , t h e n t h e r e exists a member X G II such t h a t C u { X ) > 1.

P r o o f . For each subgroup H C A and each element u G U, write

and observe that this element of U is unambiguously defined since the order of the factors in the product is irrelevant because U is abelian. Also , since the action by an element h € H s imply permutes the factors of uH, it follows that ( u H ) h = uH for a l l h e H , and thus uH G C V { H ) . Since we can assume that C u ( X ) = 1 for each member X G II, we have u x = 1 for a l l u G U and al l X G n .

N o w fix u G U, and compute that

x e n

where the second equality follows because as a runs over a l l of the elements of a l l of the subgroups X G II, each nonidenti ty element of A occurs exactly once, while the identity of A occurs a to ta l of )H| times. If we choose u G U such that u l 1 1 ! " 1 ^ 1, it follows that uA + 1, and since uA G C V ( A ) C C r j ( X ) for a l l X G II, the proof is complete. •


6.9 . T h e o r e m . L e t A be a F r o b e n i u s c o m p l e m e n t . T h e n n o s u b g r o u p of A i s a n e l e m e n t a r y a b e l i a n p - g r o u p of o r d e r e x c e e d i n g p , a n d n o s o l v a b l e s u b g r o u p of A i s a F r o b e n i u s g r o u p . A l s o , every s u b g r o u p of A h a v i n g o r d e r p q , w h e r e p a n d q a r e ( p o s s i b l y e q u a l ) p r i m e s i s c y c l i c .

P r o o f . Let A act on N , where the act ion is Frobenius and N > 1. Suppose that A has a subgroup that is either an elementary abelian p-group of order p e > p or is a solvable Frobenius group. Since the act ion of every subgroup of A on N is Frobenius, we can replace A by the appropriate subgroup and assume that A itself is either elementary abelian of order p e or is solvable and Frobenius. A l so , i n the elementary abelian case, we can replace A by a subgroup if necessary, and so we can assume that e = 2.

In either case, A is solvable and acts coprimely on N , and so i f we choose any prime divisor r of \ N \ , Theorem 3.23 guarantees the existence of an A¬invariant Sylow r-subgroup R of N . T h e n R > 1, and so Z ( R ) > 1, and since Z ( R ) is characteristic i n R, this subgroup is A-invar iant . Since the action of A on Z ( R ) is Frobenius, we can replace N by Z ( R ) , and hence we can assume than N is abelian.

Since A is either elementary abelian of order p 2 or is a Frobenius group, A has a par t i t ion II of cardinal i ty 1 + n , where n is a divisor of \ A \ . (If A is elementary of order p 2 , then n = p , and if A is Frobenius, then n is the order of its Frobenius kernel.) Since n divides \ A \ , it is coprime to | N | , and thus since N is nontr iv ia l , some element « e J V satisfies un ^ 1. (In fact, we can take u to be any nonidenti ty element of N . ) It follows by L e m m a 6.8 that C N ( X ) > 1 for some member X G II. B u t since \X\ > 1, this contradicts the fact that the action of A on N is Frobenius, and we conclude that as claimed, the original group A cannot contain an elementary abelian p-group of order exceeding p or a solvable Frobenius group.

W h a t remains is to show that if B C A , where \ B \ = p q , and p and q are possibly equal primes, then B is cyclic . F i r s t , if p = q, then \ B \ = p 2 , and since we have shown that B eannot be elementary abelian, the only other possibi l i ty is that it is cycl ic , as required. F ina l ly , assume that p > q. T h e n B must have a normal Sylow p-subgroup P of order p. If Q G S y l , ( B ) and Q acts (by conjugation) nont r iv ia l ly on P , we have seen that the action of Q on P must be Frobenius. Thus B is a solvable Frobenius group, which we know is impossible. It follows that Q centralizes P , and it is easy to see in this case that B = P Q is cyclic . T h i s completes the proof. •

6 .10 . C o r o l l a r y . L e t P G Sy lp (A) , w h e r e A i s a F r o b e n i u s c o m p l e m e n t . T h e n P c o n t a i n s a t most one s u b g r o u p of o r d e r p .

P r o o f . We can assume that P > 1. T h e n Z ( P ) > 1, and we can choose Z C Z { P ) w i t h \ Z \ = p . We argue that Z is the only subgroup of order p

6 B 189

i n P . Otherwise, let U C P be a second subgroup of order p, and observe that Z U is a subgroup of order p 2 since Z < P . A l so , Z P is not cyclic since it contains two subgroups of order p , and this contradicts Theorem 6.9. •

We digress now to classify the p-groups that have at most one subgroup of order p, as in Corol la ry 6.10. Some obvious examples are the cyclic p-groups, and if p = 2, also the generalized quaternion 2-groups. These, we recall, are the 2-groups P of order at least 8 that have a cyclic subgroup C = (c) of index 2 and an element a of order 4 i n P - C such that ca = c~\ It is not hard to see that if P is generalized quaternion and C is cyclic of index 2 as above, then every element of P - C has order 4. The subgroup of order 2 i n C, therefore, is the unique subgroup of order 2 i n P .

In fact, cyclic p-groups and generalized quaternion 2-groups are the only p-groups that have just one subgroup of order p.

6.11. Theorem. L e t P be a p - g r o u p c o n t a i n i n g a t most one s u b g r o u p of o r d e r p . T h e n e i t h e r P i s c y c l i c , o r else p = 2 a n d P i s g e n e r a l i z e d q u a t e r n i o n .

W i t h only a l i t t le extra work, we can prove a stronger theorem, and so al though the next result is not direct ly relevant to Frobenius actions, it seems reasonable to present it here. We shall see that Theorem 6.11 is an immediate consequence.

It is easy to see that an a b e l i a n p-group w i t h at most one subgroup of order p must be cyclic. (This is immediate from the fundamental theorem of abelian groups.) It follows that i f P is any p-group that has at most one subgroup of order p, then a l l abelian subgroups of P are cyclic. In fact, if p ^ 2, the apparently much weaker assumption that a l l n o r m a l abelian subgroups of P are cyclic is sufficient to guarantee that P is cyclic . For p = 2, however, there are some possibilities other than cyclic and generalized quaternion groups.

6.12. Theorem. L e t P be a p - g r o u p i n w h i c h every n o r m a l a b e l i a n s u b g r o u p i s c y c l i c . T h e n e i t h e r P i s c y c l i c , o r e l s e p = 2 a n d P i s d i h e d r a l , g e n e r a l i z e d q u a t e r n i o n o r s e m i d i h e d r a l .

Reca l l that a 2-group P is dihedral if it has a cyclic subgroup C = ( c ) of index 2 and an involut ion a e P - C such that ca = c~\ L i k e dihedral and generalized quaternion 2-groups, a semidihedral 2-group also has a cyclic subgroup of index 2. Specifically, a 2-group P of order at least 16 is semidihedral if it has a cyclic subgroup C = (c) of index 2, and there is an involut ion a i n P - C such that ca = zc~1, where z is the unique element of order 2 i n C. (Note that we need to assume that \ P \ > 16 because i f | P | = 8, then c has order 4, and so zc~l = c. In that case, a would centralize C and P


would be abelian.) It is not hard to see that most dihedral 2-groups, as well as a l l generalized quaternion and semidihedral 2-groups actually do satisfy the hypotheses of Theorem 6.12. (The exception among dihedral groups is D 8 , of order 8, which has two noncyclic normal abelian subgroups of order 4-)

In a dihedral or semidihedral group, there is by definition at least one involut ion outside of the cyclic subgroup of index 2. These groups, therefore, have more than one subgroup of order 2, and so they do not satisfy the hypothesis of Theorem 6.11. It follows that Theorem 6.11 is a consequence of Theorem 6.12.

We begin work toward a proof of Theorem 6.12 w i t h the following.

6 .13 . L e m m a . L e t P be a n o n a b e l i a n 2 - g r o u p h a v i n g a c y c l i c s u b g r o u p C = (c) of i n d e x 2, a n d l e t a G P - C. If ca = c~l, t h e n P i s d i h e d r a l o r g e n e r a l i z e d q u a t e r n i o n , a n d if ca = zc~l, w h e r e z i s t h e u n i q u e i n v o l u t i o n i n C, t h e n P i s s e m i d i h e d r a l .

P r o o f . Since P is nonabelian and P = (a,c), the elements a and c cannot commute. In the case where ca = c~\ it follows that the order o(c) > 4 and i n the case where ca = zc~\ we have o(c) > 8. N o w if ca = c T 1 , then a inverts every element of C, and if ca = zc'1, then ( c 2 ) a = ( c 2 ) " 1 , and so a inverts every nongenerating element of C. In both cases, therefore, every element of order 4 i n C is inverted by a . Since a 2 G C is centralized by a, the group (a 2 ) can contain no element of order 4, and thus either o ( a ) = 2 or a 2 = z and o ( a ) = 4. If ca = c " 1 , therefore, P is either dihedral or generalized quaternion, respectively, and if ca = zc~l and o ( a ) = 2, then P is semidihedral. In the remaining case, a 2 = z and ca = zc'1, and thus (ca) 2 = c a c a = c a 2 c a = 1. T h e n o(ca) = 2 and since = ca, we can replace a by c a to see that P is semidihedral. •

A n easy consequence is the following description of the nonabelian groups of order 8, which we w i l l need i n what follows.

6 .14 . C o r o l l a r y . L e t P be a n o n a b e l i a n g r o u p of o r d e r 8. T h e n P i s i s o m o r p h i c e i t h e r t o t h e d i h e d r a l g r o u p D 8 o r t o t h e q u a t e r n i o n g r o u p Q8.

P r o o f . Since P is nonabelian, not every nonidenti ty element is an involut ion. We can thus choose c G P of order 4, and we write C = (c), so that C has index 2 in P . If a G P - C then since P is nonabelian, a cannot centralize c, and so a induces a nontr iv ia l automorphism of C. The cyclic group C of order 4, however, has only one nontr iv ia l automorphism, and we deduce that ca = c'1. The assertion now follows by L e m m a 6.13. •

6 B 191

6.15. L e m m a . L e t T be a p - g r o u p w i t h o r d e r d i f f e r e n t f r o m 8, a n d suppose t h a t \ T : Z ( T ) | = p 2 . L e t C be a c y c l i c s u b g r o u p such t h a t Z ( T ) < C < T . Then T has a c h a r a c t e r i s t i c e l e m e n t a r y a b e l i a n s u b g r o u p of o r d e r p 2 .

Proof. Since C < T is abelian and T / C is cycl ic , we can apply L e m m a 4.6 to conclude that \ C \ = | T ' | | Z ( T ) | . It follows that \T'\ = p , and thus T' C Z ( T ) , and T has nilpotence class 2.

F i r s t , suppose that p + 2. B y Theorem 4.8, the map 9 : T - > T denned by 9 { x ) = x ? is a homomorphism, and we let K = ker(0). T h e n K = { x G P | x P = 1} is a characteristic subgroup of T, and it suffices to show that \ K \ = p 2 . Since T / Z { T ) cannot be cyclic since T is nonabelian, this factor group of order p 2 must be elementary abelian. Thus 9 ( T ) C Z ( T ) , and it follows that \ K \ = \ T \ / \ 9 { T ) \ > | T | / | Z ( T ) | = p 2 . F ina l ly , \ K : K n C\ < \ T : C\ = p , and since K n C is a cycl ic subgroup of i f , we have | i f n C\ < p . It follows that \ K \ < p 2 , and we are done i n this case.

We can now assume that p = 2. In what follows, we w i l l twice use the easy observation that if A is a noncyclic abelian 2-group that has a cyclic subgroup of index 2, then { a G A | a 2 = 1} is a characteristic elementary abelian subgroup of order 4. F i r s t , applying this to the group T/T' (which is not cyclic because T' is central and T is not abelian) we obtain a characteristic elementary abelian subgroup E/T' of order 4, and thus E is characteristic i n T and | £ | = 8.

N o w E is noncyclic, and hence E ^ C . A l so , E±T since | T | / 8, and we conclude that E 2 C. It follows that £ n C is a proper subgroup of C, which, therefore, is contained i n Z { T ) . A l so , E n C i s a cyclic subgroup of E w i t h index 2, and hence E is abelian but not cyclic . It follows that E has a characteristic elementary abelian subgroup AT of order 4, and since £ is characteristic i n T, so too is K . Th i s completes the proof. •

Next , we present an elementary number-theoretic fact.

6.16. L e m m a . L e t p be p r i m e , a n d l e t e be a p o s i t i v e i n t e g e r . Suppose t h a t i i s a n i n t e g e r such t h a t i p EE 1 m o d p e . Then one of t h e f o l l o w i n g o c c u r s .

(a) i EE 1 m o d p * - 1 .

(b) p = 2 and i ==-1 m o d 2 e .

(c) p = 2 and i EE 2 6 " 1 - 1 mod 2 e .

Proof. F i r s t , suppose that p > 2. If i = 1, then (a) certainly holds, and so we can assume that i + 1, and we write i - l = m p t where p does not divide m and i > 0. We have i = i p = 1 m o d p , where the first congruence follows

by Fermat 's l i t t le theorem, and thus t > 0. A l so ,

ip = (1 + mp'y EE 1 + m p t + 1 + ( f y r n 2 p 2 t m o d p 3 t .

Since p is odd, the b inomia l coefficient (P) is divisible by p , and we have iP EE 1 + m p t + 1 m o d p 2 i + 1 . It follows that the highest power of p d iv id ing %P - 1 is p t + 1 . B y hypothesis, however, p e divides *P - 1, and thus e < t + l . T h e n e - 1 < t, and we have i = 1 m o d p e ~ l , as required.

N o w assume p = 2, so that 2 e divides i 2 - 1 = ( i - +1). Since one of the factors i - 1 or i + 1 is not divisible by 4, the other must be divisible by 2 6 " 1 , and thus i = ± 1 m o d 2 6 " 1 . Assuming that (a) fails, we have i = - 1 m o d 2 e ~ \ and we can write i = - 1 + a 2 e " 1 for some integer a . If a is even, then i EE —1 m o d 2 e and (b) holds. Otherwise, a = 2 b + 1 for some integer 6, and so i = - 1 + (26 + l ) 2 e - 1 EE 2e~l - 1 m o d 2 e , proving (c). •

We can now assemble the pieces.

P r o o f o f T h e o r e m 6 .12 . Let C be max ima l among abelian normal subgroups of P . B y hypothesis, C is cycl ic , so we can assume that C < P . A l s o , C = C P ( C ) since otherwise, we could find a subgroup B < P w i t h C C B C C p ( C ) , where | B : C | = p. Since C is central i n B and B / C is cycl ic , it would follow that B is abelian, which contradicts the choice of C. Thus P / C = P / C P { C ) is isomorphically embedded i n A u t ( C ) , which is abelian, and thus P / C is abelian. Also , if \ C \ = 4, it follows that \ P / C \ = 2 and \ P \ = 8, and by Coro l la ry 6.14, there is nothing to prove i n this case. We can assume, therefore, that \ C \ + 4, and we write \ C \ = p e .

Let T / C C P / C have order p , and observe that T < P since P / C is abelian. A l so , T is not abelian and \T\ / 8 since \ C \ / 4. Since T cannot have a characteristic elementary abelian subgroup of order p 2 , it follows by L e m m a 6.15 that C P cannot be central in T , where C = (c). Thus ( C P) a ^ where we have chosen a e T - C .

We can write ca = c i for some integer i. Since aP G C , it follows that

c = c = c ,

and since c has order p e , we conclude that %P EE 1 m o d p e . Furthermore, C P ^ ( c P ) a = ( c P f , and since c p has order p e ~ \ we conclude that i EE 1 mod p 6 " 1 . It follows by L e m m a 6.16 that p = 2 and i = - 1 m o d 2 6 " 1 . Thus

(c 2r = ( c 2 ) - 1 . Now P / C acts faithfully on C , and by the computat ion of the previous

paragraph, every involut ion i n the abelian group P / C acts to invert the element c 2 , which has order at least 4. It follows that there is only one such involut ion because if there were two, their product would be an involut ion

6 B 193

that centralizes and does not invert c 2 . Since P / C is an abelian 2-group w i t h just one involut ion, we conclude that P / C is cycl ic .

We argue next that \ P / C \ = 2. Otherwise, the element a is a square modulo C, and it follows that i = j 2 m o d 2 e for some integer j . B u t i = - 1 m o d 2 e " \ and since e > 3, it follows that j 2 = - 1 m o d 4, and this is not possible.

N o w P = T, and by L e m m a 6.16, either i = - 1 m o d 2 e or i = 2e~l - 1 m o d 2 e . In other words, either ca = c~l or ca = zc~\ where z = c 2 " " 1 is the involut ion i n C. The result now follows by L e m m a 6.13. •

We re turn now to the study of Frobenius complements.

6.17. Corol lary. Suppose t h a t A i s a F r o b e n i u s c o m p l e m e n t . T h e n each S y l o w s u b g r o u p of A i s c y c l i c o r g e n e r a l i z e d q u a t e r n i o n .

Proof. B y Coro l la ry 6.10, a Sylow p-subgroup of A has at most one subgroup of order p . B y Theorem 6.11, such a group must be cyclic or generalized quaternion. •

Observe that by Corol la ry 6.17, a l l Sylow subgroups of an abelian Frobenius complement A are cycl ic , and so A i tself must be cycl ic .

6.18. Corol lary. L e t A be a F r o b e n i u s c o m p l e m e n t of o d d o r d e r . T h e n A ' a n d A / A ! a r e c y c l i c a n d h a v e c o p r i m e o r d e r s .

Proof. B y Coro l la ry 6.17, a l l Sylow subgroups of A are cyclic . The result now follows by Theorem 5.16. •

We know more about the structure of a Frobenius complement than what its Sylow subgroups look like, and we can use this to derive more subtle information. For example, we have the following.

6.19. Theorem. L e t A be a F r o b e n i u s c o m p l e m e n t of o d d o r d e r . T h e n A has a u n i q u e s u b g r o u p of o r d e r p f o r each p r i m e p d i v i d i n g \ A \ .

Proof. F i r s t , consider a prime divisor r of \ A ' \ . T h e n r does not divide \ A / A ' \ by Coro l la ry 6.18, and thus every subgroup R of order r i n A actual ly lies i n A ' . Since A ' is cyclic, however, it has only one subgroup of order r , and thus R is unique, and i n part icular , R < A .

N o w let q be a prime divisor of \ A \ such that q does not divide \ A ' \ . T h e n q divides the order of the cyclic group A / A ' , and so A / A ' has a unique subgroup X / A ' of order q. T h e n every subgroup Q C A of order q lies i n X , and i n fact X = A ' Q . A l so , Q G S y l g ( A ) since q does not divide \ A ' \ . To prove that Q is unique, therefore, it suffices to show that Q < X .

Let R C A ' be an arbi trary subgroup of prime order, say r . We know that R < A , and thus R Q is a subgroup of A of order r q . It follows that R Q is cyclic by Theorem 6.9, and in part icular, Q centralizes R. Thus C A > ( Q ) contains every subgroup of A ' having prime order.

Since A ! is abelian and has order coprime to q, it follows by F i t t ing ' s theorem that A ' = [ A ' , Q] x C A . ( Q ) . Since the second factor contains every prime order subgroup of A ' , the first factor can contain no such subgroup, and we conclude that [ A ' , Q ] = 1. Thus Q centralizes A ' , and so Q < A ' Q = X , as required. •

Next , we present two results that are needed in Chapter 7.

6 .20 . L e m m a . Suppose t h a t A i s a finite a b e l i a n g r o u p t h a t acts f a i t h f u l l y o n a finite g r o u p N , w h e r e \ A \ a n d \ N \ a r e c o p r i m e . Assume i n a d d i t i o n t h a t t h e a c t i o n of A o n every p r o p e r A - i n v a r i a n t s u b g r o u p of N i s t r i v i a l . T h e n A i s c y c l i c .

We w i l l deduce this as a corollary of the following.

6 . 2 1 . T h e o r e m . Suppose t h a t A i s a finite a b e l i a n g r o u p a n d t h a t i t acts o n a finite g r o u p N , w h e r e \ A \ a n d \ N \ a r e c o p r i m e . If A i s n o t c y c l i c , t h e n

N = { C N ( a ) | 1 / a € A ) .

P r o o f . The assertion is t r i v i a l i f N = 1, so we assume that TV > 1, and we proceed by induct ion o n \ N \ . Wr i t e

K = { C N ( a ) | 1 / a € A ) ,

so that our goal is to show that K = N .

If M is a proper subgroup of N that admits the action of A , then by the inductive hypothesis,

M = ( C M { a ) | 1 / a G A ) C A T ,

and so K contains every proper A-invariant subgroup of N . N o w assume that K < N , and choose a prime divisor p of \ N : K \ . Since A is certainly solvable, and its order is coprime to \ N \ , we know that there exists some A¬invariant Sylow p-subgroup P of N , and we see that P % K since p divides |TV : K \ . It follows that P = N , and so TV is a p-group, and thus N ' is a proper A-invar iant subgroup. T h e n N ' C K , and it follows that K < N . Also , K is A-invar iant since it is uniquely determined by A and N .

N o w consider the induced action of A on N = N / K . Since "fixed points come from fixed points" in coprime actions, it follows that C^ (a ) = C N ( a ) for a l l elements a <E A . B u t i f 1 ^ a <E A , then C N ( a ) C K , and so C N ( a ) is t r i v i a l . It follows that C w { a ) is t r iv ia l , and thus the action of A on the

P r o b l e m s 6 B 195

nontr iv ia l group N is Frobenius. B u t A is not a Frobenius complement since it is abelian but not cyclic. Th i s contradict ion completes the proof. •

P r o o f o f L e m m a 6 .20 . Since A is abelian, the subgroup C N ( a ) is A¬invariant for a l l a G A . (This is because C N { a ) is uniquely determined by the element a, which is invariant under conjugation by A . ) A l so , if a + 1, then C N ( a ) < N because the act ion of A on N is faithful, and thus by hypothesis, C N ( a ) C C N { A ) for a l l nonidenti ty elements a G A . If A is not cycl ic , however, then by Theorem 6.21, the subgroups C N { a ) generate N , and thus N = C N ( A ) . Since A acts faithfully, this yields A = 1, and this is a contradict ion, since we assumed that A is not cyclic. The result now follows. •

P r o b l e m s 6 B

6 B . 1 . Show that every Frobenius group contains a solvable Frobenius subgroup, and deduce that a Frobenius complement cannot have a Frobenius group as a subgroup.

6 B . 2 . Let A act faithfully on N , where A is abelian, and assume that no nonidenti ty proper subgroup of N is A invariant. Show that A is cycl ic .

H i n t . Let p be a prime divisor of \ N \ and let P G S y l p ( A ) . T h e n C N ( P ) is A-invar iant and nontr iv ia l .

6 B . 3 . Show that if the coprimeness assumption i n Theorem 6.21 is dropped, the result becomes false.

6 B . 4 . Suppose that G has a par t i t ion IT such that [ X , Y] = 1 for every two dist inct members X and Y of IT.

(a) Show that G is abelian.

(b) Show that the elements of G have equal prime orders, so that G is elementary abelian.

H i n t . For (b), consider x G X G IT and y G Y G IT, where X / Y. If o { x ) < o ( y ) , consider the element { x y ) o l - x \ In which member of IT does it lie?

N o t e . T h e hypotheses of the previous problem are automatical ly satisfied if the members of IT are a l l normal i n G.

6B.5. Suppose that G has a par t i t ion consisting of subnormal subgroups. Show that G is nilpotent.

Hint . W o r k i n g by induct ion on \ G \ , show that every subgroup H < G that is not contained i n some member of the par t i t ion is nilpotent. Deduce that every member of the par t i t ion that is not contained i n F ( G ) is normal i n G and has prime index.

6B.6. Le t C be a cyclic 2-group of order at least 8. Show that C has exactly three automorphisms of order 2.

6B.7. Let P be a nonabelian 2-group that has a cyclic subgroup of index 2. If \ P : Z ( P ) | > 4, show that P is dihedral , semidihedral or generalized quaternion.

6B.8. Let P be a 2-group of order at least 8, and assume that \ P : P ' \ = 4. Show that P is dihedral , semidihedral or generalized quaternion.

H i n t . Let Z C G ' n Z ( G ) have order 2. Work ing by induct ion on deduce that P has an abelian subgroup A of index 2 and that A is either cyclic or is the direct product of Z w i t h a cyclic group. In the latter case, and assuming that | P | > 16, deduce that Z < Z ( P ) and derive a contradict ion.

Note . Th i s problem proves a theorem of O. Taussky-Todd.

6B.9. Let G be solvable, and assume that every element of G has prime-power order. Show that | G | is divisible by at most two primes.

6 C

We now begin work toward a proof of the theorem of Thompson 's P h . D . thesis: Frobenius kernels are nilpotent. We prove this first under the addit ional assumption that the given Frobenius kernel is solvable. (This special case, which is needed for Thompson's result, appears i n a 1957 paper of G . H igman , where it is described as being already known.)

6.22. Theorem. L e t N be a s o l v a b l e F r o b e n i u s k e r n e l . T h e n N i s n i l p o t e n t .

Proof. B y hypothesis, there exists a nontr iv ia l group A having a Frobenius action on N . Since the action of every subgroup of A on N is also Frobenius, we can replace A by a subgroup of prime order, and so we can assume that \ A \ = p , & prime. Let G = N x > A , and as usual, view N and A as subgroups of G , where N < G is complemented by A .

Assuming that N is not nilpotent, we proceed by induct ion on |7V|. Since N > 1, we can choose a min ima l normal subgroup U of G, w i th U C N . The induced action of A on N / U is Frobenius by Theorem 6.2, and of course

6 C 197

N / U is solvable. B y the inductive hypothesis, therefore, N / U is nilpotent, and thus 7V°° C U. B u t N°° > 1 since N is not nilpotent, and of course, 7V°° < G. T h e min imal i ty of U, therefore, yields U = N°°, and thus U is the unique m i n i m a l normal subgroup of G contained in N . (Any other one would also have to be N°°.)

Since U is solvable and is min ima l normal i n G, L e m m a 3.11 guarantees that U must be an abelian g-group for some prime q. B u t N is not a g-group since i t is not nilpotent, and so we can choose a prime divisor r / q of \ N \ . Let R e S y \ r ( N ) , and choose R to be A-invar iant . (Al though the existence of an A-invar iant Sylow r-subgroup follows v i a Theorem 3.23, we do not really need the full strength of that theorem here. Th i s is because | A | = p and |SyL.( /V) | is not divisible by p, and so a simple counting argument suffices to find an A-invar iant Sylow r-subgroup of N . ) Observe that R is not normal i n N since, otherwise, it would be characteristic, and hence normal in G. B u t R 2 U, and so this would contradict the fact that U is the unique min ima l normal subgroup of G contained in AT.

N o w R U / U is a Sylow r-subgroup of the nilpotent group N / U , and thus R U / U < N / U and R U < N . A l so , R U admits the action of A , and so R U is a solvable Frobenius kernel. If R U < N , then R U is nilpotent by the inductive hypothesis, and hence R is characteristic i n R U , and hence R < N , which is a contradiction. We deduce that R U = N .

Since A R is a Frobenius group, it has a par t i t ion II consisting of R and the \ R \ conjugates of A in A R . T h e n | H | - 1 = \ R \ is a power of r , and so this number is coprime to \ U \ . B y L e m m a 6.8, therefore, some member X e II has nontr iv ia l fixed points i n the abelian group U. The action of every G-conjugate of A on U is Frobenius, however, and so X cannot be one of the conjugates of A . We deduce that X = R , and thus since U is abelian and R U = N , we have

1 < C u ( R ) C Z { R U ) = Z ( N ) .

Since U is the unique min ima l normal subgroup of G contained in N and Z { N ) is nont r iv ia l and normal in G, we conclude that U Q Z ( N ) , and thus R < U R = N . Th i s is a contradiction, and the proof is complete. •

Next , we state a result of Thompson that w i l l enable us to prove the theorem of his thesis, which was that a l l Frobenius kernels are nilpotent (and not just solvable Frobenius kernels, as i n Theorem 6.22). Actua l ly , the main result of Thompson's thesis was a cri ter ion for a group to have a normal p-complement, where p is an odd prime; the result about Frobenius kernels is a comparatively easy corollary. A b o u t five years after he completed his degree, Thompson found a much better version of his normal p-complement theorem w i t h a very much easier proof, and that is the result we present and

prove i n Chapter 7. Here we give only a par t ia l statement of this improved theorem.

6 .23 . T h e o r e m (Thompson) . L e t P e S y \ p { G ) , w h e r e G i s a finite g r o u p a n d p + 2, a n d assume t h a t N G ( X ) has a n o r m a l p - c o m p l e m e n t f o r eve r y n o n i d e n t i t y c h a r a c t e r i s t i c s u b g r o u p X of P . T h e n G has a n o r m a l p -c o m p l e m e n t .

Firs t , observe that this can be viewed as a strong form of Frobenius ' normal p-complement theorem, which is our Theorem 5.26. Reca l l that Frobenius ' theorem tells us that to prove that G has a normal p-complement, it suffices to show that N G ( X ) has a normal p-complement for every nonident i ty p-subgroup X of G. It is a t r iv ia l i ty (by Sylow theory) that in Frobenius ' theorem, it is enough to consider only p-subgroups X that are contained in some fixed Sylow p-subgroup P of G. The force of Theorem 6.23 is that for p ^ 2, it suffices to restrict attention to nonidenti ty c h a r a c t e r i s t i c subgroups X of P ; it is not necessaty to consider a l l nonidenti ty subgroups of P . (The normal p-complement theorem of Thompson 's thesis can also be viewed as a strong form of Frobenius ' theorem, val id for p ^ 2. It is somewhat more complicated to state, however.)

We mention that the condi t ion p ^ 2 in Theorem 6.23 is essential. C o n sider, for example, G = 5 4 , the symmetric group of order 24. A Sylow 2-subgroup P of G is isomorphic to the dihedral group D 8 , and the non-identity characteristic subgroups of P are P , the unique cyclic subgroup of order 4 i n P and Z ( P ) , of order 2. It is easy to check that the normalizer i n G of each of these subgroups is P itself, and so these normalizers a l l have normal 2-complements. The group G, however, does not have a normal 2-complement.

A s we shall see i n Chapter 7, it is not actually necessary to consider a l l nonidenti ty characteristic subgroups X of P ; two of them suffice. One of these is the center Z ( P ) , and the other is the Thompson subgroup J ( P ) , which we w i l l define. Of course, if P > 1, then Z ( P ) > 1, and as we shall see, also J ( P ) > 1.

Assuming Theorem 6.23, we now present the main result of this section.

6 .24. T h e o r e m . L e t N be a F r o b e n i u s k e r n e l . T h e n N i s n i l p o t e n t .

P r o o f . A s i n the proof of Theorem 6.22, there exists a group A of prime order p such that A has a Frobenius action on N . Proceeding by induct ion on \ N \ , suppose first that there exists a normal A-invar iant subgroup M of N w i t h 1 < M < N . T h e n the action of A on M is Frobenius, and by Corol la ry 6.2, so too is the action of A on N / M . B y the inductive hypothesis, therefore, bo th M and N / M are nilpotent, and thus N is solvable. It follows

P r o b l e m s 6 C 199

by Theorem 6.22 that N is nilpotent, and we are done i n this case. We can assume, therefore, that N contains no nont r iv ia l proper A-invar iant normal subgroup.

We argue now that N is an r-group for some prime r , and thus N is nilpotent, as wanted. Otherwise, \ N \ has at least two prime divisors, and so we can choose an odd prime r d iv id ing |7V|. Le t R be an A-invar iant Sylow r-subgroup of N . (As i n the proof of Theorem 6.22, the existence of an A-invar iant Sylow r-subgroup follows by a simple counting argument because | A | is prime.)

Le t AT be a nonidenti ty characteristic subgroup of R. Since R is A¬invariant, it follows that X is also A-invar iant . Furthermore, X < N because N is not an r-group, and so by the result of the previous paragraph, X cannot be normal in N . T h e n N N ( X ) is a proper A-invar iant subgroup of N , and so by the inductive hypothesis, N N ( X ) is nilpotent, and hence it has a normal r-complement. (Every nilpotent group has a normal r-complement.) It follows by Theorem 6.23 that N has a normal r-complement AT, and we have 1 < K < N since N is not an r-group, but it does have order divisible by r . A l so , K is characteristic i n N , and so it is A-invar iant . Th is contradicts the result of the first paragraph, and so the proof is complete. •

P r o b l e m s 6 C

6 C . 1 . Let a G A u t ( G ) , and suppose that a fixes only the identity in G .

(a) If a has prime order, show that G is nilpotent.

(b) Show by example that G need not be nilpotent if a has order 4.

H i n t . There exists an example for (b) w i t h | G | = 75.

6 C . 2 . Le t A be elementary abelian of order p 2 , and suppose that A acts on a nonnilpotent group N v i a automorphisms. Assume that C N ( A ) = 1.

(a) Show that N has has nilpotent subgroups K t > 1 for 1 < i < p + 1 such that K i n K 5 = 1 when i ^ j .

(b) N o w fix a prime q, and let Q i be the unique Sylow g-subgroup of K u

where the AT; are the subgroups of (a). Let X = ( Q i | 1 < i < p + l ) . Show that X e Syl g (7V).

H i n t . For (b), consider an A-invar iant Sylow g-subgroup Q of N . Show first that A C Q . If X < Q, show that X C Y < Q, for some A-invar iant subgroup Y that is proper i n Q. Show that the action of A on Q/Y is Frobenius.

C h a p t e r 7

The Thompson Subgroup

7 A

In this chapter, we complete the proof of Thompson 's theorem that Frobe-nius kernels are nilpotent. The missing ingredient i n our par t ia l proof i n Chapter 6 is to show that i f p ^ 2, then a sufficient condi t ion for a group G to have normal p-complement is that for every nonidenti ty characteristic subgroup X of some Sylow p-subgroup P of G, the subgroup N G ( X ) has a normal p-complement. (Since every subgroup of a group that has a norm a l p-complement must also have a normal p-complement, this is clearly a necessary condi t ion too, but, of course, that fact is far less interesting.)

Thompson 's first normal p-complement cri ter ion appeared i n his 1959 P h . D . thesis, w i th a long and subtle proof. T h e n i n 1964, Thompson published a much shorter (though s t i l l subtle) proof of a stronger theorem, and that is the theorem and proof we present i n this chapter. (Thompson's 1964 paper appeared i n the very first issue of the J o u r n a l of A l g e b r a , which was an auspicious beginning for that periodical.) Thompson showed i n his J o u r n a l of A l g e b r a paper that i f p + 2, then a group G has a normal p-complement provided that N G ( X ) has a normal p-complement for just two specific characteristic subgroups X of P , where P e S y l p ( G ) . These two cr i t ica l characteristic subgroups are the center Z ( P ) and the Thompson subgroup J ( P ) , bo th of which are nontr iv ia l i f P is nont r iv ia l . (We know, of course, that the center of a nontr iv ia l p-group is nont r iv ia l , and as we shall see when we present the definition, the Thompson subgroup of a nontr iv ia l p-group is also guaranteed to be nontrivial .) O f course, i f the normalizers

201

202 7. T h e T h o m p s o n S u b g r o u p

of a l l nont r iv ia l characteristic subgroups of P have normal p-complements, then i n part icular , N G ( Z ( P ) ) and N G ( J ( P ) ) have normal p-complements, and so Thompson 's 1964 theorem yields our Theorem 6.23, which was the key to the proof that Frobenius kernels are nilpotent i n Chapter 6. In fact, Thompson 's 1964 theorem is sl ightly stronger than we have just stated since we can replace the normalizer of Z ( P ) w i t h the centralizer of Z ( P ) . (Since C G ( Z ( P ) ) C N G ( Z ( P ) ) , the requirement that the centralizer should have a normal p-complement is weaker than, and is impl ied by, the corresponding assumption for the normalizer.)

7 . 1 . T h e o r e m (Thompson) . L e t P € S y l p ( G ) ; w h e r e G i s a finite g r o u p a n d p + 2, a n d assume t h a t C G ( Z ( P ) ) a n d N G ( J ( P ) ) h a v e n o r m a l p -c o m p l e m e n t s . T h e n G has a n o r m a l p - c o m p l e m e n t .

A s we have explained, once we complete the proof of Theorem 7.1, we w i l l have established that a l l Frobenius kernels are nilpotent. The Thompson subgroup J ( P ) has other applications too, and i n part icular, it provides a crucial ingredient i n the group-theoretic proof of Burnside 's pV-theorem that we present later i n this chapter. For that reason, we prove somewhat more about the Thompson subgroup than the m i n i m u m needed for the proof of Theorem 7.1.

Unfortunately, the literature contains several sl ightly different definitions of "the" Thompson subgroup of a p-group P , and i n general, these definitions y ie ld different subgroups, a l l of which have been referred to as J ( P ) . In part icular , the version of J ( P ) that we are about to define is not always equal to Thompson 's J ( P ) . Our definition is sl ightly easier to use than that i n Thompson 's paper, however, and we feel that this justifies the risk of confusion caused by competing definitions.

G i v e n a p-group P , let £(P) be the set of a l l of those elementary abelian subgroups of P that have the m a x i m u m possible order. T h e n the T h o m p son subgroup J ( P ) of P is the subgroup generated by a l l of the members of £{P). For example, i f P is the dihedral group of order 8, then obviously, P does not contain an elementary abelian subgroup of order 8. B u t P does contain an elementary abelian subgroup of order 4, and so £{P) is the set of a l l elementary abelian subgroups of order 4 i n P . There are two of these, and together they generate the whole group P , and thus J ( P ) = P i n this case. O f course, if P is nontr ivia l , then it must contain a nontr iv ia l elementary abelian subgroup, for instance, one of order p , and thus the set £(P) contains at least one nontr iv ia l subgroup. It follows that J ( P ) > 1, as we mentioned previously.

In his 1964 paper, Thompson d id not restrict attention to elementary abelian subgroups. Instead, he defined J ( P ) to be the subgroup generated by

7 A 203

al l abelian subgroups of P of largest possible rank. (The r a n k of an abelian p-group A is the integer r such that the unique largest elementary abelian subgroup i i i ( A ) has order p r . Equivalent ly, if the abelian p-group A is decomposed as a direct product of nontr iv ia l cyclic factors, then the rank of A is the number of such factors.) Since an elementary abelian subgroup w i t h m a x i m u m possible order in P is an abelian subgroup of m a x i m u m possible rank, we see that our subgroup J ( P ) is always contained in Thompson's , but it can be properly smaller. (The containment is proper, for example, if P is abelian, but not elementary abelian.) Yet another variat ion on the definition that has appeared i n the li terature is to consider the subgroup of P generated by a l l abelian subgroups of largest possible order.

A n immediate consequence of the definition is the following.

7 .2 . L e m m a . L e t 3 { P ) C Q C P , w h e r e P i s a p - g r o u p . T h e n J ( P ) = J ( Q ) , a n d i n p a r t i c u l a r , J ( P ) i s c h a r a c t e r i s t i c i n Q.

P r o o f . Since (£(P)) = J ( P ) C Q, it follows that every maximal-order elementary abelian subgroup of P is contained i n Q, and because Q C P , these are maximal-order elementary abelian subgroups of Q. Every maximal-order elementary abelian subgroup of Q, therefore, has the same order as the members of £{P), and hence is a member of £{P). It follows that £{Q) = £(P), and thus 3 ( G ) = J ( P ) . •

Before we can begin our proof of Theorem 7.1, we need a number of pre l iminary results. T h e first of these is a fairly technical l emma about the general linear group G L ( 2 , p ) , which, we recall , is the group of invertible 2 x 2 matrices over the field F of order p.

7 .3 . L e m m a . L e t G = G L ( 2 , p ) , w h e r e p ^ 2 i s p r i m e , a n d l e t P C G be a p - s u b g r o u p . Suppose P C N G ( L ) , f o r some s u b g r o u p L of G, w h e r e \ L \ i s n o t d i v i s i b l e by p , a n d assume f u r t h e r t h a t a S y l o w 2 - s u b g r o u p of L i s a b e l i a n . T h e n P C C G ( L ) .

The condit ion that p ^ 2 i n L e m m a 7.3 is necessary since G L { 2 , 2 ) is isomorphic to the symmetric group 5 3 , and so there is a p'-group L of order 3 normalized but not centralized by a Sylow p-subgroup P of order 2. The somewhat unnatural requirement that L should have an abelian Sylow 2-subgroup also cannot be omit ted, at least when p = 3. If G = G L { 2 , 3), then G has a subgroup L of order 8 that is normalized but not centralized by a Sylow 3-subgroup P , which has order 3. (The product L P of order 24 is the special linear group S L ( 2 , 3), which is the set of matrices w i t h determinant 1 i n G.) Th i s is not a counterexample to L e m m a 7.3, however, since L is a nonabelian 2-group, isomorphic to the quaternion group Q 8 . In fact, i t is only for p = 3 that such an example can occur, and so L e m m a 7.3 could be


strengthened by requiring the condit ion on a Sylow 2-subgroup of L only if p = 3. W h e n we apply the lemma, however, we w i l l know that L is abelian, and so we have no need for a stronger result.

We digress briefly to discuss general linear groups i n general, and also special linear groups and certain other related groups. (We saw some of these groups i n Chapter 1, and we study them further i n Chapter 8.)

If F is an arbi t rary field and n is a positive integer, then G L ( n , F ) is the group of a l l invertible n x n matrices over F , and the special linear group S L { n , F ) is the subgroup of G L { n , F ) consisting of those matrices that have determinant equal to 1. Since the determinant map is a homomorphism from G L ( n , F ) onto the mult ipl icat ive group F x of F , it follows that S L ( n , F ) , which is the kernel of this homomorphism, must be a normal subgroup. Al so , we have G L { n , F ) / S L { n , F ) = F x .

N o w suppose that F is finite, so that \ F \ = q, where q is a prime power. Since F is the unique field (up to isomorphism) of order q, it is no loss to write G L ( n , q ) in place of G L ( n , F ) , and i n fact, this notat ion is fairly common. Similar ly , we write S L { n , q ) for the corresponding special linear group, which is a normal subgroup of index | F X | = q - 1 i n G L ( n , q ) .

We can compute \ G L { n , q ) \ by counting the number of ways we can construct a n n x n mat r ix w i t h l inearly independent rows, where each entry comes from the field F of order q. O f course, there are a total of qn row vectors of length n over F . E a c h of these other than the zero row is a possibil i ty for the first row of our matr ix , and so there are exactly qn - 1 possible first rows. To keep the rows linearly independent, we must not allow the second row to be one of the q different scalar multiples of the (nonzero) first row, so after the first row is chosen, there are exactly qn - q possibilities for the second row. After the first two (linearly independent) rows are selected, we must exclude a l l of their q2 different linear combinations from appearing as the th i rd row, and this leaves qn - q2 possibilities for the th i rd row. Cont inu ing like this, we see that there are exactly q n - q k ~ 1 possibilities for the k t h row, and we deduce that

n—1 n

\ G L ( n , q ) \ = l\(qn - q l ) = qN l [ ( q i - I ) , i=0 j = l

where N = 1 + 2 + • • • + (n - 1) = n ( n - l ) / 2 . In part icular, | G L ( 2 , q ) \ = ( q 2 - l ) ( q 2 - q ) = q ( q - \ ) 2 { q + 1), and | 5 L ( 2 , ? ) | = q { q - l ) ( q + 1).

N o w consider the subgroup P C G L { n , q ) consisting of the matrices that have zeros below the diagonal, ones on the diagonal and arbi t rary elements of F i n a l l above-diagonal positions. (Note that P really is a subgroup, and that P C S L ( n , q ) . ) Since row k of an n x n mat r ix contains exactly k - 1 above-diagonal positions, we see that the total number of above-diagonal

7 A 205

positions i n an n x n mat r ix is 1 + 2 + • • • + (n - 1) = N , where N is as before. T h e n \ P \ = qN, and thus \ G L ( n , q ) \ = \ P \ m , where m is a product of numbers of the form g> - 1, and so m is relatively pr ime to q. Since q is a power of some prime, say p , we see that the subgroup P is a Sylow p-subgroup of G L { n , q ) . Note that i f n > 1, we can easily find a second Sylow p-subgroup, different from P , by tak ing lower tr iangular matrices in place of the upper triangular subgroup P .

T h e scalar matrices i n G L ( n , q ) are the scalar multiples of the ident i ty matr ix , and so there are exactly q - 1 of these, and it is easy to see that these matrices form the center Z ( G L ( n , q ) ) . The factor group G L { n , q ) / Z ( G L { n , q ) ) is the projective general linear group, and is denoted P G L ( n , q ) . The scalar matrices that lie in S L ( n , q ) are the matrices of the form a - 1 , where a n = 1, and in fact, these form the center Z ( 5 L ( 2 , g ) ) . W r i t i n g Z = Z { S L { n , q ) ) , we see that \ Z \ is equal to the number d of elements a G F x such that a n = 1. Since F x is cyclic of order q - 1, it follows that d = ( n , q - 1), the greatest common divisor. B y definition, the projective special linear group P S L ( n , q ) is the factor group S L { n , q ) / Z . For n > 2, this group is simple except when n = 2 and q is 2 or 3. (In fact, P S L ( n , F ) is also simple when n > 2 and F is an infinite field.)

Tak ing n = 2, we have \ P S L ( 2 , q ) \ = q ( q - l ) ( g + l ) / d , where d = 1 i f q is a power of 2 and d = 2 if q is odd. In part icular , for pr ime powers q = 4 , 5 , 7 , 8 , 9 , 1 1 we have \ P S L ( 2 , q ) \ = 60,60,168,504,360,660, respectively, and these account for a l l of the nonabelian simple groups of order at most 1000. (We have not omit ted the al ternating groups A 5 and A s since P S L ( 2 , 4 ) = A s = P S L { 2 , 5 ) and P S L { 2 , 9 ) = A 6 . ) We mention also that P S L { 2 , 7 ) = P S L { 2 > , 2 ) and A 8 = P 5 L ( 4 , 2 ) . In fact, these are the only isomorphisms among alternating groups and the simple groups of type P S L ( n , q ) . (The simple groups P 5 L ( 3 , 4 ) and A 8 have equal orders, but they are not isomorphic.)

F ina l ly , we mention one way that general linear groups arise i n abstract group theory. Consider an elementary abelian p-group E of order p n , and view E as an additive group. T h e n E can be identified w i t h a vector space of dimension n over the field of order p , and under this identification, we see that A u t ( F ) = G L { n , p ) . T h i s observation (in the case n = 2) is crucial to the proof of Theorem 7.1, and that is why L e m m a 7.3 is relevant.

In order to prove L e m m a 7.3, we need the following observation about the groups S L { 2 , q ) , where q is odd.

7.4. L e m m a . If q i s o d d , t h e n t h e n e g a t i v e of t h e i d e n t i t y m a t r i x i s t h e u n i q u e i n v o l u t i o n i n S L { 2 , q ) .


This can be proved by a computat ional argument along the following lines. If t is an involut ion i n S L { 2 , q ) , then the fact that t2 = 1 yields four nonlinear equations that must be satisfied by the four entries of the mat r ix t, and the fact that det(t) = 1 yields one more equation. It is not hard to show that if the characteristic is different from 2, these five equations have just two solutions, corresponding to t = I and t = - I , where I is the 2 x 2 identi ty matr ix . We prefer the following more conceptual proof, however.

P r o o f o f L e m m a 7.4. Let t € S L ( 2 , q ) be an involut ion other than - / . Wr i t e f { X ) to denote the polynomia l X 2 - 1, and observe that f { t ) = 0. Since neither the polynomia l X - 1 nor the polynomia l X + 1 yields 0 when t is substi tuted for X , it follows that f ( X ) = X 2 - l is the min ima l po lynomia l for t. The Cay ley -Hami l ton theorem asserts that the min ima l po lynomia l of an arbi t rary square mat r ix divides the characteristic po lynomia l , and i n our si tuation, where t is a 2 x 2 matr ix , we know that the characteristic polynomia l of t has degree 2, and hence f ( X ) is the characteristic po lynomia l of t. T h e n t has two distinct eigenvalues, 1 and - 1 , and so the product of the eigenvalues of M s - 1 . B u t the product of the eigenvalues (counting mult iplici t ies) for an arbi trary square mat r ix is the determinant of the matr ix , and it follows that det(t) = - 1 . B u t t 6 S L { 2 , q ) , and so det(t) = 1, and this contradict ion completes the proof. •

It follows by Theorem 6.11 that a Sylow 2-subgroup of S L { 2 , q ) (for odd q ) is either cyclic or generalized quaternion. A n elementary argument that shows that the cyclic case can never occur depends on the fact that i n a finite field F of odd order q, it is always possible to find elements a and b such that a 2 + b2 = - 1 . Assuming that a and b satisfy this condit ion, consider the two matrices:

a b b — a

These clearly have determinant 1, and so they lie i n S L { 2 , q ) . It is easy to check that x 2 = - I = y 2 and that y x = -y = y ~ \ and it follows that the subgroup ( x , y ) of S L ( 2 , q ) is isomorphic to the quaternion group Q 8 . A Sylow 2-subgroup of S L { 2 , q ) cannot be cyclic, therefore, and so it must be generalized quaternion. We mention also that since \ S L ( 2 , 3)| = 24, a Sylow 2-subgroup of S L ( 2 , 3) is isomorphic to Q8.

To see why the equation a 2 + b2 = - 1 must have a solution i n F , observe that the number of elements of F that have the form 1 + a 2 is exactly ( q + l ) / 2 . (This is so because the map a i-> 1 + a 2 is two-to-one for nonzero elements of F , and so the image of this map contains (q - l ) / 2 elements different from 1, for a total of (q + l ) / 2 elements.) Similar ly, there are exactly ( q + l ) / 2 elements of F that have the form - b 2 , and since the

0 - 1 1 0

and y

7 A 207

sum of the cardinalities of these two sets exceeds g, the sets must overlap. It follows that 1 + a 2 = - b 2 for some choice of a , b € F , as wanted.

P r o o f o f L e m m a 7 .3 . Work ing by induct ion on \ L \ , we can assume that P centralizes every proper subgroup of L that it stabilizes. If we choose a prime q d iv id ing | L : C L ( P ) | , we can find a P - invar iant Sylow g-subgroup Q of L , and since Q % C L ( P ) , we must have Q = L , and thus L is a g-group. Also , [ L , P ] is P - invar iant , and so i f [ L , P ] < L , we have [ L , P , P ] = 1, and since [ L , P , P] = [ L , P] by L e m m a 4.29, we conclude that [L, P] = 1, as required. We may assume, therefore, that [ L , P] = L , and i n part icular, L C G' C S L { 2 , p ) , where the second containment follows because G L ( 2 , p ) / S L ( 2 , p ) is isomorphic to the mult ipl icat ive group of the field of order p , and so it is abelian.

If q = 2, then by assumption, the 2-group L is abelian, and we know by L e m m a 7.4 that L contains a unique involut ion. It follows that L is a cyclic 2-group, and so A u t ( L ) is also a 2-group, and the p-group P cannot act nontr iv ia l ly on L . We can assume, therefore, that q is an odd prime, and since \ L \ is a power of q that divides \ S L ( 2 , p ) \ = p { p - l ) ( p + 1), it follows that \ L \ divides one of p - 1 or p + 1. (This , of conrse, is because the prime q cannot divide both p - 1 and p + 1.) In part icular , we have \ L \ p + 1. We conclude that \ L \ = p + 1, and so \ L \ is even. B u t \ L \ is a power of the odd prime q, and this contradict ion completes the proof. •

Next , we prove what we cal l the "normal -P theorem".

7 .5 . T h e o r e m . L e t P G S y l p ( G ) , w h e r e G i s a finite g r o u p , a n d assume the f o l l o w i n g c o n d i t i o n s .

(1) G i s p - s o l v a b l e .

(2) p + 2.

(3) A Sylow 2 - s u b g r o u p of G i s a b e l i a n .

(4) G acts f a i t h f u l l y o n some p - g r o u p V.

(5) \ V : C v ( P ) \ < p .

Then P < G .

T h e reader may wonder how Theorem 7.5, which applies only to p -solvable groups, can possibly be relevant to the proof of Thompson 's theorem (7.1), where there certainly is no p-solvabil i ty assumption. The answer is that the proof of Theorem 7.1 proceeds v i a a subtle induct ion that w i l l guarantee p-solvabili ty when we need it . (Thompson's inductive argument

208 7. The Thompson S u b g r o u p

w i l l also guarantee that the Sylow 2-subgroup is abelian when that fact is needed.)

P r o o f o f T h e o r e m 7.5 . Assume that G is a counterexample of m i n i m u m possible order. T h e n G does not have a normal Sylow p-subgroup, and we can choose Q € S y l p ( G ) w i t h Q + P . The subgroup ( Q , P ) of G has at least two Sylow p-subgroups, and it clearly satisfies a l l five hypotheses of the theorem. It too is a counterexample, therefore, and so by the min imal i ty of G, we have G = ( P , Q ) .

N o w Q = P 9 for some element g € G, and thus C V { Q ) = { C V ( P ) ) 9

also has index at most p i n V. Let U = C V ( P ) n C V { Q ) , and observe that \ V : U \ < \ V : C V { P ) \ \ V : C V ( Q ) \ < p 2 , and U < V. A l so , since both P and Q act t r iv ia l ly on U, it follows that the action of G = ( P , Q ) on U is t r iv ia l , and so [ U , G ] = 1. In part icular, U is G-invariant , and we have a natural action of G on V/U.

Let K be the kernel of the action of G on V/U, so that [V, K ] C U, and thus [ V , K , K ] C [ U , K ] = 1. Since V is a p-group, it follows that K must also be a p-group. (If Q is a Sylow g-subgroup of AT, where q ^ p, then [V, Q] = [V, Q, Q] = 1, and thus Q = 1 since the action of G on V is faithful, by assumption.)_Since K C O p ( G ) , it follows that K C P_and A ^ C Q, and thus the group G = G / K has distinct Sylow p-subgroups P and Q . Al so , G acts faithfully on V/U, and we see that P centralizes C V ( P ) / U , which has index \ V : C V { P ) \ < p i n V/U. It follows that the group G satisfies al l five conditions w i t h respect to its Sylow p-subgroup P and its action on V/U, but it does not have a normal Sylow p-subgroup. B y the min imal i ty of G , therefore, we must have K = I , and so G acts faithfully on V/U. We can thus replace V by V / U and assume that \V\ < p 2 .

Since G acts faithfully on V, it is isomorphically embedded i n A u t ( V ) . If V is cyclic, then A u t ( V ) is abelian, and so G is abelian and P < G, which is not the case. We conclude that V is noncyclic, and so it must be elementary abelian of order p 2 , and thus A u t ( V ) = G L { 2 , p ) , and we can view G as a subgroup of G L { 2 , p ) . In part icular, \P\ < p , and since P is not normal , it follows that O p ( G ) = 1. Now let L = C y ( G ) , so that P normalizes the p'-group L . Since we are assuming that a Sylow 2-subgroup of G is abelian, the same is true for L , and thus by L e m m a 7.3, we see that P acts t r iv ia l ly on L . B y the H a l l - H i g m a n L e m m a 1.2.3 (our Theorem 3.21) we have P C C G ( L ) C L , and thus since P is a p-group and L is a p'-group, we have P = 1. Th i s is a contradict ion since P is not normal . •

P r o b l e m s 7 A 209

P r o b l e m s 7 A

7 A . 1 . Show that a nilpotent max ima l subgroup of a simple group must be a 2-group.

N o t e . The simple group PSL{2,17) has a m a x i m a l subgroup of order 16, so nontr iv ia l 2-groups actually can occur as max ima l subgroups of simple groups.

7 A . 2 . Le t S = 5 L ( 2 , 3 ) and write Z = ( - / ) , so that Z is the unique subgroup of order 2 in S. Show that the group S/Z of order 12 has four Sylow 3-subgroups, and deduce that it has a unique Sylow 2-subgroup. Conclude from this that S has a normal subgroup of order 8.

H i n t . Use the fact mentioned i n the text that i f n > 2 and q is a power of the pr ime p , then G L { n , q ) has at least two Sylow p-subgroups. N o further mat r ix computations are required for this problem.

7 A . 3 . Let G = G L ( n , q ) , where q is a power of the pr ime p , and let P be the Sylow p-subgroup of G consisting of the upper t r iangular matrices w i t h ones on the diagonal. Let D be the group of a l l diagonal matrices in G.

(a) Show that D C N G ( P ) .

(b) Show that D P is the group of a l l upper t r iangular matrices i n G w i t h arbi trary nonzero entries on the diagonal.

(c) V i e w G as acting by right mul t ip l ica t ion on the space of n-dimensional row vectors over the field F of order q. Identify a l l P- invar iant sub-spaces of V, and observe that there is exactly one such subspace of dimension k for each integer k w i t h 1 < k < n .

(d) Show that a l l of the P- invar iant subspaces of V are TV-invariant, where N = N G ( P ) , and use (c) to deduce that TV = D P .

7 A . 4 . If q is a power of the prime p , show that S L { 2 , q ) and P S L ( 2 , q ) each have exactly q + 1 Sylow p-subgroups.

7 A . 5 . Le t P be a p-group, and suppose that U < P is elementary abelian. Show that U normalizes some group Ee£(P).

H i n t . Choose E e £{P) such that | P n E \ is as large as possible. If U does not normalize E , wri te P = E u , where u € U and E ^ F . Let H = ( E , F ) and Z = E n P . Show that Z ( H n U ) G £{P).

7 A . 6 . Suppose that Q is a generalized quaternion group and that | A u t ( Q ) | is divisible by an odd prime p . Show that p = 3 and \ Q \ = 8, and deduce that the hypothesis that a Sylow 2-subgroup is abelian i n L e m m a 7.3 and Theorem 7.5 is unnecessary if p > 3.

7 B

In this section, we prove the following "normal-J theorem". Th i s w i l l be used for the proof of Theorem 7.1, and also for the group-theoretic proof of Burnside 's p V - t h e o r e m .

7.6. Theorem. L e t P e S y l p ( G ) , w h e r e G i s a finite g r o u p , a n d assume t h e f o l l o w i n g c o n d i t i o n s .

(1) G i s p - s b l v a b l e .

(2) p ^ 2 .

(3) A S y l o w 2 - s u b g r o u p of G i s a b e l i a n .

(4) C y ( G ) = l .

(5) P = C G ( Z ( P ) ) .

T h e n J ( P ) < G .

Since we are assuming that G is p-solvable and that C y ( G ) - 1, it follows (assuming that G is nontrivial) that O p ( G ) > 1. T h e point of the theorem, however, is not s imply that G has a nontr iv ia l normal p-subgroup; it is that a p a r t i c u l a r p-subgroup, namely J ( P ) , is normal i n G . Note that it is not un t i l Step 7 of the following proof that we use the actual definition of the set £{P).

P r o o f of T h e o r e m 7.6. Suppose that the theorem is false, and let G be a counterexample of m i n i m u m possible order, p a r t i c u l a r , G is nontr ivial , so Op(G) > 1, and we write U = O p ( G ) and G = G / U . A l so , let I = C y (G) , where L D U . (This, of course, unambiguously defines the subgroup L < G.) We proceed in a number of steps, the first of which consists of three closely related statements, a l l of which are consequences of the Ha l l -H igman L e m m a 1.2.3.

Step 1.

(a) Z ( P ) C U.

(b) If P C P C G , then O p / ( P ) = 1.

(c) C Q ( L ) C L .

Proof. Since G is p-solvable and C y ( G ) = 1, L e m m a 1.2.3 (our Theorem 3.21) tells us that U = O p ( G ) D C G ( P ) . B u t U C P , and so Z ( P ) C C G ( U ) and (a) follows. Also , since U = O p ( G ) , we see that O p ( G ) = 1, and L e m m a 1.2.3 yields (c). For (b), let U C H C G and write M = ( V ( i f ) . Then M and U are normal i n H and M n U = 1 since U is a p-group and M is a p'-group. Thus M C C G { U ) C U, so M = M n P = 1, proving (b).

7 B 211

Step 2. There exists A G £(P) such that A % U.

Proof. Otherwise, a l l members of £(P) are contained i n P , and thus J ( P ) C P . It follows by L e m m a 7.2 that J ( P ) = J ( P ) is characteristic i n P , and since U < G , we have J ( P ) < G . Th is is a contradict ion since we are assuming that G is a counterexample.

Step 3. Let U A C H < G , and suppose that P n P G S y l p ( P ) . T h e n A centralizes P / T l L .

Proof. Observe that P satisfies the first four hypotheses of the theorem. F i r s t , P is p-solvable since it is a subgroup of the p-solvable group G . A l so , p is stilt different from 2, so the second condi t ion holds. A Sylow 2-subgroup of H is contained in a Sylow 2-subgroup of G , which is abelian, and so P satisfies condi t ion (3), and finally, O p , { H ) = 1 by Step 1(b).

N o w let S = H n P G S y l p ( P ) . Since Z ( P ) C i / C S C P b y Step 1(a), we have Z ( P ) C Z ( 5 ) , and thus C H ( Z ( S ) ) C C G ( Z ( P ) ) = P . Thus Cjy(Z (5)) is a p-subgroup of H containing the Sylow p-subgroup S, and so C H ( Z ( S ) ) = S and H satisfies the fifth hypothesis too.

Since H < G , the theorem holds for H , and hence J ( 5 ) < P . A l so , since A G £ ( P ) and A C 5 C P , it follows that A G £(S), and so A C J ( 5 ) . T h e n

[ P n L , A ] c [ P n L , J ( 5 ) ] c ( H n L ) n J ( S ) = L n J ( S ) c p ,

where the second containment holds because bo th P n L and J ( 5 ) are normal i n P , and the final containment holds because U is the unique Sylow p -subgroup of L , and thus it contains every p-subgroup of P We now have 1 = [ P n L , A] = [ P n L , A ] , as wanted.

Step 4. G = L A and P = P A .

Proof. Wr i t e P = L A , and observe that P A is a p-subgroup of P and that |P : P A | = \ L ( U A ) : U A \ = \ L : L n P A | , which divides the p ' -number | L : P | . It follows that P A G S y l p ( P ) , and thus U A = P n P . If H < G , then since L C P , Step 3 yields A C C ^ ( L ) C L , where the second containment holds by Step 1(c). Since A is a p-group and L is a p ' -group, we have A = 1, and so A C P . Th i s contradicts the choice of A , however, and we conclude that P = G , as wanted. F ina l ly , we have U A = H n P = G n P = P .

Step 5. | A | = p.

Proof. F i r s t , A is nontr iv ia l since A % U. A l so , A is elementary abelian, and so it suffices to show that it_is cyclic . N o w A j ic ts coprimely on L , and this action is faithful since C Q ( L ) C L and L n A = 1. B y L e m m a 6.20,

therefore, it suffices to show that A acts t r iv ia l ly on every A-invar iant proper subgroup of L .

Suppose that M is ^- invar iant , where M < L , and where we assume (as we may) that M D U. T h e n A C N G ( M ) , and so M A is a group, and M A D U A — P . A l so , since A is a p-group, the p'-part of \ M A \ is equal to the p'-part of | M | , and this is s t r ic t ly less than the p'-part of \ L \ since \ L : M | is a p'-number exceeding 1. It follows that M A < G, and we can apply Step 3 to deduce that A centralizes M A n L D M , proving Step 5.

N o w let V = { z G Z ( P ) | = 1}, so that V is an elementary abelian normal subgroup of G. Then G acts by conjugation on F , j m d since V C Z ( P ) , we see that J7 acts t r ivial ly . Th is yields an action of G = G / U on V.

Step 6. The action of G on V is faithful.

P r o o f . Let K - C G ( V ) , so that K is the kernel of the action of G on V. We argue that AT is a p-group by considering a Sylow g-subgroup Q of AT, where g is any prime different from p. T h e n Q acts coprimely on the abelian group Z(E7), and Q fixes a l l elements of order p i n Z ( U ) since these elements a l l lie i n V and Q C AT - C G ( V ) . It follows by Coro l la ry 4.35 that Q acts t r iv ia l ly on Z ( U ) . B u t Z ( P ) C U, and so Z ( P ) C Z([7), and thus Q C C G ( Z ( P ) ) = P . We conclude that Q = 1, and thus AT is a p-group, as claimed. B u t AT < G , so AT C O p ( G ) = 17, and thus AT = 1, as wanted.

Step 7. | V : V n A \ < p.

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Proof. Wr i t e D = U T \ A and E = V C \ A as i n the diagram, and observe that \ V : E \ = \ V : V n D \ = \ V D : D \ . Now D is an elementary abelian subgroup of P , and V is a central elementary abelian subgroup of P , and thus V D is elementary abelian. Since A 6 £(P), it follows that < | A | , and thus | V D : P | < \ A : P | = = p. We now have \ V \ E \ = \ V D : D \ < p , as wanted.

Step 8. We have a contradiction.

We want to apply the "normal -P theorem" (Theorem 7.5) to the faithful action of G on V, and so we need to check that \ V : C V ( P ) \ < p . B u t P = U A , so P = A , and of course, A centralizes V n A since A is abelian. Thus \ V : C V ( P ) \ < \ V : V n A \ < p , as wanted, and we conclude that P < G. T h e n P < G and A C P C O p ( G ) = U, which is not the case. •

7 C

In this section, we prove Theorem 7.1, thereby complet ing the proof that Frobenius kernels are nilpotent. We begin w i t h an extension of L e m m a 2.17.

7.7. L e m m a . W r i t e G = G/N, w h e r e N i s a n o r m a l p ' - s u b g r o u p of a finite g r o u p G. If P i s a p - s u b g r o u p o f G , w e h a v e

(a) N Q ( P ) = N G ( P ) a n d

(b) C Q ( P ) = C o J P ) .

Proof. Statement (a) is essentially L e m m a 2.17. A l s o , since overbar is a homomorphism, it is clear that C G ( P ) C C G ( P ) , and so to establish (b), it suffices to prove the reverse containment ._By (a), overbar defines a surjec-tive homomorphism from NG_(P ) to N ^ P ) , and so by the correspondence theorem, the subgroup C ^ ( P | C r % ( P ) is the image of some subgroup X C N G ( P ) . In other words, X = C ^ ( P ) , and we have 1 = [P , X ] = [P, X ] , and thus [ P , X ] C N . B u t X C N G ( P ) , and so we also have [P, X ] C_ P . Thus [ P , X ] C N n P = 1, and X C C G ( P ) . We conclude that C ^ P ) = A C C G ( P ) , and the proof is complete. •

We are now ready to prove Theorem 7.1, which, we recall, asserts that G has a normal p-complement if p ^ 2 and bo th N G ( J ( P ) ) and C G ( Z ( P ) ) have normal p-complements, where P is a Sylow p-subgroup of G. (Of course, i f these conditions hold for P , they w i l l also hold for every Sylow p-subgroup of G.) We repeatedly use the fact that the property of having a normal p-complement is inherited by subgroups and homomorphic images.

Observe the subtle way that the subgroup U is chosen in the following proof. T h i s is one of the keys to this very clever argument. (This tr ick appears both i n Thompson 's thesis and in his 1964 paper.)

P r o o f of T h e o r e m 7 .1 . Suppose that G is a counterexample of min ima l order. T h e n G fails to have a normal p-complement, and so by Frobenius ' theorem (Theorem 5.26), there must be some nonidenti ty p-subgroup U of G such that N G ( P ) fails to have a normal p-complement. A m o n g al l such "bad" subgroups, choose P such that | N G ( P ) | p is as large as possible. (In other words, U is chosen so that a Sylow p-subgroup of N G ( P ) has order as large as possible.) Subject to that condit ion, choose U to have order as large as possible. We proceed in a number of steps.

Step 1. P = O p ( G ) .

Proof. Let S 6 Syl p (7V), where N = N G ( P ) , and suppose that N < G. T h e n N is not a counterexample to the theorem, and since N fails to have a normal p-complement, at least one of N N ( 3 ( S ) ) or C N ( Z ( S ) ) must fail to have a normal p-complement. T h e n one of N G ( J ( S ) ) or C G ( Z ( 5 ) ) fails to have a normal p-complement, and so by hypothesis, S cannot be a full Sylow p-subgroup of G. T h e n S is properly contained in a Sylow p-subgroup, and hence we can choose a p-subgroup T such that S < T and S < T. (We are, of course, using the fact that "normalizers grow" in p-groups.)

N o w N G ( A ) fails to have a normal p-complement, where X is one of J ( S ) or Z ( 5 ) . A l so U > 1, so S > 1, and thus X > 1, and therefore X is a member of the set of bad subgroups from which we selected P . Also , X is characteristic i n S, and so X < T and T C N G ( A ) . We thus have | N G ( X ) | P > | T | > \S\ = | N G ( P ) | p , and since this contradicts the choice of P , we conclude that N = G, and so U < G and U C O p ( G ) . Now O p ( G ) is a member of the set of nonidenti ty p-subgroups whose normalizers fail to have normal p-complements, and since | N G ( O p ( G ) ) | p = \ P \ = | N G ( [ / ) | p , the choice of U guarantees that \ U \ > | O p ( G ) | , and thus U = O p ( G ) , as required.

Step 2. G / U has a normal p-complement, and so G is p-solvable.

Proof. Wr i t e G = G / U , and observe t h a t J G | < \ G [ . T h e n G is not a counterexample to the theorem, and so since P 6 S y l p ( G ) , it suffices to show that each of N G ( J ( P ) ) and C G ( Z ( P ) ) has a normal p-complement. There is nothing to prove i f p does not divide and so we can assume that P is nontr ivia l , and thus J ( P ) and Z ( P ) are nontr ivia l . Let U C X C P , where X is either J (J 5) or Z ( P ) , and observe that X > U. A l so , X | P | , and so by the choice of P , we see that X cannot be one of the bad subgroups whose

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normalizers fail to have normal p-complements. Since X contains the kernel U of o v e r b a d e have N ^ X ) = N G ( X ) by the correspondence theorem, and thus N ^ X ) has a normal p-complement. In part icular , bo th r % ( J ( P ) ) and N g ( Z ( P ) ) have normal p-complements, and therefore, G has a normal p-complement, as required.

Step 3. C y ( G ) = 1.

Proof. Let K = C y ( G ) , and write G — G/K. Now P is a Sylow p-subgroup of G , and since K is a p'-group, overbar defines an isomorphism from P onto P. It follows that J ( P ) = J ( P ) and Z ( P ) = Z ( P ) . T h e n

N G ( J ( P ) ) = N g ( J C P ) ) = N G ( J ( P ) ) and

C G ( Z ( P ) ) = C G ( Z ( P ) ) = C G ( Z ( P ) ) ,

where the second e q u a l i t y m each of these_equations follows by L e m m a 7.7. We conclude that I % ( J ( P ) ) and C ^ { Z { P ) ) _ have normal p-complements, and so i f \ G \ < | G | , it would follow that G has a normal p-complement. T h i s is not the case, however, since K = O p , { G ) and G fails to have a normal p-complement. T h e n | G | is not smaller than \ G \ , and so K = 1, as required.

Step 4. P is a max ima l subgroup of G.

Proof. Suppose that P C H < G. T h e n N H ( J ( P ) ) and C H ( Z ( P ) ) have normal p-complements, and since H is not a counterexample, it follows that H has a normal p-complement K . T h e n K < H and O P ( G ) < H , and since K n O P ( G ) = 1, we have K C C G ( O P ( G ) ) C O P ( G ) . (We are using L e m m a 1.2.3, which applies because G is p-solvable by Step 2 and C y (G) = 1 by Step 3.) T h e n K = I and P is a p-group, and hence P = P .

Step 5. C G ( Z ( P ) ) = P .

Proof. Certainly, C G ( Z ( P ) ) 5 P . B u t C G ( Z ( P ) ) < G since G fails to have a normal p-complement, and thus C G ( Z ( P ) ) = P by Step 4.

Step 6. The normal p-complement of G / P is abelian.

Proof. F i r s t , recall that G / P has a normal p-complement L / U by Step 2. Suppose that P C X C L , where X is normalized by P . Then P X is a group, and hence by Step 4, either P X = P or P X = G . If P X = P , then | X | is a power of p, and so X = P . If, on the other hand, P X = G , then | G : X | is a power of p, and hence X = L . It follows that no nonidenti ty proper subgroup of I = L / P is P- invar iant .

We can certainly assume that L is nontr ivial , and so it has a nontr iv ia l P- invar iant Sylow g-subgroup for some prime q. It follows from the result of the previous paragraph that L is a g-group, and since it is nontr ivia l , we have ( I ) ' < L . O f course, ( I ) ' is P- invar iant , and thus ( I ) ' = 1, and it follows that L is abelian.


P r o o f . We argue that G satisfies a l l five hypotheses of the normal-J theorem (Theorem 7.6). Certainly, p ^ 2, and we know that G is p-solvable, that O p > ( G ) = 1, and that P = C G ( Z ( P ) ) . A l l that remains, therefore, is to check that a Sylow 2-subgroup of G is abelian.

WriteJL7 = G / U and let Q G S y l 2 ( G ) , so that Q is a p'-subgroup of G , and thus Q is contained in the abelian normal p-complement of G. (We are using Step 6 here.) Also , Q = Q because P is the kernel of the canonical homomorphism G - > G and Q n P = 1, and thus Q is abelian, as wanted. B y Theorem 7.6, therefore, J ( P ) < G . B u t then G = N G ( J ( P ) ) has a normal p-complement, and this is our final contradiction. •

P r o b l e m s 7 C

7 C . 1 . (Thompson) Let P G S y l p ( G ) , where G is finite and p ^ 2, and suppose that N G ( A ) / C G ( A ) is a p-group for every characteristic subgroup X of P . Show that G has a normal p-complement.

H i n t . W o r k by induct ion on | G | and use the inductive hypothesis to show that if X is characteristic i n P and A > 1, then N G ( X ) has a normal p-complement.

7 D

In this section, we prove Burnside 's theorem.

7.8. T h e o r e m . L e t G be a finite g r o u p of o r d e r p a q b , w h e r e p a n d q a r e p r i m e s . T h e n G i s s o l v a b l e .

Burnside 's proof of this theorem was one of the first major applications of character theory, early i n the 20 t h century. (Characters were invented by Frobenius, who used them to prove his theorem on the existence of Frobe-nius kernels, but Burnside was one of the early developers of the theory.) Burnside 's proof appears in the second (1911) edit ion of his group theory text, of which a very substantial fraction is devoted to character and l in ear representation theory. It is interesting that the first (1897) edit ion of Burnside 's book includes no character theory at a l l , apparently because he

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felt that it was not good for much. Burnside wrote i n his preface to the first edi t ion that "it would be difficult to find a result that could be most direct ly obtained by the consideration of [linear representations]." He began the preface to the second edit ion w i t h the assertion that "Very considerable advances ... have been made since the appearance of the first edit ion of this book." (Was he th inking pr imar i ly of his own pV-theorem?) In fact, Burnside 's proof is elegant and not very difficult; it lies relatively near the surface of character theory, and it appears fairly early in most modern texts on the subject. Nevertheless, it has always seemed somewhat strange that one had to use facts about the field of complex numbers i n order to prove this theorem about finite groups.

B y the late 1960s, considerable advances (to use Burnside 's words) had been made i n pure finite group theory. In part icular , the Thompson subgroup, and Thompson's techniques of "local analysis" had been developed, and these had proved to be very powerful tools, as we have seen in this chapter. Thompson 's P h . D . student D a v i d Goldschmidt wondered whether or not these tools could be used to give a non-character proof of Burnside 's theorem. (The real point here was to demonstrate the power of T h o m p son's techniques; no one claimed that there was any deficiency in Burnside 's elegant proof.)

Goldschmidt almost succeeded: he was able to prove the pY-theorem provided that bo th p and q are different from 2. A few years later, and approximately simultaneously, H . Bender and H . M a t s u y a m a found ways to handle the remaining case. Matsuyama 's argument was easier, but Ben der's approach, when supplemented by Matsuyama 's idea, yielded an overall s implif icat ion of Goldschmidt ' s argument, and that is essentially the proof we present here. (But Bender used Glauberman 's Z(J ) - theorem, which we have decided not to include i n this book. Instead, we appeal to the normal-J theorem.)

P r o o f o f T h e o r e m 7.8. Let G be a counterexample of smallest possible order. Since the order of every proper subgroup of G also has at most two prime divisors, every such subgroup must be solvable, and also if 1 < N < G, then G / N is solvable. In particular, if AT is a nont r iv ia l proper normal subgroup of G, then bo th N and G / N are solvable, and hence G is solvable, which contradicts our assumption that G is a counterexample. It follows that G is simple.

We focus now on the max ima l subgroups of G. If AT is any nontr iv ia l normal subgroup of a max ima l subgroup M , then of course, M C N G ( A T ) , and since 1 < AT < G and G is simple, we know that N G ( A T ) < G. It follows by the maximal i ty of M that M = N G ( K ) . (This observation w i l l be used repeatedly in what follows.) A l so , if M is max ima l , then M is solvable

and nontr ivia l , and so it has a nonidentity normal subgroup of prime-power order. Thus either O p ( M ) > 1 or O q ( M ) > 1, and one of our intermediate goals is to show that it is not possible for both O p ( M ) and O q ( M ) to be nontr iv ia l . We proceed i n a number of steps.

Step 1. Suppose that K C G is nilpotent, and assume that M = N G ( A T ) is a max ima l subgroup of G . If bo th p and q divide \ K \ , then M is the unique max ima l subgroup of G containing K .

P r o o f . Assuming that the assertion is false, let AT be max ima l among counterexample subgroups, and let AT C X , where A is a max ima l subgroup of G different from M . Wr i t e K = K p x K q , where K p and K q are respectively the (nontrivial) Sylow p-subgroup and Sylow g-subgroup of AT. Since K p is characteristic i n AT, we have K p < M , and it follows that M = N G ( A T p ) .

Now M i l X = N x ( A T p ) is a p-local subgroup of X . A l so , K q < M n X , and so K q C C y (AT n X ) . Since X is solvable and X n M is a p-local subgroup of X , we can apply Theorem 4.33 to conclude that C y ( M n X ) C C y ( X ) . Th i s yields K q C C y ( X ) = O q { X ) , and so wr i t ing L q = O q ( X ) , we have K q C L q . Similar ly, K p C L p , where L p = O p ( X ) , and thus both L p and L q

are nontr iv ia l . Wr i t e L = L p L q , and note that K C L . A l so , L is nilpotent, | L | is divisible by both p and g and X = N G ( L ) is a max ima l subgroup of G. If AT < L , it follows by the choice of K that X is the unique max ima l subgroup of G that contains L . B u t ATP C L p , and thus

A ? C C G ( L P ) C C G ( A T P ) C N G ( A p ) = M ,

and similarly, L p C M . T h e n L C M , and since X is the unique max ima l subgroup containing L , we have M = X , which is a contradiction. In the remaining case, K = L , and thus M = N G ( A T ) = N G ( L ) = X , and again we have a contradiction.

Step 2. Let P e S y l p ( G ) , and suppose that P normalizes some nontr iv ia l subgroup V of G. T h e n (V, Q ) = G for every Sylow g-subgroup Q of G. In part icular, V cannot be a g-group.

P r o o f . Let Q G S y l , ( G ) , and write H = ( V , Q ) . Now | P Q | = | P | | Q | = | G | since P n Q = 1, and thus P Q = G . Let g be an arbi trary element of G and write g = xy, w i t h x G P and y G Q. Since P normalizes V , we have V9 = Vxy - Vy C P , and thus P contains a l l G-conjugates of V . It follows that H contains the nonidentity normal subgroup VG generated by the conjugates of V, and since G is simple, VG = G, and so P = G . To prove the final assertion, observe that if V is a g-group, we can choose Q to contain V, and then (V, Q ) = Q < G , contradict ing what we have proved.

Step 3. Let M be max ima l i n G . T h e n either O p ( M ) = 1 or O q ( M ) = 1.

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P r o o f . Assume that the assertion is false, and wri te Z p = Z ( O p ( M ) ) and Z g = Z ( 0 ? ( M ) ) , so that Z p and Z q are both nont r iv ia l and normal in M . A l so , Z p and Z q are abelian, and they centralize each other since they intersect t r iv ia l ly , and thus the group Z = Z p Z q is abelian. Furthermore, Z < M , and so M = N G ( Z ) , and it follows by Step 1 that M is the unique max ima l subgroup of G that contains Z . N o w if 1 ^ z G Z , then Z C C G ( z ) < G , and so CG(z) is contained i n some m a x i m a l subgroup of G that contains Z , and we conclude that C G { z ) C M .

Let 5 G S y l p ( M ) . Since S normalizes the nontr iv ia l g-subgroup Z q , it follows by Step 2 that S cannot be a full Sylow p-subgroup of G . T h e n 5 < P for some Sylow p-subgroup P of G , and hence N P ( S ) is a p-subgroup of G str ic t ly larger than S. T h e n N P ( S ) £ M , and we can choose an element g G G - M such that 5» = S.

N o w write A = ( Z p y , and observe that since Z p C O p ( M ) C 5, we have 4 = ( Z p ) 9 C S9 = S C M , and thus A acts on the normal subgroup Z q of

M . Suppose either that the action of A on Z q is not faithful or that A is not cyclic . In bo th cases, we argue that

Z q = ( C Z q { a ) \ l ^ a e A ) .

This is clear if the action is not faithful since i n that case, there exists a nonidenti ty element a G A such that C Z q ( a ) = Z q . If A is not cyclic, then since it is abelian, Theorem 6.20 applies to yie ld the assertion.

We argued previously that CG(z) C M for a l l nonidenti ty elements z G Z . T h e n C G { z 9 ) C Af» for a l l nonidenti ty elements z9 G Z 9 , and i n part icular, since A = ( Z p ) » C 2 J , w e have C G ( a ) C M » i f l ^ o e A If the action of A on Z 9 is not faithful or if A is not cycl ic , therefore, we deduce that Z q C M f . A l so Z p C S = S « C M » , and thus Z = Z p Z ? C M » . B u t M is the unique max ima l subgroup containing Z , and therefore M = M 9

and we have 5 G N G ( M ) = M , which is not the case.

We deduce that A is cyclic and that it acts faithfully on Z q . A l so , since A = ( Z p y , we see that Z p is cyclic . B y similar reasoning, w i t h the roles of p and q interchanged, we deduce that Z q is cycl ic . B u t A is a nontr iv ia l p-group that acts faithfully on Z q , and so p divides | A u t ( Z , ) | , and since Z q is cyclic, we see that p divides q - 1, and so p < q. Exp lo i t i ng the symmetry between p and q once again, we have q < p , and this, of course, is a contradict ion, which completes the proof of Step 3.

If M is an arbitrary max ima l subgroup of G , we have observed that at least one of O p ( M ) or O q ( M ) is nontr iv ia l , and now by Step 3, we know that exactly one of these subgroups is nont r iv ia l . In other words, G has two flavors of max ima l subgroups: those for which O p ( M ) > 1 and those for which O q ( M ) > 1, and we refer to these as p - t y p e maximals and g- type

maximals , respectively. We stress that every max ima l subgroup of G has exactly one of these types.

Before we proceed w i t h the next step, we define a p - c e n t r a l element of G to be a nonidenti ty element i n the center of some Sylow p-subgroup of G, and similarly, a g - cen t r a l element is a nonidenti ty element of the center of some Sylow g-subgroup of G.

Step 4. Suppose that y G N G ( V ) , where y is g-central and V is a p-subgroup. T h e n V contains no p-central elements.

P r o o f . G i v e n a p-subgroup U C G, write U* to denote the subgroup generated by the p-central elements i n U. Since the G-conjugates of a p-central element are p-central, it follows that N G ( J 7 ) permutes the p-central elements of U, and thus N G ( J 7 ) normalizes U*. In part icular, since y normalizes V, it is also true that y normalizes V*, and our goal is to show that V* = 1.

Now V* is a p-subgroup of G that is normalized by y and is generated by p-central elements, and we let W be a p-subgroup of G, max ima l subject to the conditions that it is normalized by y and generated by p-central elements. If V* > 1, then W > 1, and we work to obtain a contradict ion.

Wr i t e TV = N G ( W ) , and let S G Syl p (TV) so that S Z } W . Since y is g-central, we can choose Q G S y l ? ( G ) such that y G Z ( Q ) . T h e n (y) is normalized by Q, and since y G TV, we have (y, S) C TV < G , where the strict inequali ty holds because 1 < W < TV.

It follows by Step 2 (wi th the roles of p and g reversed) that S cannot be a full Sylow p-subgroup of G . Since S G Syl p (TV), we can reason as before that N G ( 5 ) g and thus there exists an element g e G - TV such that

= S. In particular, W ^ ^7, but V P 9 C S9 = S C N .

Since VP is generated by p-central elements and W9 + W, there must be at least one p-central generator x of W such that x 9 0 W. Because x is p-central, we can choose P e S y l p ( G ) such that x G Z ( P ) . Now G = P Q , and so we can write g = a b , w i t h a G P and b G Q. T h e n x 5 = x a f e = x b G Wb

and x 5 G C TV, and thus x s G Wb n TV. Since W 6 n TV is a p-subgroup that normalizes the p-group W, it follows that W(Wb n TV) is a p-subgroup of G . A l so , W = W*, and so ( W ( W b n TV))* D ( W , x » ) >

Now V P 6 is normalized by y b = y . Since y G TV, it follows that Wb D TV is normalized by y, and therefore ( ^ ( V F 6 n N ) ) * is a p-subgroup that is normalized by y. Th i s subgroup is generated by p-central elements and it s t r ic t ly contains W, contradict ing the choice of W.

Step 5. Every p subgroup of G is centralized by a p-central element. Also , p-type max ima l subgroups of G contain no g-central elements.

7 D 221

Proof. G i v e n a p-subgroup V C G, choose P G S y l p ( G ) w i t h P D F . T h e n Z ( P ) C C G ( V ) , and so C G ( V ) contains a p-central element.

N o w suppose that M is a p-type max ima l subgroup, and let V = O p ( M ) , so that V > 1 and 0 9 ( M ) = 1 by Step 3. T h e n V 5 C M ( V ) = C G ( V ) by the Ha l l -H igman L e m m a 1.2.3, and thus V contains a p-central element. B y Step 4, no g-central element lies i n N G ( V ) = M , as required.

Step 6. A g-central element of G cannot normalize a nontr iv ia l p-subgroup.

Proof. Let V be a nontr iv ia l p-subgroup of G, and let M be a max ima l subgroup of G containing N G ( V ) . It follows by Step 5 that M contains a p-central element, and thus by Step 5 again, w i t h p and q interchanged, M cannot have g-type, and thus it has p-type. B y Step 5 yet again, M cannot contain a g-central element, and in part icular , N G ( V ) contains no g-central element.

Step 7. p + 2 and g ^ 2.

Proof. Suppose g = 2 and choose an involut ion t i n the center of a Sylow g-subgroup. B y Theorem 2.13, there exists an element x G G of order p such that x l = X ' 1 , and thus t G N G ((x)) . Since i is g-central, however, this contradicts Step 6. We deduce that g ^ 2, and, similarly, p ^ 2.

Step 8. Let M be a p-type max ima l subgroup of G, and let S G S y l p ( M ) . T h e n J ( 5 ) < M and 5 is a full Sylow p-subgroup of G.

Proof. We wish to apply the normal-J theorem (Theorem 7.6) to the group M , and so we proceed to check the five hypotheses. F i r s t , M is solvable, and so it is certainly p-solvable. B y Step 7, we know that p ^ 2 and that a Sylow 2-subgroup of M is t r iv ia l , and hence is abelian, and thus the second and th i rd hypotheses hold. Since M has p-type, we know that C y ( M ) = O q ( M ) = 1, and the fourth condi t ion holds.

The remaining condi t ion is that C M ( Z ( S ) ) = S, and to establish this, it suffices to show that C M ( Z { S ) ) is a p-group. Otherwise, C M ( Z ( S ) ) contains some subgroup Y of order q, and thus Z ( S ) normalizes Y. Th is w i l l contradict Step 6 (with p and q interchanged) if we can show that Z ( S ) contains a p-central element.

Let S C P G S y l p ( G ) , and observe that S = M n P . Since M is a p-type max ima l subgroup, we can write M = N G ( V ) for some p-subgroup V, and we have V C S C P . T h e n Z ( P ) C N G ( V ) f~]P = M C \ P = S, and it follows that Z ( P ) C Z { S ) . Thus Z ( S ) contains p-central elements, and so C M ( Z ( 5 ) ) = S, as wanted. We see now that M satisfies the hypotheses of the normal -J theorem, so we have J ( 5 ) <J M .

Final ly , i f S is not a full Sylow p-subgroup of M , we can write S < T , where T is a p-subgroup of G properly containing S. T h e n T Ç N G ( J ( 5 ) ) = M , and this is a contradict ion since T > S and 5 is a full Sylow p-subgroup of M .


P r o o f . O f course, the p-part and the g-part of \ G \ are unequal, and so we can assume that \ G \ P > \ G \ q . If S,T G S y l p ( G ) , therefore, we have

| G P | 2 > \ G p \ \ G q \ = \ G \ > \ST\ = = . f i l l , , |o n T\ \s n T |

and so | S T l T | > 1.

Since O p ( G ) = 1 and J ( P ) > 1 for P G S y l p ( G ) , the Thompson subgroups of the Sylow p-subgroups of G cannot a l l be equal, so we can choose S,T G Sylp(G) w i t h J (5) ^ J ( T ) . Do this so that D = S n T is as large as possible, and observe that D < S since S ^ T. B y the computat ion of the first paragraph, D > 1, and hence N G ( D ) < G , and we can choose a max ima l subgroup M D N G ( D ) . Now N G ( D ) contains a p-central element by Step 5, and thus by Step 5 w i t h p and q reversed, M cannot be of g-type. T h e n M is of p-type, and so by Step 8, the Sylow p-subgroups of M are full Sylow p-subgroups of G .

Choose U G S y l p ( M ) w i t h U D M n 5 , and observe that

f / n S = [ / n M n S = M n S D N 5 ( P ) > D.

Since P G Sylp(G) , it follows by the maximal i ty of \ D \ that J ( 5 ) = J(17), and similarly, we can choose V G S y l p ( M ) such that 3 ( V ) = J ( T ) . Now J ( P ) and J ( V ) are conjugate in M since the Sylow subgroups P and F are conjugate. B y Step 8, however, J ( P ) < M , and we deduce that J ( 5 ) = J ( P ) = J ( V ) = J ( T ) . Th i s contradicts the choice of S and T, and the proof is complete. •

C h a p t e r 8

Permutation Groups

8 A

In this chapter, we return to the study of group actions on abstract sets. Let G act on a nonempty finite set ft, and recall that such an action defines a natural homomorphism (the permutat ion representation associated wi th the action) from G into the symmetric group Sym(ft) . We say that G is a p e r m u t a t i o n g r o u p on ft i f the act ion is faithful, or equivalently, if the permutat ion representation is an injective homomorphism. If G is a permutat ion group on ft, it is common to identify G w i t h its isomorphic copy i n Sym(ft) , so one can view permutat ion groups on ft s imply as subgroups of Sym(ft) .

O f course, if we intend to use a permutat ion representation of G to make statements about the group G itself, it is essential that the corresponding action should be faithful, but for many purposes, and, i n particular, for the basic definitions of the subject, faithfulness is irrelevant. For that reason, and despite the tit le of this chapter, we w i l l generally not assume that our actions are faithful except when such an assumption is really necessary. We w i l l always assume in this chapter, however, that the sets being acted on are finite, al though much of the theory works more generally.

Reca l l that if G acts on ft, then ft is par t i t ioned into G-orbits , and that the action is transitive if there is just one orbit: ft itself. Equivalently, the action of G on ft is transitive i f for every choice of points a , ¡5 G ft, there exists an element g G G such that a - g = ¡3.

The following describes the (necessarily nonempty) set of al l elements of G that carry a to 0 i n a transitive action. (It is essentially a restatement of Theorem 1.4, which underlies the fundamental counting principle.)

223

224 8. P e r m u t a t i o n G r o u p s

8.1. L e m m a . Suppose t h a t a g r o u p G acts t r a n s i t i v e l y o n ft. F i x a p o i n t a G ft, a n d l e t H = Ga be the s t a b i l i z e r of a i n G. F o r a n a r b i t r a r y p o i n t P G ft, l e t TT{P) = { x G G I a-x = P ) . Then TT{P) i s a r i g h t coset of H i n G, a n d IT i s a b i j e c t i o n f r o m ft o n t o the set A = { H x \ x G G } of a l l r i g h t cosets o f H .

Proof. G i v e n 0 G ft, choose g G G such that a-g = p , and observe that since the elements of H stabilize a , every element of the coset H g carries a to P , and hence H g C i r ( P ) . Conversely, i f x G i r ( 0 ) , then a-x = p , and so a - { x g ~ l ) = a . Then , xg~l G H , and hence x G H g , and it follows that i r ( p ) = H g G A .

The map TT is injective since if i r { P ) = T T { ^ ) , then an element of this set carries a to P and also to 7, and thus p = 7. F ina l ly , TT is surjective since for each coset H y G A , we have H y = i r ( a - y ) . U

O f course, if G is an arbi trary finite group and H C G is any subgroup, then G acts t ransi t ively on the right cosets of H i n G. In the si tuat ion of L e m m a 8.1, therefore, G acts on A , and it is interesting to compare this action w i t h the given action of G on ft. In fact, the bijection TT : ft -> A of L e m m a 8.1 is a permutation isomorphism, which means that i r ( P - g ) = • n { P ) - g for a l l p G ft and g G G. (The "dot" on the right of this equation, of course, denotes the right mul t ip l ica t ion action of G on right cosets of H . ) To check that the right cosets of H on each side of the equation i r ( 0 - g ) = i r ( P ) - g are the same, it suffices to observe that each of them contains xg, where x G G is an arbi t rary element that carries a to p .

It should now be apparent that the study of the transitive actions of a group is equivalent to the study of the actions of the group on the sets of right cosets of its various subgroups. Furthermore, we know that the kernel of the action of G on the set of right cosets of a subgroup H is exactly c o r e G ( A 7 ) , and thus the study of transitive permutat ion groups (that is, faithful transitive actions) is equivalent to the study of subgroups having t r i v i a l core. In other words, much of permutation-group theory can be subsumed w i t h i n "pure" group theory. Nevertheless, the permutat ion point of view provides useful insights and suggests interesting theorems, and that is why we study it . Also , one should remember that historically, permutat ion groups (as subgroups of symmetric groups) appeared before Cayley gave the modern axiomatic definition of an abstract group. For many years, the permutat ion group point of view was the only one available i n group theory.

Reca l l that i f G acts on ft and g G G and a G ft, then ( G a ) 3 = Ga.g. In a transitive action, therefore, the point stabilizers are a l l conjugate, and, i n part icular, if some point stabilizer is t r iv ia l , then all of them are t r iv ia l .

S A 225

If the point stabilizers in a transit ive action of G on ft are t r iv ia l , it follows by L e m m a 8.1 that for each choice of a,P G ft, there is exactly one element of G that carries a to (3. In this si tuation, where for a l l a and 13, there is a unique element g G G such that a-g = P, we say that G is regular or sharply transitive on ft. (Conversely, of course, if G is regular on ft, then there is a unique group element that carries a to a , and so the point stabilizer Ga is the t r iv ia l subgroup.) Note that regular actions are automatical ly faithful.

B y the fundamental counting principle, we see that if G is transitive on ft, then | G | is a mult iple of |ft|. (The integer |ft| is the degree of the action.) In part icular, \ G \ > |ft|, and G is regular if and only if equality holds. We mention that a not-necessarily-transitive act ion i n which a l l point stabilizers are t r i v i a l is said to be semiregular.

The not ion of t rans i t iv i ty can also be refined i n a different direction. Suppose that G acts on a set ft containing at least k points, where k is some positive integer, and consider the set £>fc(ft) of ordered fe-tuples ( a i , a 2 , • • •, a k ) of distinct points on G ft. (The condi t ion |ft| > k is needed so that O f c(ft) w i l l be nonempty.) The act ion of G on ft defines a natural componentwise act ion of G on O k { Q ) v i a the formula

(ai, a 2 , a k ) - g = { { a ^ - g , ( a 2 ) - g , ( a k ) - g ) .

(It is a t r iv ia l i ty that this is a genuine group action.) We say that G is A;-transitive on ft if the associated componentwise action of G on O f c(ft) is transitive. Equivalently, and continuing to assume that |ft| > k, an act ion of G on ft is fc-transitive if given any two ordered fe-tuples { a x , a 2 , . . . , a k ) and { ( 3 i , ( 3 2 , . . . , ( 3 k ) of distinct points of ft, there exists an element g G G such that ( a i ) - g = P i for 1 < i < k. (Note that al though we require that the a i are distinct and that the p i are dist inct , there is no requirement that the sets and {#} should be disjoint.)

Observe that i f G is fc-transitive on ft, then G is automatical ly m -transitive for a l l integers m w i t h 1 < m < k. In part icular, a fc-transitive act ion is necessarily 1-transitive, which s imply means that it is transitive, and thus i n general, /c-transitivity is stronger than transit ivi ty. A n act ion that is fc-transitive w i t h k > 1 is said to be mult iply transitive, and if k = 2, the action is doubly transitive. In part icular, mul t ip ly transitive actions are always doubly transitive.

If |ft| = n , it is easy to see that |O f c ( f t ) | = n ( n - 1) • • • (n - k + 1), where there are exactly k factors i n the product. If G is A;-transitive on ft, therefore, then because it is transitive on O f c ( f t ) , the fundamental counting principle guarantees that | G | must be a mult iple of |O f c ( f t ) | . In part icular, if G is doubly transitive of degree n , then n ( n - 1) must divide | G | .

W h a t are some examples of mul t ip ly transitive group actions? F i rs t , it is easy to see that the full symmetric group Sn is n-transitive on ft = { 1 , 2 , . . . , n } . In fact, given two ordered n-tuples (OH) and (#) of distinct points of ft, there is obviously a unique element g G Sn that carries o n to P i for a l l i. In general, if an act ion is A;-transitive, and if for each pair of fc-tuples of dist inct points there is a unique element of G carrying one to the other, we say that the action is s h a r p l y fc-transitive. In other words, G is sharply fc-transitive on ft precisely when it is sharply transitive ( i . e . , regular) on the set Ojt(ft). In particular, if G is sharply fc-transitive on ft, where |ft| = n , then | G | = \Ok(ù)\ = n ( n — l ) ( n - 2 ) • • • ( n — k + 1).

The al ternating group A n is ( n - 2)-transitive on ft = {1, 2 , . . . , n } . To see this, observe that the stabilizer of the (n - 2)-tuple t = (1, 2 , . . . , n - 2) i n the symmetric group Sn contains only the identity and the transposit ion exchanging the points n - 1 and n . The stabilizer of t i n A n , therefore, is t r iv i a l , and so by the fundamental counting principle, the A n - o r b i t of t has cardinal i ty equal to \ A n \ . B u t \ A n \ = n! /2 = | 0 n _ 2 ( f t ) | , and it follows that A n is transitive on 0 n _ 2 ( f t ) as wanted. In fact, since a point stabilizer in the act ion of A n on 0 n _ 2 ( f t ) is t r iv ia l , A n is sharply transitive on this set of (n - 2)-tuples, and so A n is sharply (n - 2)-transitive on ft.

We mention that A n and Sn are the only subgroups of Sn that are (n - 2)-transitive on ft = { 1 , . . . , n} . One way to see this is to observe that if G Ç Sn is (n - 2)-transitive, then |<D n_ 2(ft)| = n! /2 must divide \ G \ , and thus G has index at most 2 in Sn. It follows that G contains every element of order 3 i n Sn, and, i n particular, it contains a l l 3-cycles. It is not hard to see, however, that the 3-cycles i n Sn generate A n , and thus A n Ç G. In other words, G is A n or Sn, as claimed.

We have seen that Sn and A n are respectively sharply n-transitive and (n - 2)-transitive of degree n . In fact, unless k is tiny, there are only a few other examples (sharp or not) of fc-transitive permutat ion groups. One of these is the M a t h i e u group M 1 2 , which is one of the 26 sporadic simple groups. It is sharply 5-transitive on 12-points, and so

\ M 1 2 \ = 12-1M0-9-8 = 95,040 = 2 6 - 3 3 - 5 - l l .

The stabil izer of a point i n M 1 2 is the sporadic simple group M n , which is sharply 4-transitive on the remaining 11 points. (In general, we w i l l see that i f k > 1, then a point stabilizer in a fc-transitive action on n points necessarily acts ( k - In t rans i t ive ly on the remaining n - 1 points, and it is easy to check that sharpness is inherited i n this situation.) O f course M n

has index 12 i n M 1 2 , and so

| M n | = 1 M 0 - 9 - 8 = 7,920 = 2 4 - 3 2 - 5 - l l .

8 A 227

In fact, M n is the smallest of the sporadic simple groups. A point stabilizer i n M n is sharply 3-transitive on the remaining 10 points, but it is not simple. Th i s group, which is sometimes called M w , has order 7920/11 = 720. It has a normal subgroup of index 2, isomorphic to the al ternating group A 6 , but M i o is n o t isomorphic to S 6 . In fact, there are three nonisomorphic groups of order 720 that contain copies of A 6 ; these are S 6 , M w and P G L { 2 , 9 ) . (Recal l that P G L { n , q ) is the projective general linear group; it is the factor group G L { n , q ) / Z , where *Z = Z ( G L ( n , q ) ) is the group of scalar matrices i n G L { n , q ) . ) E a c h of these three groups of order 720 happens to be a permutat ion group of degree 10. B o t h M w and P G L ( 2 , 9 ) are 3-transitive i n their 10-point actions, but S6 is only 2-transitive and not 3-transitive. (One way to dist inguish these three groups is v i a their Sylow 2-subgroups: a Sylow 2-subgroup of S6 is a direct product of D 8 w i t h a cyclic group of order 2; a Sylow 2-subgroup of P G L { 2 , 9) is D m and a Sylow 2-subgroup of M i o is the semidihedral group S D m . )

The M a t h i e u group M 2 4 (which is another of the 26 sporadic simple groups) is 5-transitive on 24-points. B u t it is not sharply 5-transitive; the stabilizer i n M 2 4 of a 5-tuple of distinct points has order 48. It follows that

| M 2 4 | = 4 8 | C 5 | = 48-24-23-22-21-20 = 244, 823,040 = 2 1 0 -3 3 -5-7-11.23 .

The stabilizer of a point i n M 2 4 is the M a t h i e u group M 2 3 , which is also one of the sporadic simple groups; it is 4-transitive on the remaining 23 points. Clearly, M 2 3 has index 24 i n M 2 4 , and so | M 2 3 | = 10,200,960. The stabilizer of a point i n M 2 3 is the simple M a t h i e u group M 2 2 , which is 3-transitive on the remaining 22 points and has order 443, 520. The stabilizer of a point i n M 2 2 is simple, but it is not one of the sporadic simple groups; it is isomorphic to P 5 L ( 3 , 4 ) . (As a consequence of L e m m a 8.29, we shall see that F 5 L ( 3 , 4 ) is a 2-transitive permutat ion group of degree 21.)

For k > 4, we have named four fe-transitive permutat ion groups other than Sn and A n i n their natural actions. These four groups are M 2 4 and M 1 2 , which are 5-transitive, and M 2 3 and M u , which are 4-transitive. In fact, there are no other examples. The only known proofs of this, however, require an appeal to the classification of simple groups. (But there is an elementary argument due to C . Jordan that shows that a s h a r p l y fe-transitive group w i t h k > 4 must either be a full symmetr ic or al ternating group, or else k is 4 or 5 and the degree is 11 or 12, respectively.) We shall see, however, that there are infinitely many 3-transitive permutat ion groups other than symmetric and alternating groups, and i n fact there are infinitely many s h a r p l y 3-transitive groups. (See P rob lem 8A.17.)

8.2. L e m m a . Suppose t h a t a g r o u p G acts t r a n s i t i v e l y o n ft. L e t 7 6 f t , a n d fix a n i n t e g e r k < |ft|. T h e n G i s k - t r a n s i t i v e o n ft if a n d o n l y if t h e p o i n t s t a b i l i z e r G 7 i s ( k - \ ) - t r a n s i t i v e o n ft - {7}.


P r o o f . F i r s t , assume that G is fc-transitive on ft. Let

a = { a 1 , a 2 , . . . , a k . 1 ) and b = ( J 3 U p 2 , • • •, Pk-i)

be (fc-l)- tuples of distinct points of ft-{7}, and let A and B be the fc-tuples constructed by appending 7 to a and b , respectively. The entries in a and i n b a r e distinct and lie i n ft - {7}, and hence the entries in each of A and B are dist inct . B u t G is fe-transitive on ft, so there exists an element g G G that carries A to B . T h e n g carries a to b and 7 to 7, or, in other words, a - g = b and 5 G G 7 . It follows that G 7 is ( k - Intransitive on ft - {7}, as wanted.

Conversely, suppose that G1 is ( k - Intransitive on ft - {7}, and let

A = ( « i , a 2 , • • • ,«fe) and B = {Pi, p 2 , . . . , Pk)

be ordered fc-tuples of distinct points of ft. Since the action of G on ft is transitive, we can choose elements x , y G G such that (a f c )-s = 7 and

= 7. T h e n A - x and are fc-tuples of distinct points of ft, and the last entry of each of them is 7.

Le t a and b be the ( fc- l ) - tuples obtained from A - x and B - y , respectively, by deleting the last entry. The entries of each of a and b , therefore, are dist inct and lie i n ft - {7}, and thus there exists an element g G G1 that carries a to b . Since A - x and B - y can be obtained from a and b by appending 7, which is fixed by g , we see that { A - x ) - g = B - y , and thus xgy~l is an element of G that carries A to B . Th i s completes the proof. •

We w i l l use L e m m a 8.2 to construct an infinite family of 3-transitive permutat ion groups, but we start w i t h a family of 2-transitive group actions. Reca l l that if n > 1 and q is a prime power, then the general linear group G L { n , q ) is the group of invertible n x n matrices over the field F of order q , and the subgroup S L { n , q ) , the special linear group, is the set of determinant-one matrices in G L { n , q ) .

Since G L { n , q ) is isomorphic to the group of invertible linear transformations of an n-dimensional vector space V over F , it is customary to identify the mat r ix group G L { n , q ) w i t h the corresponding group of linear transformations. One often speaks, therefore, of the "natural action" of G L { n , q ) on V, and the "natural action" of its subgroup S L { n , q ) on V. B u t , of course, we should remember that the isomorphism between G L { n , q ) and the group of linear transformations of V is not really natural ; it depends on the choice of a basis for V.

8.3. L e m m a . L e t V be a n n - d i m e n s i o n a l v e c t o r space o v e r a field F of o r d e r q , w h e r e n > 2. T h e n f o r e v e r y p r i m e - p o w e r q , t h e n a t u r a l a c t i o n of S L { n , q ) o n t h e set of a l l o n e - d i m e n s i o n a l subspaces o f V i s d o u b l y t r a n s i t i v e .

8 A 229

P r o o f . F i r s t , observe that invertible linear transformations of V carry sub-spaces to subspaces and preserve dimensions, and thus S L ( n , q ) actually does permute the set of one-dimensional subspaces of V. A l so , since n > 2, there are at least two of these, and so it makes sense to ask if the action is 2-transitive.

Now let 01,02,61,62 e V be nonzero, and consider the one-dimensional subspaces F o i , F a 2 , F h and F b 2 . Assuming that F a i + F a 2 and that F b i + F b 2 , we must show that there exists an element g G S L ( n , q ) that carries F a x to F b i and F a 2 to F6 2. Since F a x ^ F a 2 , the vectors a x and a 2 are l inearly independent, and so there exist vectors 03, o 4 , . • -, a n such that { a i I 1 < i < n } is a basis for V, and similarly, we can expand the set {61,62} to a basis { b i I 1 < i < n } .

Since the a; form a basis, there is some (unique) linear transformation t on V that maps a* to bi for 1 < i < n , and since the h form a basis, t must be invertible. Thus t G G L ( n , q ) , and t carries F o i to F61 and F a 2

to F6 2. The proof is not yet complete, however, because t may not have determinant 1.

Let 5 = det(t), so that 0 ^ S G F , and let e = 8~l. Now {e&i, 62, ...,&„} is a basis for V, and so, reasoning as before, there is a unique invertible linear transformation x on V that carries 61 to e&i and 6< to 6 for 2 < z < n . The mat r ix corresponding to x w i t h respect to the basis { b i } is diagonal, w i t h (1, l ) -entry equal to e and a l l other diagonal entries equal to 1. Thus det(x) = e, and hence det( ix) = det(t)det(x) = e5 = 1 and t x G S L ( n , q ) . Now t x carries 01 to e&i, and so it carries F a x to F61. It also carries o 2 to 62, and hence it carries F a 2 to F6 2. Th i s completes the proof. •

Clearly, scalar mul t ip l ica t ion by a nonzero field element on a vector space fixes a l l subspaces. It follows i n the si tuat ion of L e m m a 8.3 that the subgroup Z of S L ( n , q ) consisting of a l l determinant-one scalar matrices is contained in the kernel of the action on the set of one-dimensional subspaces of V, and thus the factor group S L { n , q ) / Z = P S L { n , q ) acts on this set. It is not hard to see that i n fact, Z is the full kernel of the action, and hence P S L { n , q ) acts faithfully, and so it is a doubly transit ive permutat ion group. We prove this in L e m m a 8.29.

8.4. C o r o l l a r y . L e t n > 2. T h e n t h e g r o u p G L { n , 2 ) i s d o u b l y t r a n s i t i v e o n t h e set of n o n z e r o v e c t o r s i n t h e n - d i m e n s i o n a l v e c t o r space V o v e r t h e field of o r d e r 2.

P r o o f . Since the field has order 2, a one-dimensional subspace of V contains exactly one nonzero vector, and each nonzero vector, of course, lies i n exactly one one-dimensional subspace. The action of G L ( n , 2 ) = S L { n , 2 ) on the


nonzero vectors of V, therefore, is permutation-isomorphic to its action on the one-dimensional subspaces of V, which we know is doubly transitive. •

Next , we need the following easy result.

8 .5 . L e m m a . Suppose t h a t a g r o u p G acts t r a n s i t i v e l y o n ft, a n d l e t A = Ga, w h e r e a G ft. T h e n a s u b g r o u p N Ç G i s t r a n s i t i v e o n ft if a n d o n l y if A N = G, a n d N i s r e g u l a r o n ft if a n d o n l y if i n a d d i t i o n , A n N = 1. A l s o , i f N i s r e g u l a r a n d n o r m a l i n G, t h e n t h e c o n j u g a t i o n a c t i o n of A o n t h e n o n i d e n t i t y elements of N i s p e r m u t a t i o n i s o m o r p h i c t o t h e a c t i o n of A o n t h e set ft - { a } .

P r o o f . F i r s t , A N = G, if and only if every right coset of A in G has the form A n for some element n € N , and this happens if and only if N acts transit ively by right mul t ip l ica t ion on the right cosets of A in G. The right mul t ip l ica t ion action of G on these cosets, however, is permutat ion isomorphic to the action of G on ft, and so N is transitive on the right cosets of A if and only if it is transitive on ft. Th is shows that as claimed, N is transitive on ft if and only if A N = G. A l so , assuming that N is transitive, we know that it is regular if and only if the point stabilizer N a = A n N is t r iv i a l .

Now suppose that N is regular and normal , and let TT : N -> ft be the map defined by i r ( n ) = a-n for n G N . To show that TT is injective, let x,y G N , and suppose i r ( x ) = i r { y ) . Then a-x = a-y, and thus x = y since by regularity, x is the unique element of N that carries a to a-x. For surjectivity, observe that since N is transitive, every element of ft has the form a-n = 7r(n) for some element ne N . Thus TT is a bijection from N to ft, and since TT(1) = a , it follows that vr defines a bijection from N - {1} onto ft - {a} . To show that this is a permutat ion isomorphism of the actions of A on these two sets, we must check that yr(n a) = yr(n)-a for a G A and nonidenti ty elements n G N . We have

vr(n a) = a - i a ^ n a ) = ( a - n ) - a = i r ( n ) - a ,

as required, where the second equality holds because aeA = Ga. U

8.6. C o r o l l a r y . L e t A be a g r o u p t h a t acts v i a a u t o m o r p h i s m s o n a finite g r o u p N , a n d assume t h a t t h e r e s u l t i n g a c t i o n of A o n t h e set of n o n i d e n t i t y elements of N i s k - t r a n s i t i v e f o r some p o s i t i v e i n t e g e r k. L e t G = N x A , a n d v i e w N a n d A as s u b g r o u p s of G. T h e n t h e right-multiplication a c t i o n of G o n t h e set of r i g h t cosets of A m G i s ( k + 1 ) - t r a n s i t i v e .

P r o o f . Let ft be the set of right cosets of A i n G, and observe that A — Ga, where a G ft is the point a = A . Since G is certainly transitive on ft, it suffices by L e m m a 8.2 to show that A is A;-transitive on ft - { a } . B u t

8 A 231

A N = G and i f l i V = 1, and hence TV is a regular normal subgroup of G by L e m m a 8.5. A l so by L e m m a 8.5, the act ion of A on ft - { a } is permutat ion isomorphic to its action on N - {1}, and by hypothesis, this latter action is /c-transitive. T h i s completes the proof. •

F ina l ly , we can construct the promised family of 3-transitive permutat ion groups.

8.7. C o r o l l a r y . L e t V be a n e l e m e n t a r y a b e l i a n 2 - g r o u p of o r d e r 2n w i t h n > 2 , a n d w r i t e G = V x G L ( n , 2 ) , w h e r e t h e a c t i o n of G L ( n , 2 ) i s d e f i n e d by i d e n t i f y i n g V w i t h a v e c t o r space of d i m e n s i o n n o v e r a field of o r d e r 2. T h e n G i s a 3 - t r a n s i t i v e p e r m u t a t i o n g r o u p of d e g r e e 2n.

P r o o f . Wr i t e A = G L ( n , 2 ) . B y Coro l la ry 8.4, the action of A on the nonidenti ty elements of V is 2-transitive, and hence by Corol la ry 8.6, the act ion of G on the right cosets of A is 3-transitive. The degree of this action is \ G : A \ = \V\ = 2n, and so a l l that remains is to show that the action of G on the right cosets of A is faithful. The kernel K of the action, however, is normal i n G, and since K C A , we have V n K = 1, and so K C C G ( V ) . T h e n K acts t r iv ia l ly on V, and since the natural action of A = G L ( n , 2) on V is faithful, it follows that K = 1. •

Actua l ly , Corol la ry 8.6 is less general than it may appear to be. In fact, the integer k in the statement of that result can almost never exceed 2.

8.8. L e m m a . L e t A be a g r o u p t h a t acts v i a a u t o m o r p h i s m s o n a finite g r o u p N , a n d suppose t h a t t h e r e s u l t i n g a c t i o n of A o n t h e n o n i d e n t i t y elements o f N i s k - t r a n s i t i v e , w i t h k > \ . T h e n

(a) N i s a n e l e m e n t a r y a b e l i a n p - g r o u p f o r some p r i m e p .

(b) I f k > 1 t h e n e i t h e r p = 2 o r \ N \ = 3.

(c) I f k > 2 , t h e n \ N \ = 4 a n d k = 3.

P r o o f . Let x G N have prime order p . Since A acts v i a automorphisms and is transit ive on the nonidentity elements of N , it follows that every nonidentity element of N has order p , and hence TV is a p-group. T h e n Z(7Y) is nontr ivia l , and so we can assume that x is central i n N , and hence every nonidentity element of N is central. Thus N is abelian, proving (a).

N o w assume that the act ion of A on the nonidenti ty elements of N is 2-transitive. If p > 2 then x and x ' 1 are dist inct nonidenti ty elements, and if |TV| > 3, we can choose a nonidenti ty element y of N different from bo th x and x ~ l . B y 2-transitivity, there is an element a G A that carries the ordered pair { x , x ~ l ) to the ordered pair ( x , y ) . T h e n x = x a and y = ( a T 1 ) " = ( x 0 - ) ' 1 = x ~ l . T h i s contradict ion proves (b).

Fina l ly , assume that the action is 3-transitive. In particular, 3 < k < \ N - {1}| , so \ N \ > 4, and we have p = 2 by (b). Choose a nonidentity element y of N different from x , and observe that xy + 1 and that xy is different from bo th x and y . i f |7Y| > 4, choose a nonidenti ty element z different from x , y and xy, and consider the ordered triples ( x , y , x y ) and { x , y , z ) , which consist of distinct elements. B y 3-transitivity, there exists a G A such that x = x a , y = y a and z = { x y ) a = x a y a = xy, which is a contradict ion. Thus |7V| = 4, and so k < \ N - {1}| = 3. •

We close this section w i t h a variat ion on L e m m a 8.8(a). In order to conclude that N is an elementary abelian p-group in that lemma, we shall see that the assumption that A acts transit ively on the set of nonidentity elements of N is unnecessarily strong.

We say that an action of G on is h a l f - t r a n s i t i v e if a l l orbits have equal size. ( A transitive action, therefore, is certainly half-transitive.) The following is a result of D . S. Passman and the author.

8.9. T h e o r e m . L e t A be a finite g r o u p t h a t acts f a i t h f u l l y v i a a u t o m o r phisms o n a finite g r o u p N , a n d assume t h a t t h e r e s u l t i n g a c t i o n of A o n t h e n o n i d e n t i t y elements of N i s h a l f - t r a n s i t i v e . T h e n e i t h e r t h e a c t i o n i s F r o b e n i u s , o r else N i s a n e l e m e n t a r y a b e l i a n p - g r o u p f o r some p r i m e p , a n d n o n o n i d e n t i t y p r o p e r s u b g r o u p of N a d m i t s t h e a c t i o n of A .

Note that a Frobenius action of A on N is automatical ly half-transitive on the nonidenti ty elements of N because a l l A-orbi ts on this set have equal size \ A \ .

We need a prel iminary result about finite groups part i t ioned by normal subgroups.

8 .10. L e m m a . L e t G be a finite g r o u p a n d l e t U be a c o l l e c t i o n of p r o p e r n o r m a l s u b g r o u p s o f G . Suppose t h a t G = {JTL a n d t h a t d i s t i n c t members o/n i n t e r s e c t t r i v i a l l y . T h e n G i s a n e l e m e n t a r y a b e l i a n p - g r o u p f o r some p r i m e p .

P r o o f . Let X G II, and observe that i f X ^ Y G II, then X n Y = 1, and so Y C C G ( X ) . It follows that

G = X U |J y c i u CQ(X) . y e n

B u t G cannot be the union of two proper subgroups, and since X is proper, it follows that C G ( X ) = G, and hence X C Z ( G ) . Since | J II = G and every member of II is central, we have Z ( G ) = G , and thus G is abelian.

8 A 233

If x , y G G have different orders, we argue that they lie i n a common member of II. To see this, assume that o ( x ) > o(y). T h e n

(xy)°(y) = x

o i - y ) y o { y ) = x o ( y ) / 1,

where the first equality is val id because G is abelian and the inequality holds because o { y ) < o { x ) . Now let X , Z € II, w i t h x G X and xy G Z . T h e n x°(y) = {xy)°^ is a nonidentiy element i n X n Z , and thus X = Z . It follows that xy G X , and so y € X , as claimed.

Let x G G have prime order p, and let X G II w i t h We complete the proof now by showing that every nonidenti ty element y G G has order p . If y G G - X , then s and y do not lie i n a common member of II, and so o(y) = o ( x ) = p , as wanted. If, on the other hand, y is a nonidenti ty element of X , choose z G G - X . T h e n 2 does not share a member of II w i t h either x or y, and thus o(y) = o { z ) = o { x ) = p . •

P r o o f o f T h e o r e m 8.9. For nonidenti ty elements x G N , wri te G x = C N ( C A { x ) ) , and let II = { C x | 1 / 1 £ JV} be the collection of subgroups of N constructed i n this way. It is clear that x G Cx, and it follows that (jn = N . If 1 ^ y G Cx, we argue that G^ = G x . To see this, observe that C A { x ) centralizes y, and thus C A ( x ) C C ^ ( y ) . B u t C U ( x ) and C A ( y ) are subgroups of index r i n A , where r is the common size of the A-orbi t s on N - {1}, and thus equality holds, and C A { x ) = C A { y ) . T h e n

Cy = C N { C A ( y ) ) = C N { C A ( x ) ) = Cx ,

as claimed. In part icular, this shows that distinct members of the set II intersect t r iv ia l ly .

Now assume that the action of A on TV is not Frobenius, so that r < \ A \ . T h e n C A { x ) > 1 for nonidentity elements x G TV, and since the action of A on N is faithful, C A { x ) does not act t r iv ia l ly on N . In other words, Cx < N , and so II consists of proper subgroups of N .

We work now to show that the members of II are normal i n N so that we can apply L e m m a 8.10. F i rs t , observe that r divides |7V| - 1, and thus r and \ N \ are coprime. Let { x i , x 2 , . . . , x r } be an A-orb i t of nonidenti ty elements of N , and let K t be the conjugacy class of X i i n N . (Note that perhaps the classes K t are not a l l distinct.)

Since A acts v i a automorphisms on N and A permutes the elements X i transitively, it follows that A permutes the set of conjugacy classes K {

transitively. In part icular, these classes have equal cardinali ty, say k, where k is a divisor of \ N \ , and hence k is relatively pr ime to r .

N o w write r

and recall that \ K i \ = k for a l l i. Since the sets K { a r e conjugacy classes of N , distinct K i are disjoint, and it follows that \ U \ is a mult iple of k. Furthermore, since the sets K i are permuted by A , their union U is an A¬invariant subset of N . A l so , U consists of nonidentity elements, and so it is a union of A-orbi t s of size r , and we deduce that r divides \ U \ . Now k and r are coprime integers and bo th divide \ U \ , and it follows that r k divides \ U \ , and in part icular, r k < \ U \ . B u t U the union of the sets K i for 1 < i < r , and each of these sets has cardinali ty k, and thus \ U \ < r k . Thus \ U \ = r k , and so the sets K i must a l l be different. Th i s shows that a nonidentity class of N cannot contain distinct members of an A-orb i t .

We argue next that every nonidentity conjugacy class of N is contained in one of the subgroups Cx in IT Since distinct members of n intersect t r ivia l ly , this w i l l show that each member of II is a union of classes of N , and thus is normal i n N , as wanted. G i v e n a nonidenti ty class K of N , let x G K . Since A permutes the classes of N and C A { x ) fixes x , it follows that C A { x ) stabilizes the class AT, and thus C A ( x ) permutes the elements of K . We know, however, that distinct members of K cannot lie i n a common A-orb i t , and this shows that, in fact, C A { x ) fixes every member of K . Thus K C C N ( C A { x ) ) = Cx, as claimed, and hence we can apply L e m m a 8.10 to conclude that N is an elementary abelian p-group for some prime p .

Fina l ly , if H is a proper A-invariant subgroup of TV, we work to show that H = 1. Let { x i , x 2 , . . . , x r } be an A-orb i t outside of H , and observe that A permutes the set of cosets H x u and so the set

r

V = \ j H x i i = i

is A-invar iant . Since V C G - H , it consists of nonidenti ty elements, and so it is a union of A-orbi t s of size r . Thus r divides \V\, and also, since V is a union of cosets of H , we see that \ H \ divides | V | . B y coprimeness, r \ H \ divides | V | , and so | V | > r \ H \ , and it follows that the cosets H X i are dist inct . Th i s shows that no right coset of H other than H can contain dist inct elements of an A-orb i t .

If H > 1, then H can be contained i n at most one member of II. We w i l l show that, i n fact, H is contained in some member C G II, and that C also contains every element x G N - H . It follows that C = N , which is a contradict ion, and this w i l l prove that H = 1, as wanted. It suffices, therefore, to show that if x G N - H , then H C Cx.

Since A permutes the right cosets of H , it follows that C A { x ) stabilizes H x , and so C A { x ) permutes the elements of H x . B u t H x cannot contain distinct elements of an A-orb i t , and thus C A { x ) fixes every element of H x .

P r o b l e m s 8 A 235

We deduce that H x C Cx, and since Cx is a subgroup that contains x , we conclude that H C Cx. A s we have seen, this completes the proof. •

Note that in the exceptional case of Theorem 8.9, where the action of A on TV is Frobenius, we can conclude that N is nilpotent by the theorem of Thompson 's thesis. Thus, in a l l cases, i f A acts nontr iv ia l ly on N and half-transitively on the nonidentity elements, then N is nilpotent.

P r o b l e m s 8 A

8A.1 . If A and B are nonisomorphic groups w i t h | A | = show that there exists a permutat ion group G that has regular subgroups isomorphic to each of A and B . A l so , decide whether or not a permutat ion group can have nonisomorphic regular n o r m a l subgroups.

H i n t . For the first part, take G to be an appropriate symmetric group.

8A.2 . Let H be a transitive subgroup of some permutat ion group G. Show that C G { H ) is semiregular. Deduce that an abelian transitive permutat ion group must regular.

8A.3 . G i v e n a finite group G w i th Z ( G ) = 1, show that there exists a permutat ion group that has two dist inct regular normal subgroups, each isomorphic to G.

8 A . 4 . Suppose that U and V a r e regular normal subgroups of some permuta t ion group G. If U n V = 1, show that U and V are isomorphic and have t r i v i a l centers.

H i n t . It is no loss to assume G = U V .

8A.5 . Let G act fc-transitively on some set ft, and let H = Ga, where a € ft. Let A be the set of points in ft that are fixed by H . Show that N G ( 7 f ) acts r- t ransi t ively on A , where r is the m i n i m u m of k and | A | .

8A.6 . Let G act t ransit ively on ft, and let H = Ga, where a € ft. Let P G S y l p ( / f ) , and let A be the set of points i n ft that are fixed by P . Show that N G ( P ) acts transit ively on A .

8 A . 7 . Suppose that an abelian group A acts faithfully on a group N , and assume that its act ion on the nonidenti ty elements of N is half-transitive. Show that the action is Frobenius, and deduce that A is cyclic .

8A.8 . Let G act transit ively on ft, and suppose that N < G . Show that G permutes the A - o r b i t s in ft transitively, and deduce that N is half-transitive on ft.

8A.9 . Let G act 2-transitively on ft, and suppose that N < G and that N acts nontr ivial ly. Show that N acts transitively.

Note . In fact, 2-transit ivi ty is more than is really needed here. The right condi t ion is pr imi t iv i ty , which we w i l l discuss in the next section.

8 A . 10. If G is a solvable 4-transitive permutat ion group, show that G is isomorphic to the symmetric group 5 4 .

Hint . Show that a min ima l normal subgroup N of G is regular, and consider the conjugation action of a point stabilizer Ga on N .

Note. In fact, 5 4 and S3 are the only solvable 3-transitive permutat ion groups, but this is somewhat harder to prove.

8 A . 11. Show that for every prime power q > 1, there exists a solvable sharply 2-transitive permutat ion group of degree q.

H i n t . Consider a field of order q.

8 A . 12. Let G act t ransit ively on ft, and let X be the corresponding permutat ion character. (Recall that by definition, X is the function defined by the formula X ( g ) = \ { a G ft | a - g = a } \ . ) Show that G is 2-transitive on ft if and only if the average value of X { g ) 2 for g G G is 2.

H i n t . F i n d an action of G whose permutat ion character is X

2 , and recall that, i n general, the average value of a permutat ion character is equal to the number of orbits. (See P rob lem 1A.6 and the note following it.)

8A.13. Let G act 2-transitively on ft, and let X be the corresponding permuta t ion character. F i n d a positive integer m such that G is 3-transitive on ft i f and only i f the average value of X ( g ) 3 for g G G is m .

8A.14. Let G act t ransit ively on ft, and suppose H i s a subgroup of G w i t h \ G : H \ = m . Show that H has at most m orbits on ft, and that i f H contains no point stabilizer i n G, then H has at most m/2 orbits on ft.

Hint . W h a t is the smallest possible size of an i f -o rb i t ?

8A.15. If H C G, we can define an action of H x H on G by setting g - ( x , y ) = x ~ l g y for g G G and x,y G H . Show that the action of G on the right cosets of H is 2-transitive if and only i f H x H has exactly two orbits on G.

8A.16. Let G be 2-transitive on a set ft, where |ft| = n . Suppose that H C G is transitive, and that \ G : H \ is relatively pr ime to n - 1. Show that H is 2-transitive.

8 B 237

8 A . 1 7 . Let q be a power of 2, and let S L ( 2 , q ) act on the q + 1 one-dimensional subspaces of a two-dimensional vector space over a field of order q. Show that this action is sharply 3-transitive.

8 B

Suppose that G acts transit ively on ft, and let A C ft be a nonempty subset. If g G G, we write A - g = { a - g \ a G A } , and we refer to A - g as a translate of A . The nonempty set A is said to be a block i f every one of its translates other than A itself is disjoint from A . In other words, A is a block if and only if A n A - g = 0 whenever g G G and A - g + A . We stress that we speak of "blocks" only for transitive actions, and that, by definition, blocks are nonempty.

For example, consider the group G of symmetries of square A B C D , and view G as a (transitive) permutat ion group on the vertex set ft = { A , B , C, D } . (Note that G = D 8 , and, i n fact, G is a full Sylow 2-subgroup of the symmetric group on ft.) Now choose a diagonal of the square, say A C , and let A = { A , C } , the set consisting of the ends of that diagonal. E a c h symmetry g G G carries the diagonal A C to a diagonal, and so the only translates of A are A itself and the set { B , D } of endpoints of the other diagonal. In part icular, if A - g ^ A , then A - g n A is empty, and so A is a block.

For a more general example, suppose H C G, and let ft = { H g \ g e G } , so that as usual, G acts transit ively by right mul t ip l ica t ion on ft. Suppose that H C K C G, where K is a subgroup, and let A = { H k \ k G K } . T h e n A is a nonempty subset of ft, and i f g G G, i t is easy to see that the translate A - g is exactly the set of those right cosets of H i n G that happen to be contained i n the coset K g of K . If K g = K , then A - g = A , and if K g ^ K , then K g n K = 0, and so no member of A - g can be a member of A . We have A - g D A = 0 in this case, and so A is a block.

O f course, i n an arbi trary transitive action, the whole set ft is a block, and so is every one-point subset of ft. These are referred to as tr ivial blocks, and we say that the given transitive act ion of G on ft is primitive i f the t r i v i a l blocks are the only blocks; otherwise, the action is imprimitive. Thus, for example, the action of the group of symmetries of a square on the four vertices is impr imi t ive , and the act ion of a group G on the right cosets of a subgroup H is impr imi t ive if there exists a subgroup K such that H < K < G. (The condi t ion H < K guarantees that the block A = { H k | k G K } contains more than one point, and the condi t ion K < G guarantees that A is not a l l of ft.)


8.11. L e m m a . L e t G be a g r o u p t h a t acts t r a n s i t i v e l y o n a set ft, a n d suppose t h a t A C ft i s a b l o c k . T h e n | A | d i v i d e s |ft| a n d A has e x a c t l y | f t | / | A | dzsimci t r a n s l a t e s . These t r a n s l a t e s a r e d i s j o i n t b l o c k s ; t h e i r u n i o n i s ft, a n d they a r e t r a n s i t i v e l y p e r m u t e d by G.

Proof. Let a,b G G, and suppose that A - a + A - b . A p p l y b~l to deduce that A - i a b ' 1 ) + A , and thus A ^ a b ' 1 ) n A = 0 since A is a block. Now apply 6 to obtain A - a n A - b = 0. Th i s shows that the dist inct translates of A are a l l disjoint, and, in particular, a l l translates of A are blocks. O f course, G permutes the set of translates of A transitively.

The union of the dist inct translates of A is a nonempty G-invariant subset of ft, and since G is transitive on ft, this union must be a l l of ft. Since the distinct translates of A are disjoint, have equal cardinal i ty and cover ft, there must be exactly | f t | / | A | of them, and so this number is an integer. •

8.12. Corollary. A t r a n s i t i v e g r o u p a c t i o n o n a set of p r i m e c a r d i n a l i t y i s p r i m i t i v e .

Proof. Let ft be the relevant set, where |ft| is prime. If A C ft is a block, then | A | divides |ft|, and so | A | = |ft| or | A | = 1, and hence either A = ft or A consists of a single point. In either case, A is a t r i v i a l block, and thus the given act ion is pr imit ive . •

The following lemma gives a general, and fairly complete description of the blocks of a transitive action, and it leads to a useful necessary and sufficient condi t ion for pr imi t iv i ty . To state our result, we introduce a bit of notat ion. If G acts on ft and a G ft, then for each subgroup AT C G, we write a - K = { a - k | k G K } . In other words, a-AT is the AT-orbit containing the point a .

8.13. L e m m a . L e t G be a g r o u p t h a t acts t r a n s i t i v e l y o n a set ft, a n d l e t H = Ga, w h e r e a G ft. / / AT D H i s a s u b g r o u p , t h e n a - K i s a b l o c k c o n t a i n i n g a a n d \ a - K \ = \ K : H \ . A l s o , t h e m a p K i-> a - K i s a b i s e c t i o n f r o m t h e set of s u b g r o u p s { K | H C K C G } o n t o t h e set of a l l b l o c k s t h a t c o n t a i n a .

Proof. Suppose that g G G and that the sets a - K and ( a - K ) - g are not disjoint. We argue that g G K . To see this, observe that there exist elements x , y e K such that a-x = ( a - y ) - g . T h e n ygx'1 G Ga = H C AT, and since x and y are i n K , it follows that g G AT, as claimed.

The set a - K is certainly AT-invariant, and the result of the previous paragraph shows that K is the full stabilizer of this set in G. Thus a - K uniquely determines AT, and i t follows that the map K H-> a - K is injective

8 B 239

on the set { K \ H C K C G } . The result of the previous paragraph also shows that the set A = a - K is a block. To see this, observe that if A is not disjoint from A - g , then g G K , and thus A - g = A . Furthermore, the block A certainly contains a , and since it is the Af-orbit containing a , we have | A | = \ K : H \ by the fundamental counting principle.

We must now show that given an arbi t rary block A containing a , there exists a subgroup K D H such that A = a - K . Let K be the stabilizer of A i n the action of G on subsets of ft, and observe that a - K C A since K stabilizes A and a G A . To prove the reverse containment, let 5 G A . Since G is transitive on ft, we can choose g G G w i t h a - g = S. T h e n 8 G A n A - g , and thus A = A - g because A is a block. Thus g e K , and 5 = a - g G a - K , and this shows that A = a - K , as required. A l so , we can take S = a and g e H m the above argument, and since we showed that g € K , it follows that H C K . Th i s completes the proof. •

8 .14. C o r o l l a r y . L e t G be a g r o u p t h a t acts t r a n s i t i v e l y o n a set ft c o n t a i n i n g m o r e t h a n one p o i n t , a n d l e t H = Ga, w h e r e a G ft. T h e n G i s p r i m i t i v e o n ft if a n d o n l y if H i s a m a x i m a l s u b g r o u p o f G .

P r o o f . F i r s t , assume that H is max ima l . T h e n by L e m m a 8.13, the only blocks that contain a are a - H = { a } and a - G = ft. If A is an arbi trary block, then some translate A - g of A contains a , and A - g is a block by L e m m a 8.11. It follows that A - g is either { a } or ft, and thus A is a t r i v i a l block. Th i s shows that G is pr imi t ive on ft.

Conversely, i f G is pr imit ive, then a l l blocks containing a are t r iv i a l , and thus there are precisely two such blocks: ft and { a } . B y L e m m a 8.13, therefore, there are exactly two subgroups of G that contain H , and thus H is max ima l i n G. •

Note that Corol la ry 8.14 renders Coro l la ry 8.12 superfluous since i f |ft| is prime, then a point stabilizer has prime index, and hence it is a max ima l subgroup. It follows by 8.14 that G is pr imi t ive .

The following result gives a useful connection between p r imi t iv i ty and normal subgroups.

8 .15 . L e m m a . L e t G be a g r o u p t h a t acts t r a n s i t i v e l y o n a set ft, a n d l e t N < G. T h e n each o r b i t of t h e a c t i o n of N o n ft i s a b l o c k f o r G. I n p a r t i c u l a r , if G i s p r i m i t i v e o n ft, t h e n t h e a c t i o n of N i s e i t h e r t r i v i a l o r t r a n s i t i v e .

P r o o f . Let g G G, and observe that if A is an orbit for the action of N , then A - g is an orbit for the action of N 9 = N . (This is immediate from the fact that if a-n = (5 w i th a , (5 G ft and n G N , then ( a - g ) - ( n 9 ) = p - g . ) B u t distinct TV-orbits in ft are disjoint, and thus if A - g + A , then A - g n A = 0.


In other words, A is a block for G, as wanted. If G is pr imit ive , it follows that either A = ft, or else | A | = 1. In other words, if TV is not transitive on ft, then a l l TV-orbits have size 1, and thus TV acts t r iv ia l ly . •

A n alternative argument for the final assertion of L e m m a 8.15 is as follows. If the action of G is pr imi t ive and A 7 is a point stabilizer, then H is a max ima l subgroup of G by Corol la ry 8.14. Since H C H N C G, there are just two possibilities. E i ther H N = G, in which case it is easy to see that TV is transitive on ft, or else H N = H . In the latter si tuation, TV C H , and thus TV is contained in every conjugate of H , and it follows that TV fixes a l l points in ft.

Next , we relate p r imi t iv i ty and mult iple transi t ivi ty.

8.16. L e m m a . Suppose t h a t G i s a g r o u p w i t h a d o u b l y t r a n s i t i v e a c t i o n o n ft. T h e n G i s p r i m i t i v e o n ft.

P r o o f . If G is impr imi t ive , there exists a nontr iv ia l block A , and since A does not consist of just one point, we can choose distinct points a , 0 G A . Al so , A is not a l l of ft, and so we can choose 7 G ft — A , and, i n particular, 7 ^ a . B y 2-transitivity, there exists an element g£G that carries the pair (a,/3) to the pair (a , 7). In part icular, a - g = a , and thus a G A n A - g . B u t A is a block, and this forces A - g = A , and since 0 G A , we have 0-g G A - g = A . Th i s is a contradiction, however, because 0-g = 7 and 7 0 A . •

It is easy to find examples of pr imi t ive actions that are not 2-transitive. For example, consider the dihedral group D 2 p of order 2 p , where p > 3 is prime. B y Corol la ry 8.12 or Corol lary 8.14, this group acts pr imi t ive ly on the p cosets of a subgroup of order 2. B u t p - 1 does not divide | D 2 p | , so the act ion cannot be 2-transitive. In this context, we mention a remarkable result of W . Burnside that shows that a (necessarily primit ive) transitive permutat ion group of prime degree p must be 2-transitive except possibly when a Sylow p-subgroup is normal . Burnside 's proof of this fact was character theoretic, but a comparatively easy non-character proof was found by H . Wie land t . (Wielandt ' s proof can be found i n the book P e r m u t a t i o n G r o u p s by D . S. Passman.)

A s Burnside 's theorem on pr imit ive permutat ion groups of prime degree suggests, once one knows that a group acts pr imi t ive ly on some set, it sometimes does not require a great deal of addi t ional information to deduce that the action is mul t ip ly transitive. For example, suppose that G is a pr imi t ive permutat ion group on ft, and assume that G contains at least one transposi t ion. (Recall that a transposit ion is a permutat ion that exchanges two

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points and fixes a l l others. A l so , recall the notat ion ( a , (5) for the transposi t ion exchanging a w i t h /?.) If G is pr imi t ive and contains a transposit ion, a theorem of C . Jordan asserts that G is the full symmetric group on ft, and so, of course, i t is mul t ip ly transit ive. We w i l l obtain this result as a corollary of a much more general theorem, also due to Jordan, but there is also an entirely elementary and rather geometric proof that we cannot resist presenting first. (We also cannot resist mentioning that we are following in Jordan's footsteps here: he was the author of the very first group-theory book, published in 1870. Hi s Traité des S u b s t i t u t i o n s , of course, deals only w i t h permutat ion groups since Cayley 's modern, abstract point o f view was not enunciated un t i l 1878.)

8 .17. T h e o r e m (Jordan). L e t G be a p r i m i t i v e p e r m u t a t i o n g r o u p o n ft, a n d suppose t h a t G c o n t a i n s a t r a n s p o s i t i o n . T h e n G i s t h e f u l l s y m m e t r i c g r o u p o n ft.

P r o o f . Since every element of Sym(ft) is a product of transpositions, it suffices to show that G contains a l l transpositions on ft. To do this, consider the (undirected) graph w i t h vertex set ft, i n which dist inct points a and p are joined by an edge precisely when the t ransposi t ion ( a , ¡3) is an element of G.

If x = { a , ¡3) and g G G is arbitrary, it is easy to see that x 9 is the transposit ion ( a - g , p - g ) , which exchanges a - g and p - g . O f course, i f x G G, then also x 9 G G, and this shows that i f a and P are two points joined by an edge i n our graph, then their images under g are also joined by an edge. In other words, the permutations i n G define automorphisms of the graph; they carry points to points and edges to edges.

N o w consider the connected components of the graph, viewing each component as a nonempty set of vertices. (In other words, no edge joins a point i n one component to a point in a different component, and wi th in each component, any two distinct points can be joined by a path consisting of one or more edges.) If A is such a component, then since g G G induces a graph automorphism, i t follows that A - g must also be a component. Now if A meets A - g nontr ivial ly, then A U A - g is a connected set that contains the components A and A - g . Th i s set, therefore must be equal to each of these components, and thus A = A - g . In other words, i f A and its translate A - g are not disjoint, they cannot be distinct, and hence A is a block. The act ion of G is pr imit ive , however, and hence A must be a t r i v i a l block, and so either A is a l l of ft or A consists of just one point. B y hypothesis, our graph has at least one edge, and so there is a component A consisting of more than one point. T h e n A = ft, which means that the graph is connected.

E a c h two points a , /? G ft are joined by a path, and we define the distance d(a , p ) to be the length of the shortest path from a to p , where we set d(a , a ) = 0. Thus d(a , /?) = 1 precisely when a ^ /3 and there is an edge jo in ing a to /?. Equ iva len t^ , d (a , 0) = 1 if and only i f the group G contains the transposit ion ( a , P ) . It suffices, therefore, to show that d ( a , p ) < 1 for al l a , P e n .

Suppose (for a contradiction) that there exist a and 0 in ft w i t h d = d { a , P ) > 1, and choose these points so that d > 1 is as smal l as possible. Since there is a path of length d jo in ing a and p , there must exist a point 7 G ft such that d (a , 7) = d - 1 and d ( 7 , 0 ) = 1. Now d - 1 > 1, and so by the min imal i ty of d, it follows that d ( a , 7 ) = 1, and thus the transposit ion x = (a , 7) is an element of G , as is the transposit ion y = (7,/?). T h e n x f = (a-y, 7-y) = (a , /?), and of course x y G G . T h e n a and /? are joined by an edge, and this is a contradict ion since d ( a , P ) = d > 1. •

To state the more general theorem of Jordan, we introduce some addi t ional notat ion. If G acts on ft and A C ft is an arbi trary subset, we write

G A = { 9 G G I a - g = a for a l l a G A } , so that G A is the pointwise stabilizer of A . Th i s should be distinguished from the setwise stabilizer,

G ( A ) = { g G G I A - g = A } .

Note that G ( A ) acts on A , and G A is the kernel of this action, so that G A < G ( A ) . A l so , note that G ( A ) = G ( Q _ A ) , and thus G Q _ A < G ( A ) .

8.18. T h e o r e m (Jordan). L e t G be a g r o u p w i t h a p r i m i t i v e a c t i o n o n a set ft, a n d l e t A C ft w i t h | A | < |ft| - 2. Suppose i / i a i G A acis p r i m i t i v e l y o n ft - A . Taen ifte ac i ion of G o n Q i s ( |A| + l ) - i r a rwt i t«e .

To see that Theorem 8.18 actually does imp ly Theorem 8.17, suppose that G is a pr imi t ive permutat ion group on ft, and assume that the transposi t ion (a , P) is an element of G . Take A = ft - { a , P } , so that | A | = n - 2, where n = |ft|. Since ( a , P ) G G A , it follows that G A is transitive on { a , P } = Q - A , and, in fact, this action is pr imit ive since ft - A has prime cardinal i ty 2. B y Theorem 8.18, therefore, G is (n - Intransitive on ft, and so

| G | > | 0 n _ i ( f t ) | = n - ( n - 1) 3-2 = n! = |Sym(ft) | .

Thus G is the full symmetric group on ft, as is asserted by Theorem 8.17.

We also have the following closely related result.

8.19. Corol lary. Suppose t h a t G i s a p r i m i t i v e p e r m u t a t i o n g r o u p o n ft, a n d assume t h a t G c o n t a i n s a 2 , - c y c l e . T h e n G i s e i t h e r t h e f u l l s y m m e t r i c g r o u p o r t h e a l t e r n a t i n g g r o u p o n ft.

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P r o o f . Suppose that ( 0 , ^ , 7 ) is a 3-cycle contained i n G, and let A = ft - {a,/3,7>. T h e n GA contains the 3-cycle (a , £ , 7 ) , and hence GA is transit ive on ft - A , and it is pr imi t ive since 3 is prime. W r i t i n g n = |ft|, we have | A | + 1 = n - 2, and it follows by Theorem 8.18 that G is (n - 2)-transitive on ft. A s we have seen, this guarantees that G is either the full symmetric group or the alternating group on ft. •

We begin work toward a proof of Theorem 8.18 wi th the following sufficient condi t ion for pr imi t iv i ty .

8 .20. L e m m a . L e t G be a g r o u p a c t i n g t r a n s i t i v e l y o n ft, a n d l e t H Ç G be a s u b g r o u p . L e t X Ç ft be a n H - o r b i t , a n d assume t h a t H i s p r i m i t i v e o n X a n d t h a t \X\ > | f t | /2 . T h e n G i s p r i m i t i v e o n ft.

P r o o f . Let A be a block for the action of G on ft. We argue that if A < ft, then A n X contains at most one point. To see this, let h G H , and observe that ( A n X ) - h = A - h n X . If this set is different from A n X , then clearly A - h ^ A , and since A is a block for G, it follows that A - h n A = 0. If ( A n X ) - h ^ ( A n l ) , therefore, we have ( A n X ) n ( A n X ) - h = 0, and so if A n X is nonempty, it is a block for the act ion of H on X . B y assumption, however, H is pr imi t ive on X , and thus i f A n A is nonempty, it must either be a l l of X , or else it contains just one point . If A n X = X , then A D A , and so | A | > | A | > | f t | /2 . B y L e m m a 8.11, however, | A | is a divisor of |ft|, and hence i n this case, A = ft, which is contrary to assumption. It follows that | A n A I < 1, as claimed.

N o w suppose that A is an arbi trary block for the action of G on ft. We must show that A is a t r iv i a l block, and so assuming that A < ft, our goal is to show that | A | = 1. B y L e m m a 8.11, the translates of A are blocks, and, of course, none of them is a l l of ft. There are exactly | f t | / | A | such translates, and every point of ft is i n one of them. B y the result of the previous paragraph, each translate of A contains at most one point of X , and thus the number of translates is at least equal to \X\. T h e n | f t | / | A | > \X\ > | f t | /2 , and hence | A | < 2. Thus | A | = 1, as required. •

Suppose that G acts transit ively on ft and let X < ft. We shall say that A is a J o r d a n set i f the pointwise stabilizer G x is transit ive on ft - X , and that X is s t r o n g l y J o r d a n if, i n fact, Gx is pr imi t ive on ft-X. We caution the reader that this nomenclature is not standard; we have introduced it only for the material leading up to the proof of Theorem 8.18.

It should be obvious that if G acts on a set ft and X Ç ft is an orbit for some subgroup H Ç G, then for each element g € G, the translate X - g is an orbit for the conjugate subgroup H 9 . Furthermore, i f H is pr imi t ive on X , then H 9 is pr imi t ive on X - g . A l so , if X Ç ft and g G G, then

( G x ) 9 = Gx-g. It follows from al l of this that translates of Jordan sets are Jordan, and translates of strongly Jordan sets are strongly Jordan.

8.21. L e m m a . Suppose t h a t a g r o u p G acts t r a n s i t i v e l y o n ft, a n d l e t X a n d Y be J o r d a n subsets such t h a t X U Y < ft. T h e n X n Y i s J o r d a n , a n d if b o t h X a n d Y a r e s t r o n g l y J o r d a n , t h e n X n Y i s s t r o n g l y J o r d a n .

Proof. F i r s t , observe that Gx and GY are contained i n GXnY, which acts on ft-(XnY). Thus each G ^ - o r b i t and each G y - o r b i t on the set ft-(XnY) is contained i n a single GXnY orbit . N o w ft - ( X n Y) = (ft - X ) U (ft - Y ) , and since X and Y are Jordan, it follows that ft-X is a G x - o r b i t and ft-Y is a G y - o r b i t . To show that GXnY is transitive on ft - ( X n Y ) , therefore, it suffices to show that ft-X and ft-Y are contained i n the same G X n y - o r b i t . T h i s follows, however, since (ft - X ) n (ft - Y) = ft - { X U Y) is nonempty. Thus G x n Y is transitive on ft - ( X n Y ) , or in other words, X n Y is a Jordan set, as desired.

Suppose now that both X and Y are strongly Jordan, and assume, as we may, that |X| < \Y\. T h e n |ft - X | > |ft - Y\, and since the sets ft - X and ft - Y are not disjoint, it follows that |ft - X| is more than half the cardinal i ty of their union, which is ft - ( X n Y ) . Now Gx stabilizes the set ft - X and acts pr imi t ive ly on i t , and thus by L e m m a 8.20, the action of G x n Y on ft - ( X n Y ) (which we know to be transitive) is actual ly pr imit ive . It follows that X n Y is strongly Jordan, as required. •

Suppose that G is transitive on ft. If X ç ft is a subset obtained by deleting one point, then obviously X is a Jordan set. A l so , the empty subset of ft is guaranteed to be Jordan. B u t if G is pr imi t ive on ft and there exists a nonempty Jordan set of cardinal i ty smaller than |ft| - 1, we get a strong conclusion: the action of G on ft is doubly transitive.

8.22. Theorem. L e t G be a g r o u p a c t i n g p r i m i t i v e l y o n ft, a n d suppose t h a t X C f t i s a J o r d a n set w i t h 0 < |X| < |ft| - 1. If a G ft i s a r b i t r a r y , t h e n Ga i s t r a n s i t i v e o n ft - { a } , a n d so G i s 2 - t r a n s i t i v e o n ft. F u r t h e r m o r e , if X i s s t r o n g l y J o r d a n , t h e n i n f a c t , Ga i s p r i m i t i v e o n ft - { a } .

Proof. G iven the existence of the Jordan set X w i t h 0 < | X | < | f t | - l , we show that every one-point subset of ft is Jordan. Th i s tells us that for each point a G ft, the stabilizer Ga is transitive on ft - {a} , as wanted, and since G is transitive, L e m m a 8.2 guarantees that G is doubly transitive i n this si tuation. We w i l l also show that i f X is strongly Jordan, then every one-point subset of ft is strongly Jordan, and this w i l l complete the proof.

We can assume that X is m in ima l among nonempty Jordan subsets (or min ima l among nonempty strongly Jordan subsets) and we work to show that |X| = 1. Since translates of Jordan sets are Jordan, and translates of

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strongly Jordan sets are strongly Jordan, it w i l l follow that every one-point subset of ft is Jordan (or is strongly Jordan) as wanted.

F i r s t , assume that |X| > |ft|/2, and let y = ft - X . Now \Y\ > 1 since \X\ < |ft| - 1, and we have | Y | < |ft|/2 < |ft|. Since G is pr imi t ive on ft, it follows that Y is not a block, and hence there exists an element g G G such that Y-g ^ Y and Y n Y-g ^ 0. N o w

X U X - g = (ft - Y) U (ft - = ft - ( Y n y -y ) ,

and this is a proper subset of ft. Since X is Jordan, so too is X - g , and thus L e m m a 8.21 tells us that X n X -# is Jordan. A l s o , i f X is strongly Jordan, then by similar reasoning, X n X - 5 is strongly Jordan.

N o w \X\ > |ft|/2 > \ X U A - 5 | / 2 . The cardinal i ty of each of X and X-<y, therefore, exceeds half the cardinal i ty of their union, and it follows that X H X - g is nonempty. Furthermore, X ^ X - g since y ^ Y - 5 , and thus A n X - g < X . Thus X n X - g is a nonempty Jordan (or strongly Jordan) set that is s t r ic t ly smaller than X , and this contradicts the min imal i ty of X .

We now have \X\ < |ft|/2, and assuming that \X\ > 1, we work to derive a contradiction. Since 1 < \X\ < |ft| and G is pr imi t ive , X cannot be a block, and thus there exists g G G such that X ^ X - g and X n X - g ^ 0. Since | X U X 9 | < 2|X| < |ft|, we see that X U X 9 < ft, and so by L e m m a 8.21, the nonempty set X n X » is Jordan (and it is strongly Jordan if X is). T h i s set is properly smaller than X , however, and as before, this contradicts the min imal i ty of X . •

Fina l ly , we are ready to prove Jordan's theorem.

P r o o f of T h e o r e m 8.18. We can assume that |ft| > 2 and that | A | > 0, and we proceed by induct ion on |ft|. B y hypothesis, the set A is strongly Jordan, and since | A | < |ft| - 2, we can apply Theorem 8.22 to deduce that Ga is pr imi t ive on ft - { a } , where we take a G A .

N o w GA is the pointwise stabilizer of the set A - { a } in Ga, and by hypothesis, GA is pr imi t ive on ft - A = (ft - a ) - (A - a ) . A l s o , |A - { a } \ < |ft - { a } | - 2, and so we can apply the inductive hypothesis to the action of Ga to ft - {a} to deduce that it is |A|- t ransi t ive. B y L e m m a 8.2, therefore, G is ( |A| + Intransitive on ft, as wanted. •

Another appl icat ion of Theorem 8.18 i n the spirit of Theorem 8.17 and Coro l la ry 8.19 is the following. Its proof is somewhat more difficult, however.

8.23. Theorem. L e t G be a p r i m i t i v e p e r m u t a t i o n g r o u p o n a set ft, a n d assume t h a t G c o n t a i n s a p - c y c l e , w h e r e p i s p r i m e a n d p < |ft| - 3. T h e n G i s e i t h e r t h e a l t e r n a t i n g g r o u p o r t h e f u l l s y m m e t r i c g r o u p o n ft.

We need an easy prel iminary result.

8.24. L e m m a . L e t x be a n n - c y c l e i n t h e s y m m e t r i c g r o u p Sn. T h e n ( x ) i s t h e f u l l c e n t r a l i z e r of x i n Sn.

Proof. A l l n-cycles in Sn are conjugate, and it is easy to see that there are exactly (n - 1)! of them. Since the size of the conjugacy class of x i n Sn is \Sn : C\, where C is the centralizer of x , we have (n - 1)! = n \ / \ C \ , and thus \ C \ = n . Since ( x ) C C and |(x>| = n , the result follows. •

P r o o f of T h e o r e m 18.23. Let A be the set of points fixed by the given p-cycle, and observe that by Corol la ry 8.19, we can assume that p £ 3. Now |ft - A | is the prime number p , and since GA contains a p-cycle, it follows that G A is transitive on ft - A , and i n fact it is pr imi t ive on this set. B y Theorem 8.18, therefore, G is | A (-transitive on ft. (Actual ly , it is ( |A| + 1)-transit ive, but we do not need that extra information.) It follows that each permutat ion i n S y m ( A ) is induced by some element of G, or i n other words, G ( A ) / G A is natural ly isomorphic to S y m ( A ) .

Now let P be the subgroup of order p generated by the given p-cycle. T h e n P C GA, and since GA acts faithfully on the p points of ft - A , we know that | G A | divides p \ . It follows that P is a full Sylow p-subgroup of G A , and since G A < G ( A ) , we can apply the Fra t t in i argument to deduce that G ( A ) = N G A , where N = N G ( A ) ( P ) . We conclude that N / N A = G ( A ) / G A = S y m ( A ) , and thus N induces the full symmetric group on A .

Since | A | = |ft| - p > 3, it follows that Sym(A) contains a 3-cycle, and it is easy to see that such a 3-cycle must lie i n the derived subgroup of the symmetric group. There exists, therefore, an element x e N ' such that the permutat ion induced by x on A is a 3-cycle.

The automorphism group of P is abelian, and thus N / C is abelian, where C = C N ( P ) . T h e n x e N ' C C , and so x commutes w i t h the given p-cycle. T h e permutat ion of ft - A induced by x , therefore, commutes wi th a p-cycle on these p points. B y L e m m a 8.23, this permutat ion has order d iv id ing p , and hence x p acts t r iv ia l ly on ft - A . N o w x induces a 3-cycle on A , and since 3 does not divide p , it follows that x p also induces a 3-cycle on A . A t this point, we know that x p fixes the points of ft - A and it induces a 3-cycle on A . Since G acts faithfully on ft, we see that x p actually is a 3-cycle, and the result follows by Corol la ry 8.19. •

We close this section w i t h a result of A . Bochert that also relies on Coro l la ry 8.19. Bochert 's theorem shows that a pr imi t ive permutat ion group that is not either the alternating group or the full symmetric group must actually be a fairly small subgroup of the symmetric group. We begin wi th

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an elementary (but somewhat tedious) computat ion that enables us to find 3-cycles i n permutat ion groups.

8.25. L e m m a . L e t x a n d y be p e r m u t a t i o n s o n a set ft, a n d assume t h a t t h e r e i s e x a c t l y o n e p o i n t a e ft m o v e d by b o t h x a n d y . T h e n t h e c o m m u t a t o r [ x , y ] = x - ^ y - ^ x y i s a 3 - c y c l e .

Proof. F i r s t , we observe that a , a-x and a-y are three dist inct points. Certainly, a + a-x and a ^ a-y since x and y move a . A l s o , since a ^ a-x, we have a-y + { a - x ) - y = a-x, where the equality holds since a-x is different from a and is moved by x , and so it is fixed by y .

Next , we show that if 0 G ft is not one of the three points a , a-x or a-y, then p-ix'^y-1) = / ^ ( y ^ x " 1 ) . There is nothing to prove if bo th x ' 1 and y " 1 fix p, so by symmetry, it is no loss to assume that x " 1 moves P, and thus x moves p. Since x moves 0 and 0 + a , we know that y fixes 0, and so y " 1 also fixes 0, and we have 0-{y-lx~l) = 0-x~l. A l so , since x moves 0, i t also moves 0-x~l, and since 0-x'1 + a , it follows that y fixes 0-x~l, and so y " 1 also fixes this point. T h e n 0-(X-1y-1) = 0-x'1, and this shows that 0-(x-1y~1) = 0-{y-lx~l), as claimed. It follows that

0 . [ x , y) = 0-{x-1y~lxy) = 0-{y~lx'1xy) = 0 .

We have now shown that [ x , y ] fixes every member of ft other than a , a-x and a-y.

Now

{ a - x ) - { x , y \ = a - { y - l x y ) = a - { y - x y ) = a

where the second equality holds because a-y'1 is moved by y and is different from a , and so it is fixed by x . It follows by interchanging the roles of x and y that [ y , x ] carries a-y to a , and hence [ x , y ] = [ y , x ] ~ l carries a to a-y. T h e points a-x, a and a-y are distinct, and we know that [ x , y] carries the first of these to the second and the second to the th i rd . Since [ x , y] fixes a l l other points i n ft, it must carry a-y back to a-x, and so [ x , y] is the 3-cycle (a-x, a , a-y). •

8.26. T h e o r e m (Bochert) . L e t S = Sym(ft) , w h e r e |ft| = n , a n d suppose G i s a p r o p e r s u b g r o u p of S t h a t i s p r i m i t i v e o n ft. If G + A l t ( f t ) , t h e n \ S : G \ > [(n + l ) / 2 ] ! .

Proof. Since the pointwise stabilizer Sa is t r i v i a l , there certainly exists a subset A C ft such that SA n G = 1. Choose A w i t h this property so that TO = | A | as small as possible. N o w SA acts faithfully on ft-A, and since S is the full symmetric group on ft, it follows that SA induces the full symmetric group on ft - A , and thus | 5 A | = (n - TO)!.

We have

\ S : G \ > \SA :GnSA\ = \SA\ = ( n - m ) ! ,

and so it suffices to show that m < n/2, since then n - TO > n / 2 , and thus n - m > [ (n+ l ) / 2 ] .

If TO > n / 2 , we w i l l show that G contains a 3-cycle. T h e n since G is pr imit ive , Corol la ry 8.19 asserts that G is either the alternating or symmetric group. Th i s is contrary to assumption, however, and that w i l l complete the proof.

Assuming that m > n / 2 , we have |ft - A | = n - TO < TO = | A | , and so by the min imal i ty of | A | , it follows that G n S n - A is nontr iv ia l . Choose a nonidenti ty element x G G n S n - A , and let a G ft be a point moved by x . T h e n a does not lie in ft - A , and so a € A .

N o w | A - {a} | < | A | , and so by another appeal to the min imal i ty of | A | , we can choose a nonidentity element y G G n S A _ { a } . Since G n 5 A = 1, it follows that y does not lie i n SA, and thus y must move a . Thus x and y both move a , but there can be no other point moved by both x and y since if 0 7^ a , then either 0 G ft - A , and so 0 is fixed by x , or else 0 G A - {a} and /? is fixed by y . B y L e m m a 8.25, the commutator [ x , y ] is a 3-cycle, and the proof is complete. •

P r o b l e m s 8 B

8B.1 . Let G be transit ive on ft, and let a G ft. F i x a nonempty subset X C ft and let A be the intersection of a l l of those translates of X that contain a . Show that A is a block.

8B.2. Let G be pr imi t ive on ft, and let a,0 G ft w i t h a ^ 0. If X C ft is nonempty and proper, show that there exists an element g G G such that a - g G X and 0-g 0 X .

8B.3. Let G be a pr imi t ive permutat ion group on ft, and suppose that M and N are distinct m in ima l normal subgroups of G. Prove that the following hold.

(a) M and N are regular subgroups.

(b) M = C G ( N ) and N = C G { M ) .

(c) M = N .

H i n t . Use P rob lem 8A.4 .

8B.4. Let G be a solvable pr imi t ive permutat ion group. Show that the degree of G is a prime power and that G has a unique min ima l normal subgroup.

P r o b l e m s 8 B 249

8B.5. Let G be a pr imit ive permutat ion group on ft, and let H = GQ be a point stabilizer. If H has a fixed point i n ft - { a } , show that \ H \ = 1, and that G has prime order.

8B.6. Le t G be a pr imit ive permutat ion group on ft, and let H = G a be a point stabilizer. If # has an orbit of size 2 on ft - { a } , show that \ H \ = 2 and that G = /J>2p, the dihedral group of order 2 p , where p > 2 is a prime number.

8 B . 7 . Le t G be a pr imit ive permutat ion group on ft, and let H = Ga be a point stabilizer. If H has an orbit of size 3 on ft - { a } , show that \ H \ has the form 3-2 e for some integer e > 0.

Hint . Le t ¡5 lie i n an P - o r b i t of size 3 and let D = H p . Show that 0 2 ( £ > ) < 0 2 { K ) , where K = c o v e H { D ) .

Note . B y a result of C . Sims, e < 4, and so \ H \ divides 48. Sims conjectured that if a point stabilizer in a pr imi t ive permutat ion group has an orbit of size r , then the order of the point stabilizer is bounded by some function of r . Sims' conjecture was eventually proved by Cameron, Praeger, Saxl and Seitz, but unfortunately their proof (which is the only one known) relies on the classification of simple groups.

8B.8. Le t x € Sn be the n-cycle ( l , 2 , . . . , n - l , n ) . (In other words, x carries i to i + 1 for 1 < i < n , and it carries n to 1.) Le t 1 < m < n and suppose that H C Sn is a subgroup that acts t ransi t ively on { 1 , 2 , . . . , m } and fixes a l l of the remaining points. Show that the group G = ( H , x ) is pr imi t ive .

Hint . G iven a block A w i th | A | > 1, show that some translate of A contains a point i < m and also a point j > m. Deduce that that translate contains al l i w i t h 1 < i < m.

8B.9. Le t x G Sn be the n-cycle ( l , 2 , . . . , n ) , and let y be the m-cycle ( 1 , 2 , . . . , m ) , where 1 < m < n . Show that G = ( x , y ) is the full symmetr ic group Sn if either of the integers m and n is even, and otherwise, G is the alternating group.

8B.10. Suppose that G C Sn is a subgroup of index n that is not a point stabilizer. Show than n = 6.

Hint . Show that G is transitive, and then use P rob lem 8A.16 to deduce that i t is 2-transitive, and thus pr imit ive . A p p e a l to Bochert 's theorem to deduce that n > [(n + l ) / 2 } \ . Show that the only positive integers n for which this inequality is va l id are 1, 2, 3, 4 and 6.

N o t e . T h e symmetric group 5 6 actually does have a subgroup of index 6 that does not stabilize a point. To see this, observe that S5 acts transit ively on its six Sylow 5-subgroups. It is not hard to show that this is a faithful action, and thus S5 can be embedded as a transitive subgroup of 5 6 , and of course, this copy of S5 i n 5 6 has index 6 and does not stabilize a point.

Note also that i f Sn has a non-inner automorphism a , then a would carry a point stabilizer to a subgroup of index n that is not a point stabilizer, and by this problem, that cannot happen unless n = 6. It follows that a l l automorphisms of Sn are inner for n £ 6. Furthermore, it is easy to show that a subgroup H of index 6 in 5 6 that does not stabilize a point has t r iv i a l core, and this there is an injective homomorphism from S6 to S6 that maps i f to a point stabilizer. Th i s map is a non-inner automorphism of S6.

8 C

Throughout much of this book we have sought results that could be used to prove that a group is not simple, but i n this section, we w i l l use some permutation-group theory to prove that certain groups actually are simple. We begin w i t h a comparatively easy result of this type.

8 .27. T h e o r e m . If n > 5, t h e a l t e r n a t i n g g r o u p A n i s s i m p l e .

P r o o f . Assuming that 1 < N < A n , we work to show that N = A n . Since A n is (n - 2)-transitive i n its natural action, it is certainly 2-transitive, and thus it is pr imi t ive by L e m m a 8.16. B y L e m m a 8.15, therefore, N acts transitively.

Suppose first that n = 5. Since N acts transit ively on a set of cardinal i ty 5, we deduce that 5 divides \ N \ . B u t \ A 5 \ = 60, and so \ A 5 : N \ is not divisible by 5, and thus a l l 5-cycles in A 5 lie in N . The number of these 5-cycles is 24, and thus \ N \ > 24. Since \ N \ divides 60, the only possibilities are \ N \ = 30 or \ N \ = 60, and, in particular, \ A b : N \ is not divisible by 3. T h e n al l 3-cycles in A b lie i n N , and we count that there are 20 of these. Since N contains a l l 5-cycles and al l 3-cycles, we have \ N \ > 24 + 20 = 44. T h e only possibil i ty is that \ N \ = 60, and thus N = A 5 and A b is simple.

We can now suppose that n > 6, and we proceed by induct ion on n to derive a contradict ion i f N < A n . Let H be a point stabilizer in A n , and observe that H = A n - U and so H is simple by the inductive hypothesis. A l so , since N is transitive, we have N H = A n by L e m m a 8.5. T h e n H % N since otherwise, we would have N = A n , which we are assuming is not the case. It follows that N n H is a proper normal subgroup of H , and since H is simple, we have N n H = 1. Since N is transitive, it follows that it is a regular normal subgroup of A n , and by L e m m a 8.5, the conjugation

8 C 251

action of H on the nonidentity elements of TV is permutation-isomorphic to the natural action of H = A n - X on n - 1 points. T h e n | iV | = n , and the action of H on its nonidentity elements is (n - 3)-transitive, and, i n particular, it is 3-transitive. However, this is impossible since by L e m m a 8.8(c), the 3-transitive action of H on the nonidenti ty elements of TV would force |TV| = 4 . •

8 .28 . C o r o l l a r y . If n > 5, t h e s y m m e t r i c g r o u p Sn has e x a c t l y t h r e e n o r m a l s u b g r o u p s : t h e i d e n t i t y , t h e w h o l e g r o u p a n d A n .

P r o o f . Suppose TV < Sn and that TV is not one of A n or Sn. T h e n TV 2 A n , and thus TV n A n is a proper normal subgroup of the simple group A n . It follows that J V n A n = l , and thus \ N \ < \Sn : A n \ = 2. Since Sn is n-transitive on n-points, it is pr imit ive , and thus if TV > 1, then TV is transitive and |TV | > n . Th i s is not the case, however, and so TV = 1. •

We tu rn now to the projective special linear groups P S L ( n , q ) . Reca l l that by definition, P S L ( n , q ) = S L { n , q ) / Z , where Z is the central subgroup of S L ( n , q ) consisting of the scalar matrices w i t h determinant equal to 1. We begin work now toward a proof that w i t h just two exceptions, a l l of the groups P S L ( n , q ) are simple for prime powers q and integers n > 2. The exceptions are P S L ( 2 , 2 ) and P S L ( 2 , 3 ) , which are not simple. (In fact, these two groups are solvable since P S L ( 2 , 2 ) has order 6 and is isomorphic to the symmetric group S3, and PSL(2,3) has order 12 and is isomorphic to the alternating group A 4 . ) Ac tua l ly , the groups P S L ( n , F ) are also simple when F is an infinite field. A l though the proof that we w i l l present for finite fields goes through without essential modif icat ion for infinite fields too, we w i l l nevertheless l imi t our discussion to the finite case.

Assuming that n > 2 (as we w i l l , throughout this discussion), it follows by L e m m a 8.3 that S L { n , q ) acts 2-transitively on the set ft of one-dimensional subspaces of a vector space V of dimension n over the field F of order q. Since the action of a scalar ma t r ix on the vector space V is just scalar mul t ip l ica t ion , the subgroup Z acts t r iv ia l ly on ft. Thus Z is contained i n the kernel of the action, and we can view P S L ( n , q ) = S L ( n , q ) / Z as acting on ft. In fact, this action is faithful.

8 .29. L e m m a . F o r a l l i n t e g e r s n > 2 a n d a l l p r i m e p o w e r s q, t h e g r o u p P S L { n , q ) i s a 2 - t r a n s i t i v e p e r m u t a t i o n g r o u p of d e g r e e ( q n - l ) / (g - 1).

P r o o f . We know that P S L ( 2 , q ) acts on the set ft of one-dimensional sub-spaces of the n-dimensional space V over the field F of order q. Now V contains exactly qn - 1 nonzero vectors, and each of these lies i n a unique one-dimensional subspace. Since each such subspace contains exactly q - 1 nonzero vectors, it follows that |ft| - ( q n - l ) / (g - 1).

It remains to show that P S L ( n , q ) acts faithfully on ft, or equivalently, that Z is the full kernel of the action of S L { n , q ) on ft. To see this, let x G S L ( n , q ) lie i n the kernel of the action. T h e n x fixes every one-dimensional subspace of V, and thus x carries every vector v G V to some scalar mult iple of itself.

Let { v i | 1 < i < n } be a basis for V, and write (Vi)x = am for appropriate scalars o n G F . T h e n x corresponds to a diagonal matr ix , and to complete the proof, it suffices to show that a { = a , for 1 < i < j < n . Now for some scalar a G F , we have

a ( v i + Vj) = { v i + V j ) x = ViX + v j x = am + ajVj ,

and since v{ and V j a r e l inearly independent, this yields o n = a = a j , as required. •

Reca l l that a group G is said to be perfect i f G' = G. A l so , i f S C G is an arbi trary subgroup, then the normal closure of S i n G is the subgroup SG = (S9 \ g <E G ) ; it is the unique smallest normal subgroup of G that contains S. O f course, i f G is a nonabelian simple group, then G is perfect and SG = G for every nontr iv ia l subgroup S. Conversely, if G is perfect and it is the normal closure of an appropriate subgroup, and if, i n addit ion, G is a pr imi t ive permutat ion group, our next result guarantees that G is simple. Since P S L ( n , q ) is a doubly transitive permutat ion group by L e m m a 8.29, it is certainly a pr imi t ive permutat ion group, and hence the following lemma of K . Iwasawa is relevant to establishing the s impl ic i ty of this group.

8 .30. L e m m a (Iwasawa). L e t G be a p r i m i t i v e p e r m u t a t i o n g r o u p , a n d assume t h a t G' = G. L e t H = Ga be a p o i n t s t a b i l i z e r i n G, a n d suppose t h a t t h e r e exists a s o l v a b l e s u b g r o u p A < H such t h a t G = A G . T h e n G i s s i m p l e .

P r o o f . Let 1 < N < G, and observe that since G is a pr imi t ive permutat ion group, N is transitive, and thus N H = G. Now N < G , and since A < H , we see that H C N G ( N A ) . A l so N C N G ( N A ) , and thus since N H - G, we have N A < G. Now A G = G and A C N A < G, and it follows that N A = G. B u t A is solvable, so G / N must also be solvable, and hence i f N < G, we have ( G / N ) ' < G/N. Since G' = G, however, we see that ( G / N ) ' = G/N, and this contradict ion shows that N = G. Th i s completes the proof. •

We w i l l use L e m m a 8.30 to prove that a group of the form G = P S L ( n , q ) is simple (except when n = 2 and q < 3). To do this, we w i l l establish (in the non-exceptional cases) that G' = G, and we w i l l find a solvable (and, in fact, an abelian) normal subgroup A of a point stabilizer, such that A° = G. (Note that the groups P S L ( 2 , 2 ) and P S L ( 2 , 3 ) definitely are not perfect

8 C 253

since they are solvable.) The key to both of these tasks is to consider certain special matrices t ^ { a ) in S L ( n , q ) .

To define the matrices t ^ { a ) and to facilitate mat r ix computations, we recall that the matrix unit' e „ is the mat r ix whose only nonzero entry is 1, occurring m the (t,j)-positi 'on. It is t r i v i a l to check that mat r ix units mul t ip ly according to the formula

It is the s impl ic i ty of this rule that makes the mat r ix units so useful for computat ion.

G i v e n 1 < i, j < n w i th i + j , define U j ( a ) = 1 + a e i j , where, in this formula, 1 is the n x n identity mat r ix and a G F is nonzero. Clearly, t i t j { a ) is either upper or lower tr iangular (depending on whether i < j or i > j ) and its diagonal entries are a l l 1. The determinant of this matr ix , therefore, is 1, and so U A a ) G S L { n , q ) , as wanted.

We digress briefly to put our matrices t i d ( a ) into a somewhat more general context. Le t V be a vector space of dimension n > 2. A hyperplane i n V is a subspace of dimension n - 1, or equivalently, a subspace of codi-mension 1. A linear operator t on V is a transvection i f its fixed-point space is a hyperplane W, and t acts as the identi ty on V/W. It is not hard to show that a l l of our matrices U j ( a ) are transvections and also that the transvections i n G L ( n , q ) form a single conjugacy class. In part icular , a l l of the transvections in G L ( n , q ) actually lie i n S L ( n , q ) , and one can show that if n > 3, a l l of these transvections are conjugate in S L ( n , q ) .

We return now to work toward our m a i n goal, which is the proof that the groups P S L { n , q ) are simple (except when they are not).

8.31. Theorem. L e t n > 2. T h e n t h e g r o u p S L ( n , q ) i s g e n e r a t e d by t h e m a t r i c e s t i d ( a ) w i t h i ^ j a n d a ^ 0.

Proof. If a is any n x n mat r ix over the field F of order g, it is easy to see that the product mat r ix t i d { a ) a can be obtained from a by adding a times row j to row i, and, similarly, the mat r ix a t ^ { a ) can be obtained from a by adding a times co lumn i to column j . We describe an algori thm by which an arbi trary mat r ix a e S L ( n , q ) can be converted into the identity mat r ix by a sequence of row and column operations of these types, and it follows that 1 = p a q , where each of p and q is a product of matrices of the form i i j ( a ) . Th i s w i l l prove the result.

Let a G S L { n , q ) . The first co lumn of a is not zero, and since n > 2, some row operation of the above type (if necessary) w i l l convert a to a mat r ix i n which the (z, l ) -entry is a nonzero field element 5, where i > 1. Assuming

£i,v i f J' — u

0 otherwise.

that has been done, let 7 be the (1, l ) -entry of the resulting matr ix , where possibly, 7 = 0. If we add { l - - y ) / S times row i to row 1, we obtain a mat r ix w i t h (1, l ) -entry equal to 1. Now a sequence of additions of appropriate multiples of the first row to other rows w i l l y ie ld a mat r ix w i t h (1, l ) -entry equal to 1 and al l other entries i n the first column equal to 0, and then a sequence of additions of appropriate multiples of the first column to other columns w i l l y ie ld a mat r ix where the (1, l ) -entry is 1 and a l l other entries i n the first row and first co lumn are 0.

Observe that the mat r ix constructed i n the previous paragraph has determinant 1 since a and al l of the matrices t i d { a ) have determinant 1. We change notat ion now, and cal l this new mat r ix a . If n = 2, then a is diagonal , and since its (1,1) entry is 1 and its determinant is 1, the (2,2)-entry of a must also be 1, and thus a is the identity matr ix , as wanted. Assume then, that n > 3. Since some entry i n column 2 is nonzero, we can do row and column operations as before, but using only rows and columns numbered 2 or larger. Th i s w i l l not change either the first row or the first column of a , but we can arrange that the (2,2)-entry of the resulting mat r ix is 1, and that a l l other entries i n the second row and second column are 0.

If n = 3, the resulting mat r ix is diagonal, and so It is the identi ty mat r ix by the determinant argument that we used before. If n > 3, we again do row and co lumn operations, but this t ime wi th rows and columns numbered 3 or higher. Cont inuing like this, we eventually obtain the identity matr ix , and this completes the proof. •

8 .32 . T h e o r e m . / / n > 3 o r n = 2 a n d q > 3, t h e n S L { n , q ) i s p e r f e c t .

P r o o f . B y Theorem 8.31, it suffices to show that the mat r ix t i d ( a ) is a commutator, where i + j and a ^ 0. We begin w i t h the observation that if i £ j , then ( e ^ ) 2 = 0. T h e n (1 - a e i j ) ( l + a e „ ) = 1, and hence ( l + a e i j ) - 1 = l - a e i J .

Suppose first that n > 3, and let 1 < i , j , k < n , where i, j and k are dist inct . T h e n

(e i , f c ) 1 + a C f c ' J = (1 - a e k t j ) { e i i k ) ( l + aekJ) = e i ) f c + aehJ ,

and thus ( 1 + e i j k ) 1 + a e ^ = 1 + eijk + aeid .

It follows that

[(1 + e i > f c), (1 + a e k J ) ] = (1 + e l t k y l { l + e h k ) 1 + a e ^

= ( l - e i i k ) ( l + e i , k + a e i j )

= 1 + a e i j

= t i , j ( a ) >

8 C 255

and thus t i t j ( a ) is a commutator, as wanted.

We can now assume that n = 2, and we let (3,7 e F be nonzero. Le t

6 0

and c =

so that b and c lie i n S L ( 2 , q ) . T h e n

M " / T 1 0" "1 7 _ 1 7 T 1 0" "1 7 . 0 p . 0 1 0 iij 0 1

0

P - P i 0 / r 1 . .

1 7 ( 1 - / 3 2 ) 0 1

1 0

0

-7 1 j

/5-X7

Q

N o w let a be an arbi trary nonzero element of F . If q > 3, choose (3 / ± 1 . T h e n 1 - /32 ^ 0, and so we can choose 7 so that 7(1 - ( 3 2 ) = a . T h e n [ b , c ] = i i , 2 ( a ) , and so i i , 2 ( a ) is a commutator i n 5 L ( 2 , g ) , as wanted. B y taking transposes, we see that i 2 j l ( a ) is also a commutator in S L { 2 , q ) , and this completes the proof. •

We can now establish the ma in result of this section.

8.33. T h e o r e m . / / n > 3 o r n = 2 a n d q > 3, t h e n P S L ( n , q ) i s s i m p l e .

P r o o f . Us ing the bar convention, we can write P S L ( n , q ) = S = S/Z, where S = S L ( n , q ) and Z is the group of scalar matrices in S. Let V be the n-dimensional row space over F , and let { v { \ 1 < i < n } be the standard basis for V, so that vt is the row vector (0 , . . . , 0 ,1 ,0 , . . . , 0), whose only nonzero entry is i n posi t ion i. Now S acts by right mul t ip l ica t ion on V, and the corresponding action of S on the set ft of one-dimensional subspaces of V is faithful and doubly transitive by L e m m a 8.29. In part icular, S is a pr imi t ive permutat ion group on ft. A l so , since S is perfect by Theorem 8.32, we have ( S ) ' = SJ = S, and so S is perfect.

N o w fix an integer j w i t h 1 < j < n , and let H 3 be the stabilizer in S L ( n , q ) of the one-dimensional space Uj = F V j . T h e n H j acts on the vector space V / U j , and this induces a group homomorphism from H j into the group G L ( V / U j ) of invertible linear operators on V/Ur Let A j be the kernel of this homomorphism, so that A j is the group of a l l matrices i n S L ( n , q ) that stabilize Uj and act t r iv ia l ly on V / U j . In part icular , i f a € A j , then det(a) = 1, and since the linear transformation induced by a on V / U j is the identity map, it too has determinant 1. It follows that the

linear transformation of Uj induced by a also has determinant 1. The map induced by a on the one-dimensional space Uj is thus the identi ty map, and so ( V j ) a = Vj. A l so , since a fixes every element of V / U j , we see that if i ^ j then ( V i ) a = v{ + favj for some scalar fa depending on i . In other words,

a = 1 + fceid '

and it is easy to see that A j is exactly the set of matrices of this form. It follows by the mul t ip l ica t ion rule for mat r ix units that A 3 is an abelian group, and, of course, A j < H j since A j is the kernel of a homomorphism defined on H j . Not ice also that the matrices t i t j { a ) lie i n A j , for i ^ j .

Now S = S L ( n , q ) acts transit ively on the set of one-dimensional sub-spaces of V, and since the group A j is uniquely determined by the space Uj = F v j , i t follows that a l l of the abelian groups A j are conjugate i n S. (We are not saying, however, that they form a full conjugacy class of subgroups.) Every mat r ix of the form £„•(«) lies i n one of the conjugate subgroups A j , and since these matrices generate S by Theorem 8.31, it follows that ( A x ) 5 = S.

N o w H [ is a point stabilizer i n the pr imi t ive group S, and I T is a solvable (and in fact abelian) normal subgroup o f f f j . (It is irrelevant that A x usually does not contain Z . ) A l so , ( A [ ) s = { A i ) s = S. Since S is perfect, it follows by L e m m a 8.30 that it is simple. •

We mention that the s impl ic i ty of P S L ( n , q ) implies that the scalar subgroup Z C S L ( n , q ) is actually the full center of S L ( n , q ) .

P r o b l e m s 8 C

8 C . 1 . Show that A 6 has no subgroup of order 120, and deduce that a group of order 120 cannot be simple.

N o t e . There does exist a perfect group of order 120, however: S L ( 2 , 5).

8 C . 2 . Suppose that G is a permutat ion group of degree 11 and order 7920 = 1 M 0 - 9 - 8 . Prove that G is simple.

H i n t . Show that | N G ( P ) | = 55, where P G S y L l ( G ) .

N o t e . In fact, G must be isomorphic to the M a t h i e u group M u .

8 C . 3 . Suppose that G is a transitive permutat ion group of degree 12 and order 95,040 = 12-1M0-9-8 . Prove that G is simple.

H i n t . Show that G must be 2-transitive, and that a nonidenti ty proper normal subgroup would have to be regular. Observe that a group of order 12 can never be a min ima l normal subgroup of any group.

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Note . In fact, G must be isomorphic to the M a t h i e u group M 1 2 .

8 C . 4 . Le t G be a transitive group of degree 100 on a set ft, and suppose that a point stabilizer Ga is simple, and that it has orbits of size 22 and 77 on ft - {a}. Show that G is pr imi t ive , and deduce that it is simple.

Note . The Higman-Sims group H S satisfies these conditions, and, i n that case, the point stabilizer H is isomorphic to the M a t h i e u group M 2 2 .

8 C . 5. Let G be a 5-transitive group of degree 24, and assume that the setwise stabilizer of three points is simple. Prove that the stabilizer i n G of two points, the stabilizer i n G of one point and G itself are simple groups.

Note. T h i s problem describes the M a t h i e u groups M 2 2 , M 2 3 and M 2 4 . The three-point stabilizer is isomorphic to P S L ( 3 , 4 ) , acting 2-transitively on 21 = (4 3 - l ) / ( 4 - 1) points, as i n L e m m a 8.29.

8 C . 6 . Let A be an abelian group. Show that A u t ( A ) is simple i f and only if A has order 3 or A is an elementary abelian 2-group of order at least 8.

8 D

We assume throughout this section that G acts t ransi t ively on ft, and we study the orbits of the action of a point stabilizer on ft. A s we shall see, a good way to understand these suborbits is to consider the natural componentwise action of G on the set ft x ft of ordered pairs of points, where this action is denned by the formula ( a , 0 ) - g = (a-g,0-g).

The set ft x ft is part i t ioned into G-orbi ts , which are usually called orbitals. One of these is the diagonal {(a, a) | a G ft}. (The diagonal is certainly G-invariant , and it is a single G-orbi t because G is transit ive on ft.) If |ft| > 1, the diagonal obviously does not exhaust ft x ft, and so the to ta l number of G-orbits on this set (in other words, the to ta l number of orbitals) is at least 2. The number of orbitals is the rank of the action, which is defined only for transitive actions. If A is an orbi ta l , it is obvious that the set

A ' = {(/?, a) I ( a , 0 ) G A}

is also an orbi ta l , and we say that A and A ' are paired orbitals. O f course, an orbi ta l may be paired w i t h itself. For example, the diagonal orbi ta l is self-paired, and i f the rank is 2, there is only one other orbi tal , so i t too must be self-paired.

A transit ive action of G on ft has rank 2 precisely when |ft| > 1 and G acts transi t ively on the set of a l l of pairs i n ft x ft that lie outside of the diagonal. These, of course, are exactly the pairs of d i s t i n c t points i n ft, and we recall that by definition, G is doubly transit ive on ft i f and only i f it


acts transit ively on this set of pairs. In other words, a transitive action is 2-transitive i f and only if it has rank 2.

If G has rank 2 on ft, it is 2-transitive, and, in this case, we know that a point stabilizer Ga has exactly two orbits on ft, namely, { a } and ft - {a}. More generally, we shall see that the number of G a - o r b i t s on ft is always equal to the rank.

It should not be surprising that the number of orbits of Ga on ft is independent of the choice of a since by transit ivi ty, the various point stabilizers are conjugate in G. We might even expect that the sizes of these suborbits correspond as a varies. In fact, even more is true. G i v e n points a , 0 G ft, there is a natural , size-preserving bijection between the sets of G Q -o rb i t s and G/3-orbits on ft. To define this bijection, we introduce some notation.

Let A be an orbi tal . We write

A ( a ) = {0£ ft | ( a , 0 ) G A } ,

so that the orbi ta l A defines a function from ft into the set of a l l subsets of ft. These o r b i t a l f u n c t i o n s are int imately connected wi th the suborbits.

8 .34. L e m m a . L e t G be a g r o u p a c t i n g t r a n s i t i v e l y o n ft, a n d l e t a i n ft. T h e n as A r u n s o v e r t h e o r b i t a l s , t h e sets A (a) a r e d i s j o i n t , a n d they a r e e x a c t l y t h e o r b i t s of Ga o n ft. A l s o , if g G G, t h e n f o r each o r b i t a l A , w e h a v e A ( a - g ) = A ( a ) - g . I n p a r t i c u l a r | A ( a ) | i s i n d e p e n d e n t of a G ft, a n d , i n f a c t , \ A { a ) \ = | A | / | f t | .

P r o o f . Suppose that A x ( a ) and A 2 ( a ) have some point 0 m common, where A x and A 2 are orbitals. Th i s means that the pair { a , 0 ) lies i n each of the sets A x and A 2 , and since these sets are G-orbits on ft x ft and they are not disjoint, they must be equal. It follows that the sets A ( a ) are disjoint for dist inct orbitals A .

N o w let g G G, and suppose A is an orbi tal . Since A is a G-invariant subset of ft x ft, it follows that { a , 0) G A if and only if { a - g , 0-g) G A . Equ iva l en t s , 0 € A ( a ) i f and only if 0-g G A ( a - g ) , and this proves that A ( a - g ) = A { a ) - g , as wanted. A l so , since a - g runs over a l l of ft as g varies, it follows that for a fixed orbi ta l A , a l l of the sets A ( a ) have equal cardinality.

If g G Ga, we have A { a ) - g = A ( a - g ) = A ( a ) , and thus A ( a ) is a Ga-invariant subset of ft. To show that it is actually a G Q - o r b i t , let 0,7 G A { a ) . T h e n ( a , 0 ) and (a , 7) lie i n A , which is a G-orbi t on ft x ft. There exists, therefore, an element g G G that carries ( a , 0) to (a , 7). T h e n 0-g = 7, and since a - g = a , w e have g G Ga, as wanted.

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F ina l ly , for a € ft, we see that | A ( a ) | is exactly the number of pairs in A whose first entry is a , and thus

| A | = ^ | A ( a ) | . a & f l

B u t | A (a) | is independent of a , and this yields | A | = | f t | | A ( a ) | . •

A s we indicated, L e m m a 8.34 establishes a natural bijection from the set of G Q - o r b i t s on ft onto the set of G^-orbi ts on ft, where a , 0 € ft. G i v e n a G a - o r b i t X , we know that X = A ( a ) for some unique orbi ta l A , and we let X correspond to the G^-orbi t Y = A (/J). In fact, given X , it is easy to determine Y wi thout knowing the orbi ta l A . T h i s is because 0 = a - g for some element g € G , and so

Y = A ( 0 ) = A ( a - g ) = A ( a ) - g = X - g .

The cardinalities of the orbits of Ga for any given point a are called the s u b d e g r e e s of G , and this list of positive integers is independent of the choice of a . A l so , at least one of the subdegrees is the number 1 since {a} is an orbit of Ga. (Note that this t r i v i a l G a - o r b i t is A (a) , where A is the diagonal orbital.)

In fact, the corresponding orbits X and Y for Ga and G0 share more than cardinalit ies. F i x g € G carrying a to 0. T h e n the isomorphism u ^ v P from Ga to G0 and the bijection x ^ x - g from X to Y are compatible in the sense that

{ x - u ) - g = ( x - g ) - u 9 .

It is easy to see from this that a property such as pr imi t iv i ty , 2-transitivity, faithfulness, etc. holds for the action of G Q on X i f and only if it holds for the act ion of G p on Y. (We hope that the reader w i l l forgive us for omi t t ing a formal proof of this.)

A useful way to th ink about orbitals is v i a graph theory. We can view an orbi ta l A as the set of edges of a directed graph wi th vertex set ft. To do this, we imagine ft to be set of points i n space, and we draw an arrow from a t o 0 precisely when the pair ( a , 0 ) lies i n A . (This is not very interesting if A is the diagonal, but in a l l other cases, each arrow joins a pair of dist inct points.) The graph constructed in this way is the o r b i t a l g r a p h associated w i t h A , and we w i l l refer to its edges A-ar rows .

Since every ordered pair of points in ft lies i n exactly one orbi ta l , it follows that for each such pair ( a , 0 ) , there is a unique orbi ta l A such that A - a r r o w goes from a to 0. Perhaps a good way to th ink about this is to imagine the orbitals as colors. There is an arrow jo in ing a t o 0 for every choice of a , 0 e ft, and the "color" of that arrow is determined by the orbi ta l

that contains the corresponding pair (a , 0 ) . (This point of view is convenient when we need to th ink about two or more orbi ta l graphs simultaneously.)

Observe that for each orbi ta l A , the set of points reached by A-arrows leaving a is exactly A (a). A l so , and most important ly, i f there is a A-a r row from a to 0, then there is also a A-a r row from a - g to 0-g. (This , of course, is because A is a G-invariant subset of ft x ft.) It follows that the given action of G on ft defines automorphisms of each of the orbi ta l graphs. O f course, G acts transit ively on the points in the A-g raph , but it also is edge-transitive because A is a G-orbi t on ft x ft.

The following result provides a pretty connection between the graph theory and the group theory.

8.35. Theorem. L e t G be a g r o u p a c t i n g t r a n s i t i v e l y o n ft. T h e n G i s p r i m i t i v e if a n d o n l y if t h e g r a p h a s s o c i a t e d w i t h every n o n - d i a g o n a l o r b i t a l i s connected.

The statement of Theorem 8.35 is ambiguous since there are two different notions of connectivity relevant to directed graphs. A directed graph is path connected if given any two vertices a and 0, there is a sequence of vertices a = a Q , a i , . . . , a k = 0 such that for 0 a m . The graph is topolog ica l^ connected i f such a sequence of vertices exists and for 0 a m or there is an arrow

-»• o n - (Of course, a path-connected graph is topologically connected, but i n general, the converse is false.) Fortunately, i n the presence of sufficient symmetry, the two concepts coincide.

8.36. L e m m a . L e t Q be a finite d i r e c t e d g r a p h , a n d suppose t h a t Q i s t o p o l o g i c a l l y connected. If t h e a u t o m o r p h i s m g r o u p of Q i s t r a n s i t i v e o n v e r t i c e s , t h e n G i s p a t h connected.

Proof. It suffices to show that if 0 -> a i n Q, then there is a path (following arrows) from a to 0. Let g be an automorphism of Q such that 0-g = a , and suppose that g has order n . Since g is a graph automorphism, we have a = 0-g -> a-g, and thus a-gi - > a-gi+1 for a l l integers i. Th i s yields the path

a —> a - g —> a-g2 — • • • • — > • a-gn~l = a-g~l = 0,

and this completes the proof. •

In the si tuation of Theorem 8.35, the group G acts on each orbi ta l graph v i a graph automorphisms, and it is transit ive on the vertex set ft. B y L e m m a 8.36, therefore, there is no real ambiguity i n the statement of 8.35.

P r o o f of T h e o r e m 8.35. F i r s t , assume that each nondiagonal orbi ta l graph is connected, and suppose X C ft is a nontr iv ia l block. Choose distinct

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points a and 0 G X , and let A be the orbi ta l that contains the pair (a,/?), so that A is not the diagonal. B y assumption, the orbi ta l graph corresponding to A is path connected, and since X < ft, there must be a A-a r row leaving X . Let 7 -> 8 be a A-a r row w i t h 7 e I and 6 £ X . T h e n bo th (a , /?) and (7, 5) lie i n A , and hence there exists g G G carrying (a, (3) to (7, 5). Now 7 G A and also 7 = a-g G A - # , so X and X - g are not disjoint. B u t X is a block, and thus

8 = 0-g G X - g = X .

This is a contradict ion, and hence G is pr imi t ive .

Conversely, suppose that G is pr imi t ive , and fix a non-diagonal orbi ta l A . The orbi ta l graph of A can be decomposed into disjoint t o p o l o g i c a l ^ connected components. These components are permuted by the graph automorphisms induced by elements of G, and it follows that each component is a block. B u t A is not the diagonal, and hence each point is joined by a A-a r row to some other point. A connected component X of the corresponding orbi ta l graph thus contains more than one point, and by pr imi t iv i ty , it follows that X = ft. The graph is thus t o p o l o g i c a l ^ connected, and hence it is pa th connected too. •

F ina l ly , we are ready use this graph-theoretic point of view to establish some results about suborbits and subdegrees of pr imi t ive actions.

8 .37. T h e o r e m . L e t G be a g r o u p a c t i n g p r i m i t i v e l y o n ft, a n d suppose t h a t t h e subdegrees a r e 1 = mx < m2 < • • • < mr, w h e r e r i s t h e r a n k . T h e n

f o r l < i < r , w e h a v e ml+1/ml < m2.

P r o o f . Let a G ft, and let A be an orbi ta l w i t h | A ( a ) | = m2. Since we can assume that A is not the diagonal, the corresponding orbi ta l graph is pa th connected, and so for each point 0 G ft, there is a pa th (following A-arrows) from a to 0. Wr i t e d{0) to denote the length of the shortest such path, so that for example, d ( a ) = 0, and d{0) = 1 precisely when 0 G A (a).

Now let i < r . l i m x = m i + 1 , there is nothing to prove, so we can assume that m l + 1 > m i , and hence there is at least; one G a - o r b i t of size exceeding m i . Let 0 lie in one of these large G Q - o r b i t s , and choose 0 so that d = d{0) is as small as possible. Observe that d > 1 because otherwise, 0 is either equal to a or else it lies in A (a), and i n either case i t lies in a G Q - o r b i t of size at most m 2 , which does not exceed mj . It follows that there exists a point 7, where ¿(7) = d - 1 and there is a A-a r row 7 /?.

Now let X and Y be the unique orbitals such that (a , 7) G X and (a,0) G Y. B y the choice of 0, we have \ Y { a ) \ > ml, and so \ Y ( a ) \ > m l + 1 . Also , we have \ X ( a ) \ < m i by the min ima l i ty of d.

One can travel from a to 0 by first traversing the X - a r r o w from a to 7, and then the A-a r row from 7 to 0. Since Ga fixes a and induces automorphisms of both the X - g r a p h and the A-g raph , it follows that one can travel from a to an arbi trary point in the G a - o r b i t of (3 by first traversing some X - a r r o w and then some A-ar row. In other words, every point in Y { a ) can be reached in this way.

The number of X-a r rows leaving a is \ X ( a ) \ < mu and the number of A-arrows leaving each point i n ft is m 2 . It follows that the number of different points that can be reached from a by first traversing an X - a r r o w and then a A-a r row is at most m 2 m i . We now have

m + i < \Y{<*)\ < m2mi)

and the result follows. •

Note that i f m 2 = 1 in Theorem 8.37, then a l l G a - o r b i t s have size 1, which means that Ga acts t r iv ia l ly on ft. Thus if G is a pr imit ive permuta t ion group and a point stabilizer fixes a second point, then the point stabilizer is the t r iv ia l subgroup. B y pr imi t iv i ty , however, the point stabilizer is a max ima l subgroup, and thus \ G \ is prime. B u t of course, one does not really need anything as sophisticated as Theorem 8.37 to obtain this conclusion, which is P rob lem 8B.5 .

Next , we give another necessary condi t ion for a list of positive integers to be the subdegrees of a pr imi t ive action.

8 .38. T h e o r e m (Weiss). L e t G be a g r o u p a c t i n g p r i m i t i v e l y o n ft, a n d l e t n be t h e l a r g e s t s u b d e g r e e . If m i s a subdegree c o p r i m e t o n , t h e n m = l .

It is convenient to introduce a l i t t le more notat ion. Suppose that a , fie ft, where, as usual, G is transitive on ft. T h e n the pair (a,0) lies in some orbi ta l A , and in this si tuation, we have said that a -> (3 is a A-ar row. W h e n our pr imary concern is the size of the G a - o r b i t containing 0, we w i l l say that a - » 0 is an m-arrow, where m is this orbit size. Note that

m = \ G a : G a n G 0 \ = \Gp : G a n G 0 \ ,

where the second equality holds because \ G a \ = \ G p \ . If a -* 0 is an m¬arrow, therefore, then 0 -* a is also an m-arrow. Th i s symmetry is also a consequence of the fact that m = | A | / | f t | = | A / | / | ^ | , where A ' is the orbi ta l paired w i t h A .

The first conclusion of the following lemma is needed for our proof of Theorem 8.38; the second and th i rd w i l l not be used unt i l later.

8 .39. L e m m a . L e t G be a g r o u p a c t i n g t r a n s i t i v e l y o n ft, a n d l e t a,0,j G ft. Suppose t h a t a -> 0 i s a n m - a r r o w a n d a -> 7 i s a n n - a r r o w , w h e r e m

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a n d n a r e c o p r i m e , a n d l e t u be t h e i n t e g e r a s s o c i a t e d w i t h t h e a r r o w s (3 - + 7 a n d - f ^ p . T h e f o l l o w i n g t h e n h o l d .

(a) u > n .

(b) u d i v i d e s m n .

(c) G 7 G Q C G y G 0 .

Note that the subgroup products i n L e m m a 8.39(c) are subsets of G , but they need not be subgroups. Before presenting the proof, we briefly review a few elementary facts. Let X and Y be subgroups of a finite group G. T h e n \ G \ > \ X Y \ = | X | | Y | / | X n Y | , and thus \ G : Y \ > \ X : X n Y \ . A l so , \ G : X \ and \ G : Y\ divide \ G : X n Y\, and so if these indices are coprime, then \ G : X \ \ G : Y\ divides \ G : A D Y | . T h e n | G : A | | G : y | < | G : X D Y | , and we deduce that |G| < |X | |Y|/|X n Y\. Equa l i t y thus holds in a l l of these inequalities, and in particular, |G : Y\ = \ X : X n Y\ and | G : X \ \ G : Y\ — |G : X n y | . A l so , G = A Y i n this case.

A B C

P r o o f o f L e m m a 8.39. Wri te A = Ga, B = Gp and C = G 7 . T h e n \ A : A n B \ = m and \ A : A n G | = n , as indicated i n the diagram. Since m and n are coprime, it follows that

n = \ A : A n C\ = \ A n B : ( A n C ) n ( A n B ) \

= \ A n B : ( A n B ) nc\

< \ B : B D C \


proving (a). Also , since \ A \ = \ B \ , we see that u = \ B : B n C\ divides

\ B : A n B n C \ = \ A - . A n B n C \

= \ A : ( A n B ) n ( A n C ) \

= \ A : A n B \ \ A : A n C \

= r a n ,

and this establishes (b). F ina l ly , A = ( A n C ) ( A n B ) , and thus

C A = C { A r \ C ) { A r \ B ) = C { A r \ B ) C C B ,

and this is (c). •

P r o o f of T h e o r e m 8.38. Let a G ft. Assume that m ^ 1, and choose an orbi ta l A so that | A ( a ) | = m. T h e n A is not the diagonal, and hence by p r imi t iv i ty and Theorem 8.35, the corresponding orbi ta l graph is connected. Every A-a r row is an m-arrow, however, and hence there is a path consisting of m-arrows from a to every other point i n ft. For 5 G ft, let d ( 8 ) denote the length of the shortest such path, and note that d ( a ) = 0.

Now choose (3 G ft so that a - » f3 is an n-arrow. O f course, an m-arrow joins (3 to each point i n A(/3) , and since m ^ n , we w i l l have our desired contradict ion when we prove that every arrow leaving (3 is an n-arrow, or equivalents, that every arrow arr iving at ¡3 is an n-arrow. Now let 7 G ft. We show that the arrow 7 -> (3 is an n-arrow by induct ion on d = d ( j ) , and that w i l l complete the proof.

If d = 0, then 7 = a , and since a -+ ¡3 is an n-arrow, there is nothing further to prove in this case. Now suppose that d > 0, so that there exists a point 5 such that 5 -> 7 is an m-arrow and d ( 5 ) = d - 1. B y the inductive hypothesis, 8 -> /? is an n-arrow, and since m and n are coprime, it follows by L e m m a 8.37 that 7 -> /? is a it-arrow, where u > n. B u t by hypothesis, n is the largest subdegree, and thus u = n and the proof is complete. •

The following theorem about mult iple t rans i t iv i ty can also be proved by graph-theoretic techniques.

8.40. T h e o r e m (Manning) . Suppose G i s a g r o u p a c t i n g p r i m i t i v e l y o n ft, a n d l e t n be t h e l a r g e s t subdegree. Assume t h a t n > 3 a n d t h a t a p o i n t s t a b i l i z e r acts 2 - t r a n s i t i v e l y o n some s u b o r b i t of size n . Then n = |ft| - 1, a n d G i s 3 - t r a n s i t i v e o n ft.

Proof. Let a G ft. B y L e m m a 8.2, we must show that Ga is 2-transitive on ft - { a } . B u t Ga acts 2-transitively on an orbit of size n i n ft, so it suffices to show that n = |ft| - 1.

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B y hypothesis, there exists an orbi ta l A such that Ga is 2-transitive on A (a) and \ A ( a ) \ = n . Let X be an orbi ta l (allowing X = A ) such that an X -arrow joins two distinct points of A (a). Since Ga induces automorphisms of the orbi ta l graph corresponding to X and Ga is 2-transitive on A (a), it follows that every arrow jo in ing two dist inct points of A (a) (in either direction) is an X - a r r o w . In part icular the orbi ta l X is self-paired.

Let 0 G A ( a ) . T h e n X(f3) includes a l l of A ( a ) - {/?}, and so \X(0)\ >

n - l . Now n - 1 is relatively prime to n, and since n - 1 > 1 and n is the largest subdegree, it follows by Theorem 8.38 that n - 1 is not a subdegree. Thus \X(0)\ = n , and so there is a unique point 7 G X { 0 ) w i t h 7 0 A ( a ) .

Since Ga is 2-transitive on A (a), we know that [ G a ) p acts transit ively on the n - 1 points of A ( a ) - { 0 } = X(/3) - {7}. A l so , since (Ga)p is contained in Gp, it stabilizes the set X { 0 ) , and it follows that (Ga)p fixes 7. T h e n { G a ) f 3 C ( G / 3 ) 7 , and since ( G ^ is transit ive on X(/3) - {7}, we see that Gp is 2-transitive on X { 0 ) .

Now n > 3, so there are at least two points in the set X { 0 ) - {7} = A ( a ) - { p } , and since these points lie i n A ( a ) , the arrow jo in ing them must be an A - a r r o w . B u t Gp is 2-transitive on X(0), and so by the reasoning we used before, every two distinct points i n X { p ) are joined by an A - a r r o w . Furthermore, because G is transitive on ft and defines automorphisms of the orbi ta l graph corresponding to X , it follows that for an arbi trary point 8 G ft, every two distinct points of X(8) are joined by an X - a r r o w .

We now complete the proof by showing that X { p ) u { P } = ft, so that n = |ft| - 1 , as wanted. If this is false, then since the orbi ta l graph corresponding to X is connected, there exists an A - a r r o w 8 - » e, where 8 G X { P ) U { P } and e 0 X { P ) U { p } . T h e n e G X { 8 ) , and since e 0 A(/3), we have 5^0. T h e n 5 G X { 0 ) , and hence /? G X ( 5 ) because A is self paired. Also , e 7 0, and since /? and e lie i n X(<5), it follows by the previous paragraph that 0 -> e is an A - a r r o w . T h i s is a contradiction, however, since e#X(0). •

E v e n for impr imi t ive groups, these graph-theoretic techniques can be used to establish some necessary conditions on the set of subdegrees. To state our next result (which is a theorem of C . Praeger and the author) we introduce yet another graph. G i v e n a set D of positive integers, the (undirected) c o m m o n - d i v i s o r g r a p h Q ( D ) is constructed as follows. The vertex set of Q { D ) is D , and distinct integers o, b G D are joined by an edge precisely when they are not relatively prime. Note that i f 1 G D , then no edge is incident w i t h 1, and so {1} is a connected component of Q ( D ) .

8 .41 . T h e o r e m . L e t D be t h e set of d i s t i n c t subdegrees f o r some t r a n s i t i v e g r o u p a c t i o n . T h e n t h e c o m m o n - d i v i s o r g r a p h G { D ) has a t m o s t t h r e e c o n n e c t e d c o m p o n e n t s , i n c l u d i n g { ! } .

Thus, for example, the set {1 ,2 ,3 ,4 ,5} cannot be the full set of sub-degrees for a transitive group action because the corresponding common-divisor graph has four connected components: {1}, {2 ,4} , {3} and {5}.

Unfortunately, we need to introduce s t i l l more notat ion. Let G be transitive on ft, and let m be a subdegree. Reca l l that i f a , 0 G ft, we declared a -> (3 to be an m-arrow if 0 lies in a G a - o r b i t of size m , or equivalently, if a lies in a G r o r b i t of size m . For a G ft, wri te [ a ] m to denote the subset of ft consisting of a and all points that can be reached from a along a path of m-arrows. In other words, [ a ] m is the connected component containing a in what might be called the m-graph on ft, denned by the m-arrows. Since each element of G carries m-arrows to m-arrows, we see that G acts as automorphisms of the m-graph, and thus it permutes the (necessarily disjoint) connected components. It follows that [ a ] m is a block.

Now write

K m ( a ) = ( G p | 0 G [a] m >,

and observe that Ga C K m { a ) . It should be clear that if g G G, we have K m ( a - g ) = K m { a ) 3 , and thus since G acts transit ively on ft, the index km = \ K m ( a ) : Ga\ is independent of a . Of course, km does depend on m , and we stress that it is denned only when TO is a subdegree.

If a - + 0 is a 1-arrow, then Ga fixes 0, and thus Ga = G0. It follows that K x i a ) = Ga, and hence kY = 1. If a /? is an m-arrow and TO > 1, on the other hand, then Ga does not fix 0, and thus Ga ^ G0 and # m ( a ) 5 ( G a , G/j) > G Q . If m > 1, therefore, we have km > 1.

8 .42 . T h e o r e m . L e i G be a g r o u p a c t i n g t r a n s i t i v e l y o n ft. Suppose t h a t m < n a r e subdegrees, a n d assume t h a t every subdegree i s c o p r i m e t o a t l e a s t one of m o r n . T h e n

(a) km d i v i d e s kn a n d

(b) km d i v i d e s n .

P r o o f . F i x 7 G ft and consider the nonempty set S of a l l points a G ft such that a —> 7 is an n-arrow. If a 6 5 and a —> 0 is an m-arrow, we argue that 0 e S. We know that 0 -» 7 is a u-arrow for some subdegree n , and our goal is to show that u = n . B y L e m m a 8.39(a), we have u > n , and if u > n , then of course also u > m , and so u cannot divide either n or m . B u t L e m m a 8.39(b) tells us that u divides m n , and so if u is coprime to either m or n , it would have to divide the other, which is not the case. Th is u is not coprime to either m or n , which is contrary to the hypothesis. It follows that u = n , and thus 0 G 5 , as claimed.

Now fix a € S. Since the set [ a ] m is connected v i a m-arrows, it follows by the argument of the previous paragraph that [ a ] m C S. T h e n [ a ] m C [ 7] n,

8 D 267

and hence K m { a ) C K n { j ) . Since \ G a \ = | G 7 | , we have

fcn | j r w ( 7 ) : G 7 |

Km |ATm(a) : G a |

which is an integer. Th i s establishes (a).

If a /? is an m-arrow, we have G 1 G a C G 7 G / 3 by L e m m a 8.39(c). Since /? a is also an m-arrow and we have shown that 0 G S, we can interchange the roles of a and 0 to deduce the reverse containment, and thus G 7 G Q = G 7 G 0 . Since [ a ] m is connected v i a m-arrows, this reasoning shows that for a l l 5 G [ a ] m , the sets G 7 G 5 are equal, and thus G 7 G a is invariant under right mul t ip l ica t ion by each of the subgroups Gs. It follows that G y G a is invariant under right mul t ip l ica t ion by K m { a ) , which is the subgroup is generated by these Gs. W r i t i n g K = K m ( a ) and noting that Ga C K , we have

G 7 G Q = G ^ G a K = G y K ,

and thus \G-y\\G«\ _ , ^ n x _ , n „ , _ | G 7 | | K |

| G 7 n G Q | - | G 7 G q | - l G ^ K i - ] G ^ K \ • T h e n \ G a : G 7 n G a | = | K : G 7 n K \ , and this yields

n | G Q , : G a n G 7 | \ K - . G j C \ K \ \ G a \ = \ K : G a \ = \ K : G a | = | G 7 n K | = | C ? 7 : G T N K > '

where the first equality holds because 7 lies i n a G Q - o r b i t of size n , and the last one holds because \ G a \ = | G 7 | . Since the quanti ty on the right is an integer, it follows that km divides n , and this completes the proof. •

Theorem 8.41 follows easily from Theorem 8.42(b).

P r o o f o f T h e o r e m 8 .41 . Suppose that u,v,w G D lie in three different connected components of the common-divisor graph Q { D ) . In part icular u, v and w are distinct, and so we can assume that u < v and u < w. Since u and v lie i n different components, every subdegree is coprime to at least one of u or v, and so ku divides v by Theorem 8.42(b). Similar ly, ku divides w, and since v and w are coprime, it follows that ku = 1, and thus u = 1. Th i s shows that there do not exist three connected components different from {1} in Q { D ) , and so there are at most three components in total . •

Us ing Theorem 8.42(a) we can get addi t ional information about the connected components of Q { D ) .

8.43 . T h e o r e m . L e t D be t h e set of subdegrees f o r a t r a n s i t i v e a c t i o n , a n d assume t h a t t h e c o m m o n - d i v i s o r G { D ) a c t u a l l y has t h r e e c o n n e c t e d c o m p o nents: {1}, A a n d B . If B c o n t a i n s t h e l a r g e s t subdegree n , t h e n a < b f o r a l l a e A a n d be B . A l s o , every t w o i n t e g e r s i n B a r e j o i n e d by a n edge.

P r o o f . Let a G A and b G B , and suppose that a > b. T h e n fc6 divides a by 8.42(b), and also kb divides fca by 8.42(a). Furthermore, since a € A and n G B , we have a < n, and thus ka divides n by 8.42(b). Since kb divides ka, we see that A;6 divides n, and since it also divides a and fcb > 1, it follows that a and n are not coprime. Th i s is a contradiction, however, since a and n are in different connected components of the common-divisor graph.

If a G A and u, v G J3, then a 1, we conclude that u and v are joined by an edge. •

We saw previously that the set {1 ,2 ,3 ,4 ,5} cannot be the full set of subdegrees of a transitive action because it does not satisfy the conclusion of Theorem 8.41. The connected components for the set {1, 2, 3,4}, however, are {1}, {2,4} and {3}, and so this set is not el iminated by Theorem 8.41. B y Theorem 8.43, however, {1, 2, 3,4} cannot be a subdegree set because i n the notat ion of that theorem, n = 4, and thus A = {3} and B = {2,4}, and we do not have a < b for a l l a G A and b G B .

We close wi th an applicat ion of this material outside of the realm of transit ive permutat ion groups.

8 .44. C o r o l l a r y . L e t A a c t o n G v i a a u t o m o r p h i s m s , w h e r e A a n d G a r e finite g r o u p s , a n d l e t D be t h e set of sizes of t h e A - o r b i t s o n G. T h e n t h e c o m m o n - d i v i s o r g r a p h Q { D ) has a t most t h r e e c o n n e c t e d c o m p o n e n t s i n c l u d i n g {1}. A l s o , if t h e r e a r e t h r e e c o m p o n e n t s {1}, A a n d B , w h e r e B c o n t a i n s t h e l a r g e s t member of D , t h e n a < b f o r a l l a G A a n d be B , a n d every t w o i n t e g e r s i n B a r e j o i n e d by a n edge.

P r o o f . Let T = G x A , and view A and G as subgroups of T. T h e n T acts transit ively on the set ft of right cosets of A i n T, and A is the stabilizer of the "point" A e ft. The action of A on ft - { A } is permutat ion isomorphic to the action of A on the nonidenti ty elements of G, and thus the action of A on ft is permutat ion isomorphic to its original action on G. In particular, D is the set of subdegrees for the action of Y on ft, and hence Theorems 8.41 and 8.43 apply. •

A n important special case of Coro l la ry 8.44 is where A = G, acting by conjugation. In that si tuation, D is just the set of class sizes of G, and it follows that the set of class sizes of an arbi trary finite group satisfies the conclusion of Coro l la ry 8.44, and i n part icular, the common-divisor graph on this set has at most three connected components. Not surprisingly, more can be said i n this case than seems to be available from the general theory we have presented.

P r o b l e m s 8 D 269

P r o b l e m s 8 D

8 D . 1 . Let 1 = m i < m 2 < ••• < mr be the subdegrees for a pr imi t ive action of rank r . If m 2 = 2, show that m» = 2 for a l l i > 2.

8D.2. Show that a pr imit ive permutat ion group of degree 8 must be doubly transitive.

8D.3. Let G be a transitive permutat ion group of rank r , and suppose that the largest subdegree is n . Prove that \ G \ is bounded by some function of r and n .

8D.4. Let m be a subdegree for a transit ive act ion of G on ft, and let km

be the integer denned just before Theorem 8.42.

(a) Show that km > m.

(b) If km = m and a -> 0 is an m-arrow for a , 0 G ft, show that is a subgroup.

(c) If m > 1 and G is pr imi t ive on ft, show that km > m.

8D.5. Let G act t ransit ively on ft, and suppose the rank is 3. If the sub-degrees are 1 < m < n , where m and n are coprime, show that m + 1 must divide n .

H i n t . Observe that km must divide 1 + m + n .

8D.6. Let G be a pr imi t ive permutat ion group, and suppose that some prime number p is a subdegree. Prove that p 2 does not divide the order of a point stabilizer, and deduce that p 2 divides no subdegree.

H i n t . I f X c y and \Y : X \ = p , p r o v e that Q P ' { X ) < Y.

C h a p t e r 9

More on Subnormality

9 A

In this chapter, we present a few more applications of subnormali ty theory, and also a pretty result w i th in the theory for which no applicat ion seems to be known. We begin wi th a discussion of H . Bender 's "generalized F i t t i n g subgroup". Th i s characteristic subgroup and its associated "components" have become standard tools in group theory, and they have played an essent i a l role in the classification of simple groups.

Reca l l that the F i t t i n g subgroup of a group G is the unique largest normal nilpotent subgroup. If G is solvable, it follows by Prob lem 3B.14 that the F i t t i n g subgroup F ( G ) is large enough to contain its own centralizer i n G . If we make the weaker assumption that G is p-solvable for some prime p , and we assume in addi t ion that C y (G) = 1, it is easy to see that F ( G ) = O p ( G ) , and it follows by the H a l l - H i g m a n L e m m a 1.2.3 that i n this case too, F ( G ) contains its centralizer in G . B u t i n general, of course, F ( G ) can fail to contain its centralizer. T h i s certainly happens, for example, if G is a nonabelian simple group because in that case, F ( G ) = 1.

In this section, we define the generalized F i t t i n g subgroup F * ( G ) , and we prove that this characteristic subgroup always contains its centralizer in G . In general, F * ( G ) 3 F ( G ) , and we shall see that i n cases where the F i t t i n g subgroup contains its centralizer, F * ( G ) = F ( G ) . In other words, if no augmentation of the F i t t i n g subgroup is necessary in order for it to contain its centralizer, then no augmentation occurs when we construct the generalized F i t t i n g subgroup.

We begin work by recalling that a group G is said to be perfect if G' = G.

271

272 9. M o r e o n S u b n o r m a l i t y

9 .1 . L e m m a . L e t G be a g r o u p , a n d suppose t h a t G / Z { G ) i s s i m p l e . Then G / Z { G ) i s n o n a b e l i a n , a n d G' i s p e r f e c t . A l s o G ' / Z { G ' ) i s i s o m o r p h i c t o t h e s i m p l e g r o u p G / Z { G ) .

P r o o f . Wr i t e Z = Z ( G ) . If the simple group G / Z is abelian, it must be cycl ic , and thus G is abelian and Z = G, which is a contradiction. Thus G / Z is a nonabelian simple group, as claimed. In part icular, G is not solvable, and thus G'" ^ 1, so G" is not abelian, and hence G" £ Z.

N o w Z is a max ima l normal subgroup of G and G" £ Z, and thus G " Z > Z, and we conclude that G " Z = G. T h e n G / G " = G " Z / G " =i Z / { Z n G " ) , which is abelian. T h e n G / G " is abelian, and G' C G". The reverse containment is obvious, so G" = G', and G' is perfect.

F ina l ly , G ' / { Z n G ' ) = G ' Z / Z = G / Z is simple. It follows that Z n G ' is a max ima l normal subgroup of G\ and since G' is nonabelian, we see that Z n G ' must be the full center of G'. Thus G ' / Z { G ' ) = G / { Z n G') G/Z, and the proof is complete. •

A group H is said to be q u a s i s i m p l e if H / Z ( H ) is simple and H is perfect. B y L e m m a 9.1, therefore, if G / Z ( G ) is simple, then G' is quasisimple. To produce examples of quasisimple groups, we observe first that a l l nonabelian simple groups are quasisimple. A l so , the groups S L { n , q ) are quasisimple i f n > 3 or n = 2 and q > 3, and most of these groups are not simple. (See Theorems 8.32 and 8.33.)

O f course, if G is quasisimple, there is a unique nonabelian simple group G / Z ( G ) associated w i t h i t . A t first, it may seem that for each nonabelian finite simple group, there should be many (perhaps even infinitely many) associated quasisimple groups. In fact, there are only finitely many (and usually only a few). A l t h o u g h we w i l l not present a proof here, it is a fact that for each nonabelian simple group S, there is a unique largest quasisimple group G (called the Schur representation group of S) such that G / Z ( G ) = S. A l l of the quasisimple groups associated wi th S have the form G/Y, where G is the representation group of S and Y C Z ( G ) . (The abelian group Z ( G ) is the Schur mult ipl ier of S, and we refer the reader to a slightly more detailed discussion of Schur multipliers i n Chapter 5.) For example, if S is the alternating group A n w i t h n > 5, then unless n = 6 or n = 7, the Schur mult ipl ier has order 2, and so there are exactly two quasisimple groups associated wi th A n i n this case: the simple group A n itself, and the corresponding representation group, which has order 2 \ A n \ . If n is 6 or 7, the Schur mult ipl ier of A n is cyclic of order 6, and so there are exactly four associated quasisimple groups i n these two cases, w i t h orders \ A n \ , 2 \ A n \ , Z \ A n \ and 6 | A n | .

9 A 273

9.2. L e m m a . L e t G be q u a s i s i m p l e . If N i s a p r o p e r n o r m a l s u b g r o u p of G, t h e n N C Z ( G ) . A l s o , every n o n i d e n t i t y h o m o m o r p h i c i m a g e of G i s q u a s i s i m p l e .

Proof. Wr i t e Z = Z ( G ) , so that Z is a m a x i m a l normal subgroup of G , and let TV < G w i t h N < G. If iV % Z , then N Z > Z , and thus 7VZ = G by the maximal i ty of Z . T h e n G / i V = T V Z / i V = Z / ( i V n Z ) is abelian, and so G = G' C N < G . Th i s contradict ion shows that N C Z .

To prove the second_assertion, we_show that G = G / / V is quasisimple. We have (G) ' = G ^ = _ G , and thus G is perfect, as required. Also , since JV C Z , we have G / Z ^ G / Z , which is simple and nonabelian. Since Z C Z ( G ) and the factor group G / Z is a nonabelian simple group, it follows that Z must be the full center of G , and this completes the proof. •

A subnormal quasisimple subgroup of an arbi t rary finite group G is called a component of G . (Of course, it can happen that G has no components; this is the si tuat ion if G is solvable, for example.) Observe that if H is a component of G , then it is also a component of every subgroup of G i n which it happens to be contained.

9.3. L e m m a . L e t N be a m i n i m a l n o r m a l s u b g r o u p of a finite g r o u p G, a n d suppose t h a t H i s a c o m p o n e n t of G w i t h H % N . T h e n [ N , H ] = 1.

Proof. F i r s t , H D N < H , and of course H n N < H . B y L e m m a 9.2, therefore, H D N C Z { H ) . N o w H « G and N is m i n i m a l normal i n G , and so by Theorem 2.6, we know that N C N G { H ) , and thus [AT, i f ] C H . Also , [TV, #] C N since N < G, and hence [TV, #] C # n N C Z ( / f ) . T h e n [AT, H , H } = 1 and [#, A/", P ] = [A", H , H } = 1 , and so P , /V] = 1 by the three subgroups lemma. Since H = H ' , we have [ i i , A/] = [ H , H , N ] = 1, and the proof is complete. •

9.4. Theorem. L e t H a n d K be d i s t i n c t c o m p o n e n t s of a finite g r o u p G. T h e n [ H , K ] - 1.

Proof. We proceed by induct ion on | G | . If bo th H and K are contained i n some proper subgroup A of G , then H and K are distinct components of X . B y the inductive hypothesis applied in X , it follows that [ H , AT] = 1, as required, and so we can assume that no proper subgroup of G contains both H and K .

If G is simple, then since H and K are nontr iv ia l and subnormal, we have H = G = K , which is not the case. Thus G is not simple, and since it is certainly nontr ivia l , G has a min ima l normal subgroup N . A l so , N < G, and thus N does not contain both H and AT. If one of these components, say K , is contained i n N , then H % N , and thus [ H , K ] C [ H , N ] = 1, by

L e m m a 9.3, and we are done in this case. We can assume, therefore, that for every min ima l normal subgroup N , we have H £ N and K % N .

Let G — G/N, where N is m in ima l n o r m a H n G, and observe that H_ and K_ are nonidentity subnormal subgroups of G. B y Lemma_9.2, both H and K are quasisimple, and so they are components of G. 1iH_± K , then by the inductive hypothesis applied i n the group G , we have [ H , K ) = 1, and thus [ H , K ] C N . B u t [TV, H ) = 1 by L e m m a 9.3, and thus [ H , K , H ] = 1 and [ K , H , H ] = [ H , K , H ] = 1. B y the three subgroups lemma, therefore, we have 1 = [ H , H , K ] = [#,AT], where the second equality holds because H = H ' .

W h a t remains is the case H = K , or equivalently, H N = K N , and we can assume that this equation holds for every choice of a min ima l normal subgroup N of G. Since H N contains both H and K , it follows that H N = G. B y Theorem 2.6, the min ima l normal subgroup N normalizes H , and thus H < G , and s imi lar ly K < G , and hence [ H , K \ C # n K . If AT] > 1, we can choose our min ima l normal subgroup N so that N C H n K , and thus Tf = ZT/V - K A ^ = and we have a contradiction. •

It follows from Theorem 9.4 that the components of a finite group G normalize each other, and so each component is normal i n the subgroup generated by a l l of them. Th i s subgroup, denoted E ( G ) , is called the l a y e r of G . The layer of G is thus the product of the components of G , and of course, it is characteristic in G . (Of course, if G has no components, then E ( G ) - 1.)

In order to discuss the properties of the layer, it is convenient to introduce a another type of group that generalizes nonabelian simple groups. We say that a group G is s e m i s i m p l e if it is a product of nonabelian simple normal subgroups. (Unfortunately, not a l l authors agree on this definition, so readers should exercise caution.)

A s the next lemma shows, semisimple groups are actually d i r e c t products of nonabelian simple groups.

9.5 . L e m m a . Suppose t h a t a finite g r o u p G i s t h e p r o d u c t of t h e members of some c o l l e c t i o n X of n o n a b e l i a n s i m p l e n o r m a l s u b g r o u p s . T h e n t h e p r o d u c t i s d i r e c t , a n d X i s t h e set of a l l m i n i m a l n o r m a l s u b g r o u p s o f G .

P r o o f . The members of X are clearly min ima l normal subgroups of G , and thus if S E X and N is a min ima l normal subgroup of G different from S, then N n S = 1, and hence S centralizes N . It follows that the product of al l members of X different from N centralizes N .

We can certainly assume that X is nonempty. Let T G X , and let AT be the product of a l l other members of X , so that TAT = G , and by the

9 A 275

result of the first paragraph, K centralizes T. Since T is nonabelian and simple, it has t r iv ia l center, and thus T n K C Z ( T ) = 1, and it follows that G = T K = T x K . Now K is the product of the members of the collection X - { T } , so working by induct ion on \X\, it follows that this product is direct, and thus G = T x K is the direct product of the members of X .

Fina l ly , if N is m in ima l normal in G a n d N £ X , then by the result of the first paragraph, N is centralized by T\ * = G, and thus N C Z ( G ) . B u t G is a direct product of groups having t r i v i a l centers, so Z ( G ) = 1, and this contradict ion shows that X contains a l l m i n i m a l normal subgroups of G . •

Next we show how semisimple groups arise natural ly i n finite group theory.

9 .6 . L e m m a . L e t N be a m i n i m a l n o r m a l s u b g r o u p of a finite g r o u p G. T h e n e i t h e r N i s a b e l i a n o r i t i s s e m i s i m p l e .

P r o o f . Let S be a min ima l normal subgroup of N . If S is abelian, then S C F { N ) , and thus F(7V) is a nontr iv ia l normal subgroup of G contained i n N . Since N is min ima l normal i n G , we have F ( i V ) = N , and thus N is nilpotent. B u t then Z { N ) is a nontr iv ia l normal subgroup of G contained i n N , and thus Z { N ) = N and N is abelian.

We can now assume that S is nonabelian, and we argue that S is simple. If K < S, then K « G , and thus by Theorem 2.6, the min ima l normal subgroup N of G normalizes K , and thus K < N . B u t K C S and 5 is min ima l normal in N . Thus either K = 1 or K = S, and this shows that 5 is simple, as claimed.

We can now assume that S is a nonabelian simple group. T h e n each conjugate of S i n G is a nonabelian simple group that is normal i n N , and so the product SG of these conjugates is semisimple. A l so , 1 < SG < G and SG C N , and it follows by the min imal i ty of N that SG = N . Thus N is semisimple, as wanted. •

It follows by L e m m a 9.5 that a nonabelian min ima l normal subgroup of an arbi trary finite group must be a direct product of nonabelian simple groups.

We return now to the layer, E ( G ) .

9 .7 . T h e o r e m . L e t E = E ( G ) , w h e r e G i s a finite g r o u p , a n d w r i t e Z = Z { E ) . T h e f o l l o w i n g t h e n h o l d .

(a) E ' = E .

(b) E / Z i s s e m i s i m p l e .

(c) [ E , M] = 1 f o r every s o l v a b l e n o r m a l s u b g r o u p M o f G .

Proof. Let £ be the set of components of G , so that E = \[£. If H € £, then H = H ' C E ' , and thus E ' contains a l l of the members of £ , and we haveE'Z2T\£ = E , proving (a).

O f course, [ Z { H ) , H] = 1, and if K € £ w i t h K ^ H , then \ Z { H ) , K ] C [If, AT] = 1 by Theorem 9.4. Thus Z ( H ) centralizes a l l members of £ , and hence it is central i n E . T h e n Z { H ) C J? n Z , and since the reverse containment is clear, we have Z { H ) = H n Z for a l l components # . Now write £ = E / Z . If H £ £, then I f H / ( H n Z ) = H / Z ( H ) is a nonabelian simple group. A l so ,

is a product of nonabelian simple normal subgroups, and thus E is semisim¬ple, proving (b).

F ina l ly , let M < G b e solvable. T h e n M n E is a normal subgroup of the semisimple group E . B y L e m m a 9.5, the min ima l normal subgroups of E are nonabelian simple groups, and i n part icular they are nonsolvable. It follows that the solvable normal subgroup M ( l E contains no min ima l normal subgroup of E , and thus M D E = 1, and M D E C Z . T h e n [ M , £ ] C M n E a = Z ( £ ) , and hence [ M , E , E] = 1. A l so , [ E , M , E] = [ M , E , E] = 1, and hence by the three subgroups lemma, 1 = [ E , E , M ] = [ E , M [ , where the second equality follows by (a). Th i s completes the proof. •

F ina l ly , we define the generalized F i t t ing subgroup of a finite group G by the formula F * ( G ) = F ( G ) E ( G ) .

9.8. Theorem. F o r every finite g r o u p G, w e h a v e F * ( G ) D C G ( F * ( G ) ) .

Proof. Let C = C G ( F * ( G ) ) , and write Z = C n F * ( G ) , so that our goal is to show that Z = C. If this is false, then Z < C, and since Z and G are normal in G , we can choose a min ima l normal subgroup M / Z of G / Z w i th M C C. Since Z C F * ( G ) , we see that C centralizes Z , and since M C C , it follows that Z C Z ( M ) .

Now M / Z is either abelian or semisimple by L e m m a 9.6. If M / Z is abelian, then M is nilpotent, and so M C F ( G ) C F * ( G ) , and thus M C G n F * ( G ) = Z , which is a contradict ion. Thus M / Z is semisimple, and we let S/Z be a min ima l normal subgroup of M / Z . T h e n S/Z is a nonabelian simple group by L e m m a 9.5, and it follows that Z is the full center of S. L e m m a 9.1 now applies, and we deduce that 5 ' is quasisimple. Also , S' « G , and so 5 ' is a component of G , and S' C E ( G ) C F * ( G ) . T h e n 5 ' C C n F * ( G ) = Z, and thus S/Z is abelian. Th i s is a contradiction, and the proof is complete. •

P r o b l e m s 9 A 277

9.9. Corol lary. L e t G be a finite g r o u p . T h e n F * ( G ) D F ( G ) , a n d e q u a l i t y h o l d s if a n d o n l y i / F ( G ) 2 C G ( F ( G ) ) .

Proof. The containment is obvious. A l so , since F * ( G ) contains its central-izer i n G , it follows that if F * ( G ) = F ( G ) , then F ( G ) contains its centralizer.

Conversely, suppose that F ( G ) 5 C G ( F ( G ) ) . Since F ( G ) is solvable, E ( G ) C C G ( F ( G ) ) by Theorem 9.7(c), and thus E ( G ) C F ( G ) , and hence F * ( G ) = F ( G ) E ( G ) = F ( G ) , as required. •

P r o b l e m s 9 A

9 A . 1 . Suppose that F * ( G ) C H C G . Show that E ( H ) = E ( G ) .

9 A . 2 . Show that the socle of an arbi t rary finite group is the direct product of an abelian group wi th a semisimple group.

9 A . 3 . Let N < G, where G is semisimple. Show that N is the product of those m i n i m a l normal subgroups of G that it contains.

Hint . Wr i t e G = U x V, where U is the product of the min ima l normal subgroups of G that are contained i n N . Show that AT n V = 1.

9A.4 . Let E = E ( G ) and suppose that N < E . Show that N = MY, where M is the product of a l l components of G that are contained i n N and Y = M n Z ( £ ) .

9A.5 . Suppose that H < G and that C is a component of G not contained i n H . Show that [ H , C ] = 1.

Hint . Use P rob lem 9A.4 to show that \ ( H n E ) , C] = 1, where E = E ( H ) .

9A.6 . Let H < G , and suppose that H D C G ( H ) . Show that H D E ( G ) .

9 A . 7 . Let N be a nonabelian min ima l normal subgroup of G , and observe that Lemmas 9.5 and 9.6 imply that N is a direct product of a collection X of nonabelian simple groups. Prove that G acts t ransi t ively by conjugation the set X .

9A.8 . A group G is characteristically simple i f it is nont r iv ia l and its only characteristic subgroups are 1 and G . If G is characteristically simple, show that G is a direct product of isomorphic simple groups.

Hint . Consider G x A u t ( G ) .

9 B

If G is an arbi t rary group, then G / Z ( G ) is natural ly isomorphic to the group of inner automorphisms Inn(G) . If Z ( G ) = 1, therefore, G Inn(G) , and so we can identify G w i t h Inn(G) v i a the natural isomorphism, and this embeds G as a subgroup of A u t ( G ) . In fact, Inn(G) < A u t ( G ) , and so a group w i t h t r i v i a l center is natural ly embedded as a normal subgroup of its automorphism group. ( A l l of this is routine, but for completeness, the details are given i n L e m m a 9.11 below.) It is easy to show (and we w i l l do this too in L e m m a 9.11) that i f Z ( G ) = 1, then also Z ( A u t ( G ) ) = 1, and so we can repeat the process and embed A u t ( G ) as a normal subgroup of its automorphism group. Cont inuing like this, we obtain the automorphism tower associated wi th G :

G < A u t ( G ) < A u t ( A u t ( G ) ) < A u t ( A u t ( A u t ( G ) ) ) < • • • .

One of the earliest applications of H . Wie landt ' s subnormali ty theory was his beautiful automorphism tower theorem, which asserts that the automorphism tower associated w i t h a finite group having t r i v i a l center contains only finitely many different groups.

9.10. T h e o r e m (Wie landt ) . L e t G be a finite g r o u p , a n d assume t h a t Z ( G ) = 1. W r i t e G i = G, a n d f o r i > 1, l e t G{ = A u t ( G l _ i ) . T h e n u p t o i s o m o r p h i s m , t h e r e a r e o n l y finitely many d i f f e r e n t g r o u p s a m o n g t h e G i .

In this section, we present a proof of the automorphism tower theorem based on ideas of E . Schenkman and M . Pettet (and, of course, of Wie landt ) . Before we begin to develop the necessary theory, we mention that it does not seem to be known whether or not the hypothesis that G has t r iv ia l center is really necessary i n Theorem 9.10. W i t h o u t this assumption, we do not have natural embeddings of G{ in G l + l , so there is no automorphism tower, but nevertheless, it might be true that there are just finitely many different groups G{. For example, i f G is dihedral of order 8, it happens that A u t ( G ) ^ G , and so there is only one group G% in this case.

In the si tuat ion of Theorem 9.10, where Z ( G ) = 1, we have remarked (but we have not yet proved) that the subgroups G» al l have t r iv ia l centers, and so there is a natural embedding G t < G i + 1 . It follows that G is subnormal i n each of the groups Gu and this, of course, shows why subnormali ty theory is relevant here.

If we accept Wie landt ' s theorem for the moment, and we let Gr be the largest of the groups Gu then clearly Gr = G r + 1 . Since G r + 1 = A u t ( G r ) and Gr is identified w i t h I n n ( G r ) , the equation Gr = G r + 1 tells us that every automorphism of Gr is inner. We mention that a group G w i t h Z ( G ) = 1 and

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A u t ( G ) = Inn(G) is said to be c o m p l e t e , and so it follows from Wie landt ' s theorem that every finite group wi th a t r i v i a l center can be subnormally embedded in a complete group.

Conversely, if one of the groups Gr i n the automorphism tower of G is complete, then Gr = I n n ( G r ) = A u t ( G r ) = G r + i , and hence Gr = Gs for al l subscripts s > r . In this si tuation, there are just finitely many groups appearing i n the tower, and G r is the largest of these. In other words, Theorem 9.10 is equivalent to the assertion that a complete group appears somewhere i n the automorphism tower of an arbi t rary finite group wi th t r iv i a l center. Our strategy of proof, however, w i l l not be to show directly that some group i n the automorphism tower is complete. We w i l l show instead that if Z ( G ) = 1, then there is some integer n , depending only on G , and such that |G* | < n for a l l terms G% i n the automorphism tower of G . Th i s , of course, w i l l show that there are only finitely many different groups d occurring, as required.

It is not completely t r iv ia l to find examples where Z ( G ) = 1 and A u t ( G ) is not complete. B u t if there were no such example, then automorphism towers would never contain more than two groups, and Wie landt ' s theorem would not be very interesting. Before we begin working toward a proof of the automorphism tower theorem, therefore, it seems appropriate to mention an example where there are three different groups in the automorphism tower. Let G = S3xS3, the direct product of two copies of the symmetric group of degree 3, and observe that Z ( G ) = 1. T h e n | G | = 36, and one can compute that | A u t ( G ) | = 72 and | A u t ( A u t ( G ) ) | = 144. T h i s group of order 144, is complete; it can be constructed as the semidirect product of an elementary abelian group E of order 9, acted on by a semidihedral group of order 16, which is a full Sylow 2-subgroup of A u t ( £ ) ^ G L ( 2 , 3 ) .

Parts (a), (b) and (d) of the following establish the basic results about automorphisms to which we referred.

9 . 1 1 . L e m m a . L e t A = A u t ( G ) a n d I = Inn (G) , w h e r e G i s a n a r b i t r a r y g r o u p , a n d f o r g G G , w r i t e r g t o d e n o t e t h e i n n e r a u t o m o r p h i s m of G i n d u c e d by g . T h e f o l l o w i n g t h e n h o l d .

(a) T h e m a p g ^ T9 i s a h o m o m o r p h i s m f r o m G o n t o I , w i t h k e r n e l Z ( G ) .

(b) I < A .

(c) / / Z ( G ) = 1, t h e n C A ( I ) = l .

(d) 7 / Z ( G ) = I , t h e n Z { A ) = 1.

P r o o f . Since (x°)h = x ^ h ) for elements x , g , h £ G , it is immediate that the map g i-> T g is a homomorphism from G onto I . N o w r 9 = 1 i f and

only if X 9 = x for a l l x G G , and this is true if and only if g G Z ( G ) . It follows that the kernel of the homomorphism g ^ r g is exactly Z ( G ) , and this proves (a).

Now let a G A and g G G . T h e n for a l l x G G , we have

( a O ( f s ) a - ( a O a T ^ a = ( ( ( x ) Q " 1 ) » ) a = ( ( x ^ a ) ^ " = ( x ) ^ a ,

and so ( T 9 ) Q = T ( F L ) Q , which lies in I . It follows that I < A , proving (b).

Assume now that Z ( G ) = 1, and suppose that a G C A ( I ) i n the previous calculat ion. If g G G , we have

T9 = (T<?)a = T ( g ) a ,

and since the map g i-> r 5 is injective in this case, it follows that # = (#)a. T h i s holds for a l l g G G , and so a = 1. Th i s establishes (c), and (d) is an immediate consequence. •

Suppose that Z ( G ) = 1, and let G = G i < G 2 < G 3 < • • • be the automorphism tower of G . B y L e m m a 9.11(d), each of the groups G{ has a t r i v i a l center, and thus by 9.11(c), each of them has a t r i v i a l centralizer in its successor. T h e following result shows that in this si tuation, CGi (G) = 1 for a l l z .

It is convenient now to change notation. We write S i n place of G , so that we can use the symbol " G " to denote the largest group under consideration.

9.12. L e m m a . L e t

S = S i < S 2 < S 3 < • • • < Sr = G,

w h e r e G i s a g r o u p , a n d assume t h a t C S l + 1 { S t ) = 1 f o r a l l i w i t h l 2, and we proceed by induct ion on r . Wri te C = C G { S ) , and observe that G n S r _ i = 1 by the inductive hypothesis applied in the group S r _ i . Now S2 normalizes S, and so S2 normalizes G , and we have [ S 2 , C ] C C. A l so S2 C S r _ i < G , and thus [ S 2 , C ] C [ S r _ i , G ] C S r _ i . T h e n [ S 2 , C ] C G n Sr-i = 1, and so C centralizes S2. B y the inductive hypothesis i n G , w i th S2 in place of S and r - 1 i n place of r , we conclude that G C C G ( S 2 ) = 1. •

N o w let G have t r i v i a l center, and suppose that its automorphism tower is G = G i < G 2 < • • •. Reca l l that our goal is to show that there exists an integer n , depending only on G , and such that |G*| < n for a l l i. B y L e m m a 9.12, we know that C G i { G ) = 1, and thus the automorphism tower theorem w i l l be an immediate consequence of the following result.

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9 .13 . T h e o r e m . L e t S « G, w h e r e G i s a finite g r o u p , a n d assume t h a t CG(S) = 1. T h e n t h e r e exists a n i n t e g e r n d e p e n d i n g o n l y o n t h e i s o m o r p h i s m t y p e ofS, a n d such t h a t \ G \ < n .

O u r proof relies on two applications of the following easy lemma.

9 .14 . L e m m a . L e t N be a n o r m a l s u b g r o u p of a finite g r o u p G, a n d suppose t h a t N D C G ( N ) . T h e n \ G \ d i v i d e s |Z(7V) | |Aut(7V) | . I n p a r t i c u l a r , \ G \ d i v i d e s \ N \ \ .

P r o o f . T h e conjugation action of G on iV defines an isomorphism from G / C G ( N ) into Aut(TV), and since C G ( N ) = Z(7V), it follows that | G | d i vides |Z(JV) | |Aut (7V) | , as claimed. N o w A u t ( i V ) acts faithfully on the set of nonidenti ty elements of AT, and so |Aut(AT)| divides (|7V| - 1)!. Since | Z ( A ) | divides |AT|, the final assertion follows. •

Reca l l that in an arbi t rary finite group S, there is a unique normal subgroup min ima l w i t h the property that the corresponding factor group is nilpotent. Th i s subgroup, denoted S ° ° , is the final term of the lower central series of S. W i t h this notat ion established, and before we plunge into the details, we can give a brief outline of the proof of Theorem 9.13

We show first that if S is an arbi t rary subnormal subgroup of G , then the generalized F i t t i n g subgroup F * ( G ) normalizes S°°. If we can show that N G ( S ° ° ) has bounded order, this w i l l y ie ld an upper bound on | F * ( G ) | , and since F * ( G ) contains its centralizer i n G, the result w i l l follow by L e m m a 9.14. In order to obtain an upper bound on | N G ( S ° ° ) | , we use the hypothesis that C G { S ) = 1 to show that CG(S°°) C S°°. (This result of Schenkman seems to be the hardest step.) T h e n L e m m a 9.14 yields | N G ( 5 ° ° ) | < | Z ( 1 S ° ° ) | | A u t ( 1 S 0 O ) | , and so this is an upper bound for | F * ( G ) | , as needed.

We begin work now w i t h a number of pre l iminary results.

9 .15 . L e m m a . L e t G be a finite g r o u p , a n d suppose t h a t G = SF, w h e r e S «G a n d F < G, a n d assume t h a t F i s n i l p o t e n t . T h e n G°° = S°°.

P r o o f . Since there is nothing to prove if S = G , we can assume that S < G , and we proceed by induct ion on | G | . Let S C M < G w i t h M < G. B y Dedekind 's lemma, M = S ( F n M ) , and so by the inductive hypothesis applied to M , w i t h F n M in place of F , we have M ° ° = S°°. It thus suffices to show that M ° ° = G ° ° .

N o w M°°< G, andwe w r i t e G = G / M ° ° . Observe that G = M F since M contains S, and thus G = M F is a product of nilpotent normal subgroups, so it is contained i n F ( G ) , and hence is nilpotent. It follows that G / M ° ° is

nilpotent, and so G°° C M ° ° . The reverse containment is clear since M/G°° is nilpotent, and thus M ° ° = G ° ° . This completes the proof. •

9.16. Corol lary. L e t S « G, w h e r e G i s a finite g r o u p . T h e n F ( G ) C N G ( S ° ° ) .

Proof. B y L e m m a 9.15, we have S°° = ( F ( G ) S ) ° ° , and since this subgroup is characteristic in F ( G ) S , it is certainly normalized by F ( G ) . •

We must also show that the layer E ( G ) normalizes S°° for every subnormal subgroup S of G. Together w i t h Corol la ry 9.16, this w i l l show that F * ( G ) normalizes S°°, as wanted.

9.17. L e m m a . L e t S « G , w h e r e G i s a finite g r o u p a n d S i s n o n a b e l i a n a n d s i m p l e . T h e n t h e n o r m a l c l o s u r e SG i s a m i n i m a l n o r m a l s u b g r o u p of G, a n d s o S C S o c ( G ) .

Proof. Reca l l that SG is the subgroup generated by the members of the set X = { S 9 | g G G } . Since the conjugates of S are simple, they are certainly quasisimple, and since they are subnormal, they are components. It follows by Theorem 9.4 that the members of X normalize each other, and so SG = n X and SG is semisimple. To show that SG is m in ima l normal in G , suppose that 1 < N C SG w i t h N < G. T h e n N contains some min ima l normal subgroup of SG. B y L e m m a 9.5, this min ima l normal subgroup is a member of X , and thus N contains some conjugate of S. Since N is normal in G , it contains a l l conjugates of S, and it follows that N = SG. Th i s shows that SG is m in ima l normal i n G , as required. •

9.18. Corol lary. L e t S « G, w h e r e G i s a finite g r o u p . T h e n E ( G ) n o r m a l i z e s S°°.

Proof. Let Z = Z ( E ( G ) ) , and write G = G/Z. T h e n E ( G ) is semisimple, and by L e m m a 9.17, each of its simple factors is contained i n Soc(G) . T h e n E ( G ) C Soc(G) , and we conclude by Theorem 2.6 that E ( G ) normalizes the subnormal subgroup S = ~SZ. It follows that E ( G ) normalizes SZ, and so it normalizes ( S Z ) 0 0 = S°°, where the equality follows by L e m m a 9.15. •

Our next goal w i l l be to show that if S « G , and we assume i n addi t ion that C G { S ) = 1, then CG(S°°) C S°°. Th i s is a result of Schenkman, and we begin w i t h the following easy fact, which can also be obtained as a consequence of P rob lem ID.15 .

9.19. L e m m a . L e t N < G, w h e r e G i s a finite g r o u p , a n d suppose t h a t N C $ ( G ) , t h e F r a t t i n i s u b g r o u p . I f G / N i s n i l p o t e n t , t h e n G i s n i l p o t e n t .

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Proof. B y Theorem 1.26, it suffices to show that M < G, where M is an arbi t rary max ima l subgroup of G . B u t M Z} <&(G) Z2 N , so by the correspondence theorem, M / N is max ima l i n the nilpotent group G/N. It follows that M / N < G/N, and thus M < G. •

9.20. L e m m a . L e t N be a n o r m a l s u b g r o u p of a finite g r o u p G, a n d assume t h a t G / N i s n i l p o t e n t . T h e n t h e r e exists a n i l p o t e n t s u b g r o u p H C G such t h a t N H = G.

Proof. Since obviously, N G = G, the set of subgroups H C G such that N H = G is nonempty, and we show that a min ima l member of this set must be nilpotent. We argue that if H is m i n i m a l such that N H = G, then N n H C $ ( i f ) . Otherwise, there exists a m a x i m a l subgroup M of H such that iV n H % M , and thus since iV n H < H , we see that ( N n i f ) M is a subgroup of 7f that properly contains M . T h e n (AT n i f ) M = 7f, and we have G = N H = N ( N C ) H ) M = N M . Th i s contradicts the min imal i ty of H , and thus N n H <Z $ ( i f ) , as claimed. N o w H / { N n i f ) = i V i f / i V = G / i V , and this is a nilpotent group. It follows that by L e m m a 9.19 that i f is nilpotent. •

9.21. T h e o r e m (Schenkman). L e t S « G, w h e r e G i s a finite g r o u p a n d CG(S) = 1. T h e n C G ( S ° ° ) C S°°.

Firs t , we prove the case of Schenkman's theorem where S = G.

9.22. Theorem. L e t G be a finite g r o u p , a n d suppose t h a t Z { G ) = 1. T h e n C G ( G ° ° ) C

Proof. Let C = CG(G°°), and note that C < G. We argue that if i f C G is a nilpotent subgroup such that G°°H = G, then C n H = 1. Otherwise, G n i f is a nontr iv ia l normal subgroup of the nilpotent group i f , and so it contains a nontr iv ia l central subgroup Z of H . T h e n H C C G ( Z ) , and since Z C C , also G°° C C G ( Z ) . T h e n G = G°°H C C G ( Z ) , and this is a contradict ion, since Z ( G ) = 1 by hypothesis.

We w i l l show that it is possible to choose a nilpotent subgroup H C G such that G ° ° i f = G and so that in addi t ion, G = ( G n G ° ° ) ( G n i f ) . Since G n i f = 1 by the result of the previous paragraph, it w i l l follow that C = C n G°° , and this w i l l complete the proof.

B y L e m m a 9.20, there exists a nilpotent subgroup K C G such that G°°A: = G . W r i t i n g T = C K , we argue that T ° ° C C n G ° ° . To see this, observe first that each term of the lower central series of T is contained i n the corresponding term of the lower central series of G , and thus T ° ° C G ° ° . A l so , T / G = C K / C is nilpotent because AT is nilpotent, and thus T ° ° C G , and this shows that T°° C C D G ° ° , as wanted. B y L e m m a 9.20 applied

284 9 . M o r e o n S u b n o r m a h t y

in the group T, therefore, we can find a nilpotent subgroup H such that T = ( G n G°°)H. Since C n G ° ° C C C T , Dedekind's lemma yields c = ( c n G ° ° ) ( C n i / ) . A l so ,

G ° ° P = G ° ° ( G n G ° ° ) / f = G ° ° T D G ° ° A : = G ,

and thus G ° ° t f = G , so H has the desired properties. T h i s completes the proof. •

P r o o f o f T h e o r e m 9.21. Let G = C G ( S ° ° ) , and observe that C D S C S ° ° by Theorem 9.22. Since S normalizes S°°, it also normalizes C, and hence SC is a group, and it is no loss to assume that SC = G. We proceed by induct ion on | G | .

Suppose that S C H < G. T h e n S « tf, and since CH(S) = 1 and H < G, the inductive hypothesis yields P n G C S ° ° . B y Dedekind's lemma, however, P = S ( H n G) C S S 0 0 = 5 , and so H = S. It follows that there are no subgroups P w i t h S < H < G, and since 5 « G , we conclude that S < G , and G / 5 has no nonidenti ty proper subgroups. T h e n G/S is cycl ic , and since C / { C n S ° ° ) = G / ( G f l S ) = SC/S = G/S, we see that C/(CnS°°) is cyclic . B u t G n S ° ° C Z ( G ) , and we deduce that G is abelian. T h e n G ° ° = ( S G ) ° ° = S°° by L e m m a 9.15.

N o w Z ( G ) C C G ( 5 ) = 1, so we can apply Theorem 9.22 to G to deduce that C G ( G ° ° ) C G ° ° . Since G ° ° = S°°, the result follows. •

We can now prove Theorem 9.13, thereby completing the proof of W i e -landt 's automorphism tower theorem.

P r o o f o f T h e o r e m 9.13. Since S°° < N G ( S ° ° ) and Theorem 9.21 guarantees that S°° contains its centralizer in N G ( 5 ° ° ) , it follows by L e m m a 9.14 that

| N G ( 5 ° ° ) | < WS^WAutiS00)].

N o w F * ( G ) = F ( G ) E ( G ) , and each factor is contained i n N G ( S ° ° ) by Coro l laries 9.16 and 9.18, and thus | F * ( G ) | < |Z(S 0 O ) | |Aut (S 0 O )|. Since F * ( G ) contains its centralizer i n G by Theorem 9.8, L e m m a 9.14 yields

| G | < | F * ( G ) | ! < ( I Z ^ 0 0 ) ! ^ ^ ^ 0 0 ) ! ) ! ,

and thus | G | is bounded in terms of S, as required. •

P r o b l e m s 9 B

9B.1. Let S < G, where S is complete. Show that G = S x T for some subgroup T.

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9B.2. Suppose that Z ( G ) = 1, and let G = G i < G 2 < • • • be the automorphism tower of G. If G < Git show that i < 2.

9B.3. If G is semisimple, show that A u t ( G ) is complete, and thus the automorphism tower for G contains at most two different groups.

9B.4. Let G be dihedral of order 2n, where n is odd. Observe that Z ( G ) = 1, and show that the automorphism tower of G contains at most two different groups.

9B.5. Let G = A B , where A , B « G . Show that G°° = A ^ B 0 0 .

H i n t . F i r s t do the case where bo th A and B are normal , and then work by induct ion on | G | .

9 C

We mentioned the Sims conjecture i n Chapter 8. Restated i n purely group-theoretic language, this conjecture asserts that there exists some function /(TO), defined for positive integers TO, such that the following holds. Let H C G be a max ima l subgroup, and assume that c o r e G { H ) = 1. (In this si tuation, the subgroup H is said to be corefree.) Let g e G w i t h g $ H , and let TO = \ H : H (~l H 9 \ . T h e n \ H \ < /(TO).

It is a t r iv ia l i ty that i f TO = 1 then \ H \ = 1, and if TO = 2, then \ H \ = 2. If TO = 3, then \ H \ is a divisor of 48 by a 1967 theorem of C . Sims, but no general bounds for integers TO exceeding 3 were known to exist un t i l 1983, when P. Cameron, C . Praeger, J . Saxl and G . Seitz proved the full conjecture i n a paper relying on the classification of simple groups.

There is, however, an elementary argument that seems to come close to proving the Sims conjecture, and which is the start ing point of the paper by Cameron et a l . Th i s result, proved by Thompson and simplified by Wie land t , is an applicat ion of subnormali ty theory, and it is the subject of this section. Thompson proved that there is a function t { m ) such that i n the context of the Sims conjecture, \ H : O p ( i f ) | < t ( m ) for some prime p . In other words, al though Thompson d id not show that \ H \ is bounded, he d id prove that if H is unboundedly large, then it must be "mostly" a p-group.

9.23. T h e o r e m (Thompson) . L e t H be a c o r e f r e e m a x i m a l s u b g r o u p of a finite g r o u p G. L e t g e G - H , a n d w r i t e m — \ H : H n H 9 \ . T h e n f o r some p r i m e p , w e h a v e \ H : O p { H ) \ < ((TO!)2)!.

We prove the following somewhat more general form of this theorem. (The diagram should help the reader to keep track of the many subgroups involved.)

9.24. Theorem. L e t H a n d K be d i s t i n c t s u b g r o u p s of a f i n i t e g r o u p G. W r i t e D — H n K , a n d assume t h a t no n o n i d e n t i t y s u b g r o u p of D i s n o r m a l i n any s u b g r o u p of G t h a t p r o p e r l y c o n t a i n s e i t h e r H o r K . L e t M = c o v e H { D ) a n d N = c o v e K ( D ) , a n d l e t E = M C \ N . Then a t l e a s t one of t h e s u b g r o u p s U = c o r e H { E ) a n d V = c o r e K { E ) i s a p - g r o u p f o r some p r i m e p .

Under the hypotheses of Theorem 9.24, let a = \ H : D \ and b = \ K : D \ . B y the n!-theorem, we have \ H : M | < a! and \ K : N \ < b\. It follows that \ D : E \ < \ D : M \ \ D : N \ < { a - 1)!(6 - 1)!, and thus \ H : E \ < a \ b \ and \ K : E \ < a \ b \ . B y the n!-theorem once again, \ H : U\ < { a \ b \ ) \ and \ K : V\ < ( a \ b \ ) \ . Theorem 9.24 asserts that one of U or V is a p-group for some prime p , and since U < H and V < K , we deduce that either \ H : O P { H ) \ < { a \ b \ ) \ or \ K : O p { K ) \ < ( a \ b \ ) \ .

To see why Theorem 9.23 follows from Theorem 9.24, assume the hypotheses of 9.23, and suppose that H > 1. Since H is max ima l and has t r i v i a l core, we have H = N G ( H ) , and thus g does not normalize H , and we can take K = H 9 , so that H and K are distinct. B y the maximal i ty of H again, the only subgroup of G that properly contains either H or K is G itself, and since H is corefree, no nonidenti ty subgroup of D = H n K is normal in G. The hypotheses of Theorem 9.24 are thus satisfied, and i n the above notat ion, a = m = b. Since H and K are isomorphic, it

K

H

V

9 C 287

does not matter which of U or V is a p-group since in either case, we have \ H : O p ( H ) \ < ((m!) 2)!, as asserted by Theorem 9.23.

9.25. L e m m a . L e t H C G, w h e r e G i s a finite g r o u p a n d F ( G ) = I , a n d assume t h a t E ( G ) C H . T h e n E { G ) = E { H ) .

Proof. Since E ( G ) C H , each component of G is contained in H , and so the components of G are components of H , and we have E ( G ) C E { H ) . To establish equality, it suffices to show that a l l components of H are contained i n E ( G ) , so suppose that U is a component of H and U % E ( G ) . T h e n certainly U is not a component of G , and so each component of G is a component of H different from U. B y Theorem 9.4, distinct components of H centralize each other, and it follows that U centralizes a l l of the components of G , and thus U C C G ( E ( G ) ) . B u t F ( G ) - 1, so F * ( G ) = E ( G ) , and thus U C C G ( F * ( G ) ) C F * ( G ) = E ( G ) , and this is a contradict ion. •

Next , we prove an analog of L e m m a 9.15.

9.26. L e m m a . L e t G = SP, w h e r e G i s a finite g r o u p , a n d w h e r e S « G a n d P < G , a n d P i s a p - g r o u p f o r some p r i m e p . T h e n O P ( G ) = O p ( S ) .

Proof. If S = G , there is nothing to prove, so we can assume S < G, and we choose M w i t h S C M < G and M < G. T h e n M = S { M n P ) , and so working by induct ion on | G | and applying the induct ive hypothesis to M , we have O p { M ) = O p ( S ) , and it suffices to show that O p ( M ) = O p ( G ) .

Now O p { M ) < G , and^we wri_te_G = G / O p ( M ) . Since S C M , we have G = M P , and thus G = M P , which is a p-group. It follows that O p ( G ) C O p ( M ) . The reverse containment holds because M / O p ( G ) is a p-group, and this completes the proof. •

9.27. Corol lary. L e t G be a finite g r o u p . Suppose t h a t S « G , a n d l e t P < G be a p - g r o u p . T h e n P n o r m a l i z e s O p ( S ) .

Proof. B y L e m m a 9.26, we have 0^(5) = O p { S P ) . Th i s subgroup is characteristic in SP, so it is normalized by P . U

P r o o f of T h e o r e m 9.24. Since V < K , we have V < M < H , so V « H , and similarly, U « K . A l so , we can certainly assume that H > 1, and since F * ( H ) 2 C G ( F * ( / /)) , it follows that F * ( P ) > 1, and thus either F ( H ) > 1 or E ( i f ) > 1, and similarly, we can assume that either F(AT) > 1 or E(AT) > 1.

Suppose first that F ( H ) = 1 and F(AT) = 1, so that E ( H ) > 1 and E(AT) > 1. We show in this case that either U = 1 or V = 1, and so U or V is a p-group, as required. Assuming that U > 1 and V > 1, we see that neither U nor V is nilpotent, and thus U°° and V°° are nont r iv ia l subgroups of D ,

normal i n H and AT, respectively. T h e n H = N G ( P ° ° ) and AT = N G ( V ° ° ) because nontr iv ia l subgroups of D cannot be normal i n subgroups properly containing H or AT. Since V « P , it follows by Coro l la ry 9.18 that E ( P ) C N G ( V r ° ° ) = AT. T h e n E ( P ) C P>, and L e m m a 9.25 yields E ( P ) = E(P>). Similar ly , E(AT) = E ( D ) , and thus E ( P ) = E ( K ) is a nontr iv ia l subgroup of D , normal in bo th H and AT. T h e n i f = N G ( E ( P ) ) = N G ( E ( A T ) ) = AT, which is a contradict ion.

We can now assume that either F ( H ) > 1 or F ( K ) > 1, and by symmetry, we can suppose that F { H ) > 1. T h e n O p ( P ) > 1 for some prime p , and we write P = O p { H ) . Let X = ( y ( U ) and Y = O p ( V ) , and observe that it suffices to show that at least one of X or Y is t r i v i a l . Assuming that X > 1 and Y > 1, we have H = N G ( A ) and AT = N G ( Y ) , and we work to derive a contradiction.

Since V « H and P < # , it follows by Corol la ry 9.27 that P normalizes Y = O P ( V ) , and so P C N G ( Y ) = AT. T h e n P C P , and in fact, P < D since P < H .

Now let k E K . We have P , and by Coro l la ry 9.27 applied in the group D , it follows that P normalizes O P ( P f c ) = X k . T h e n P C N G ( A f c ) = ( N G ( A ) ) f c = H k , and we deduce that P f c _ 1 C P . Since also P f c _ 1 C AT, we have P ^ 1 C D , and thus P C D k for a l l k E K . T h e n P C c o r e ^ ( P ) = A , and so P < A ' , and we have O p ( H ) = P C O p (AT) C O p ( K ) , where the last containment follows since N < AT.

N o w 1 < P C O p (AT) , and so we can interchange the roles of P and K i n the previous argument. We deduce that O p { K ) C O p ( P ) , and so equality holds. Thus P is a subgroup of D normal in both P and K , and since P > 1, we have H = N G ( P ) = K , which is a contradiction. Th i s completes the proof. •

P r o b l e m s 9 C

9 C . 1 . Suppose that bo th P and K are subnormal in G i n Theorem 9.24. Prove that either P = 1 or V = 1.

H i n t . Otherwise, O p ( G ) > 1 for some prime p . Show that Z ( O p ( G ) ) C H by considering separately the cases where P is or is not a p-group.

9 C . 2 . Let G = A B , where A , B « G . Show that O P { G ) = O P ( A ) O P ( B ) .

H i n t . A s w i t h P rob lem 9B.5 , first handle the case where A and P are both normal , and then proceed by induct ion on \ G \ .

9 D 289

9 C . 3 . Let G = ( A , B ) , where A , B « G, and suppose that \ A : A ' \ and \ B : B ' \ are coprime. Prove that G — A B by assuming that G is a min ima l counterexample to this assertion and carrying out the following steps.

(a) Le t N be min ima l normal i n G. Show that A N B = G.

(b) Show that ( N n A ) ( N n B ) is normalized by bo th A and B .

(c) Show that N f] A = 1 — N f] B , and N has prime order, say p .

(d) N o w assume that p does not divide \ B : B ' \ . Use the previous problem to show that Q P ( G ) = Q P { A ) B .

(e) Derive a contradict ion using the fact that A Q P ( G ) is a group.

9 D

In the previous three sections, we presented applications of subnormali ty to general group theory. We close this chapter w i t h a pretty result of D . Bartels that lies w i th in subnormali ty theory itself. Our goal is to prove an analog for subnormal subgroups of the obvious fact that i f X C G, then the normal closure of X i n G, which by definition is the unique smallest normal subgroup of G that contains X , is exactly the subgroup generated by all of the conjugates of X i n G.

Firs t , we observe that for an arbi t rary subgroup X C G, there is a unique smallest subnormal subgroup of G that contains X . Th is subgroup is the subnormal closure of X i n G, and it exists because intersections of subnormal subgroups are subnormal, and so the intersection of al l subnormal subgroups containing X has the desired properties. B y analogy w i t h the normal closure X G , the notat ion X " G is sometimes used to denote the subnormal closure of X in G. Observe that we always have X " G C X G , and X « G i f and only i f X = X " G . Since X G is generated by a l l of the G-conjugates of X and X " G C X G , it is perhaps natural to guess that the subnormal closure X " G is generated by some of the conjugates of X .

G i v e n X , Y C G , we say that X and Y are strongly conjugate in G i f X and Y are conjugate in the group ( X , Y ) , or equivalently, X and Y are conjugate in every subgroup of G that contains both of them. (But note that strong conjugacy is n o t usually an equivalence relation; it is reflexive and symmetric , but i n general, it is not transitive.) We saw i n P rob lem 2A.10 that X « G if and only if X has no strong conjugates i n G distinct from itself, and this suggests that perhaps i n general, the subnormal closure of X is the subgroup generated by a l l of the strong conjugates of X i n G . T h i s is, in fact true, and it is easy to see that P rob lem 2A.10 is an immediate consequence.

9.28. T h e o r e m (Bartels) . L e t X C G , w h e r e G i s a finite g r o u p . T h e n

X " G = ( Y \Y is strongly conjugate to X ) .

We write X ^ to denote the subgroup generated by a l l of the strong conjugates of X in G. Thus Barte ls ' theorem asserts that X ( G ) = X " G , and so once we establish the theorem, the notat ion X ^ becomes superfluous.

We begin work by noting some elementary properties of the subgroup X ^ G \ F i r s t , observe that if X and Y are strongly conjugate in G and g € G, then X 9 and Y9 are strongly conjugate, and thus ( X ( G ) 9 = ( X 9 ) { G ) . The following lemma establishes a few slightly less t r iv i a l properties.

9.29. L e m m a . L e t I C G , w h e r e G i s a finite g r o u p .

(a) I f X C S « G, t h e n A< G ) C S.

(b) I f Y C X C G, t h e n Y ^ C X ^ G \

(c) I f X C K C G, t h e n x W C X ^ .

(d) If X ^ C i f C G, t h e n X™ = X ^ , a n d i n p a r t i c u l a r , if K = X ^ \ t h e n K = X ^ l

Proof. We prove (a) by induct ion on | G | . Let X C S « G, and observe that since there is nothing to prove if S = G, we can assume that S < G. T h e n there exists a subgroup M < G w i t h S C M < G, and a l l conjugates of X in G are contained in M . The strong conjugates of X i n G , therefore, are exactly the strong conjugates of X in M , and thus X ( G ) = X ^ C 5 , where the containment holds by the inductive hypothesis applied in the group M .

For (b), let Y C X , and suppose that Y and Z are strongly conjugate i n G . T h e n 2 = 7 « C 1 « for some element g G (Y, Z ) C ( X , X 9 ) , and thus X and X 9 a r e strongly conjugate in G , and C X ^ G \ Therefore Z = Y9 C X 9 C X ( G ) , and since Y ( G ) is generated by a l l such subgroups Z , it follows that Y ( ° ) C X ( G ) , as wanted.

N o w let X C AT C G . The strong conjugates of A in AT are exactly those strong conjugates of A in G that are contained in K . Thus X ^ C X ( G \ and this proves (c). If X < G ) C AT, then AT contains a l l strong conjugates of X in G , so A ^ ) = X ( G \ proving (d). •

The following is somewhat less routine.

9.30. L e m m a . L e t N be a n o r m a l s u b g r o u p of a finite g r o u p G, a n d w r i t e

G = G/N. T h e n X ( G ) = X { G ) f o r a l l s u b g r o u p s X o f G .

Proof. It is clear that the canonical homomorphism_g i-> p_maps the set of conjugates of A i n G onto the set of conjugates of X i n G . We w i l l show

9 D 291

that the set of_strong conjugates of X in G maps onto the set of strong conjugates of X i n G, and the result then follows.

F i r s t , suppose that Y is strongly conjugate-to X in G. Then Y = X 9 for some element g G ( X , Y ) , and we_have_Y = ( X ) 9 and g G ( X , Y ) = ( X , Y ) . Thus Y is strongly conjugate to X in G , as required.

Conversely, consider a strong con}ugate of X in G . Since this subgroup is conjugate to X , it has the form Y for some (not uniquely determined) conjugate Y of X in G. Let }> be the set of a l l conjugates Y of X in G such that Y is the given strong conjugate of X , and observe that it suffices to show that some member of y is strongly conjugate to X .

Choose Y _ G y so that the subgroup ( X , Y ) has the smallest possible order. Since X and Y are strongly conjugate, they are conjugate in ( X , Y ) = ( X , Y ) , and thus there is some element g G ( X , Y) such that Y = ( X ) 9 . T h e n NX9 = ( N X ) 9 = NY, and thus X 9 G y . A l so , by the choice of Y , we have \ ( X , X 9 ) \ > \ ( X , Y ) \ . Since 3 G ( X , Y ) , however, we see that X 9 C ( X , Y ) , and thus ( X , X ^ ) C ( X , Y ) . We deduce that ( X , X 9 ) = ( X , Y ) , and since g is an element of this subgroup, X 9 is strongly conjugate to X . B u t X 9 G X and so the proof is complete. •

We also need the following easy general fact.

9.31. L e m m a . L e t S « G a n d P G S y l p ( G ) , w h e r e G i s a finite g r o u p . T h e n P n S G S y l p ( S ) .

Proof. If S = G, there is nothing to prove, so we assume that S < G, and we work by induct ion on | G | . Let S C M < G w i t h M < G, and observe that \ M : M n P \ = \ P M : P \ is coprime to p . T h e n M n P e S y l p ( M ) , and thus S n P = S n ( M C ) P ) e S y l p ( S ) by the inductive hypothesis. •

P r o o f of T h e o r e m 9.28. B y L e m m a 9.29(a), we have X ^ C X " G . We must prove the reverse containment, and for this purpose, it suffices to show that X ( ° ) «3 G. Suppose that this is false, and assume that | G | is as smal l as possible for a counterexample. A l s o , assume that | X | is as smal l as possible among subgroups X C G such that X < G ) is not subnormal. Note that X ( ° ) < G , and hence X < G . We proceed in a number of steps.

Step 1. Suppose that Y and Z are conjugates of X and that y W = for some subgroup H containing Y and Z . T h e n Y ^ = Z ^ G \

Proof. Wr i te K = Y^G\ and let U = Y^H\ so that also U = Z ^ H \ We have U = Z M and K = Y ^ by L e m m a 9.29(d), and U = Y ^ C Y ^ = AT by 9.29(c). Since Y is conjugate to X , we see that K = Y ^ is conjugate to

< G, and thus K < G. Thus Z(*> « K by the min imal i ty of G, and we have

Y C P = Z ( c / ) C Z { K ) « K ,

where the second containment follows from 9.29(c) because U C K . T h e n y ( K ) ^ Z { K ) B Y 9.29(a), and we have

y(G) = ^ = y W c z<*> c z ^ ,

where the last containment follows by yet another applicat ion of 9.29(c). We have just proved that y(G> C Z ^ G \ and since the reverse containment follows symmetrically, we have equality, as required.

Step 2. X is a p-group for some prime p.

Proof. Otherwise, since X is certainly generated by its Sylow subgroups, it follows that X = ( Y | y < X ) . Now write U = ( Y ( G ) | Y < X ) and observe that since Y C y(G), we have X C P , and also, [7 C X ^ by 9.29(b). B y the min imal i ty of X , each of the subgroups Y ^ for y < X is subnormal in G, and thus [/ <^ G. B u t I C [ / , so we have X ^ C [ / b y 9.29(a). Since we established the reverse containment previously, we have X ^ = U«G. This is a contradiction, and we conclude that X is a p-group, as claimed.

Step 3. Let Y C H < G, where Y is conjugate to X , and let P € Sy l p (7 f ) .

T h e n there exists Z C P w i t h Z conjugate to X , such that Y ^ = Z ^ G \

Proof. Since H < G, we have Y ^ 7f, and thus P n Y ( i i ) G Syl p (y( i i )) by L e m m a 9.31. A s Y is a p-subgroup of Y ( i i ) , we can write Yh C P n Y ( H )

for some element / i € Y ( i i ) . W r i t i n g Z = Yh,we have Z C P and

Z(H) = { Y h ) { H ) = ( y ^ f = y W ,

and thus Y ( G ) = Z ( G ) by Step 1.

Step 4. Let M be a max ima l subgroup of G containing X , and let P € S y l p ( M ) . T h e n P G S y l p ( G ) , and M is the unique max ima l subgroup of G containing P.

Proof. For subgroups H C G , wri te /C(P) to denote the set of al l subgroups of the form Y^G\ where Y C H and Y is conjugate to A i n G . If h G P and Y C H is conjugate to X , then ( Y ^ ) ) ' 1 = ( Y / l ) ( G ) , and it follows that H acts by conjugation on the set /C(P). A l so , Z(P) is contained in the kernel of this action.

Now G acts t ransi t ively on J C ( G ) , and M stabilizes (setwise) the subset I C ( M ) < Z K ( G ) . The full stabilizer i n G of J C { M ) is thus a subgroup of G

9 D 293

that contains the max ima l subgroup M , so it is either M itself, or else i t is a l l of G.

Suppose first that G stabilizes / C ( M ) . Since G acts t ransi t ively on £ ( G ) , it follows that J C { M ) = K ( G ) . B y Step 3, we also have J C { M ) = /C (P ) , and thus K ( G ) = /C (P ) . Now Z ( P ) acts t r iv i a l ly on /C(P) = I C ( G ) , and it follows that the action_of G on K { G ) has a nont r iv ia l kernel P . O f course, K < G, and we write G = G/K, so that | G | < | G | . B y L e m m a 9.30 and the min imal i ty of G, we have

and so A T A T ^ « G. A l so , X ( G ) < K X ^ because AT stabilizes X < G ) , and thus « G. Th i s is a contradict ion, and it follows that G does not stabilize / C ( M ) .

T h e n M is the full stabilizer of J C ( M ) = /C (P ) , and this set is stabilized by N G ( P ) . Thus N G ( P ) Ç M , and since P G S y l p ( M ) , it follows easily that P must be a full Sylow p-subgroup of G, as required. A l so if P Ç N , where N is max ima l i n G, then P G Sylp(7V), and similar reasoning shows that N is the full stabilizer of /C(P) . Thus A ' = M , as wanted.

Step 5. X is contained i n a unique m a x i m a l subgroup of G.

P r o o f . Otherwise, let M and N be distinct max ima l subgroups such that X Ç M n N , and let X Ç S1 G S y l p ( M n A"). Choose M and N so that | 5 | is as large as possible. If S G S y l p ( G ) , then Step 4 yields M = A , which is a contradict ion. A l so , observe that X « S, so if S < G, then X « G. T h e n X ^ = X by 9.29(a), so A ( G ) is subnormal, which is a contradict ion. Thus N G ( 5 ) is proper, and so it is contained in some m a x i m a l subgroup R of G.

N o w let S Ç P G S y l p ( M ) and S Ç Q G Sy l p (A^) , and note that P and Q are full Sylow p-subgroups of G by Step 4. T h e n S S and P n Q > S, and thus Sylow p-subgroups of M n P and N d R are larger than S. B y the choice of M and TV, it follows that M = R = N , which is a contradiction.


P r o o f . We have A ^ G ) Ç X G < G, and thus A ^ G ) cannot be subnormal in A G , and it follows by the min imal i ty of \ G \ that X G = G.

Let M be the unique max ima l subgroup of G that contains X . T h e n A G g M , so some conjugate Y of A is not contained i n M . If ( A , Y ) is contained i n some max ima l subgroup N , then X Ç N , and so A = M , and this is a contradict ion because Y g M . Thus ( A , Y) = G and Y is strongly conjugate to X . T h e n G = (X, Y ) Ç X ( G ) < G, which is a contradiction. •


P r o b l e m s 9 D

9D.1. Just as subnormal closures exist, so too do subnormal cores. Prove that i f l C G , then there exists a unique largest subgroup of X that is subnormal i n G. Show that this subnormal core of X in G is normal in X .

9D.2. Let I C G . Prove that the intersection L of a l l strong conjugates of X in G is normal in X and subnormal in G.

Hint . Show that C X .

9D.3. Prove that the subnormal core of X in G is not necessarily the in tersection of any collection of conjugates of X .

H i n t . Let G = S 4 , the symmetric group of degree 4, and take X to be cyclic of order 4.

9D.4. A converse of L e m m a 9.31 asserts that U S Q G , and for a l l primes p and a l l Sylow p-subgroups P of G, the intersection P n 5 i s a Sylow p -subgroup of S, then S is subnormal in G. Suppose that S C G yields a counterexample to this assertion w i t h \S\ + \ G \ as smal l as possible. Show that G and S are nonabelian simple groups. In part icular, this converse of 9.31 holds if S is solvable.

Hint . Consider a min ima l normal subgroup of G.

Note. Th i s converse to L e m m a 9.31 was conjectured by O . Kegel . It was eventually proved by P . K l e i d m a n using the classification of simple groups.

C h a p t e r 1 0

More Transfer Theory

1 0 A

We begin w i t h a brief review of some mater ial from Chapter 5. Le t P be a Sylow p-subgroup of a finite group G , and let v : G -* P / P ' be the transfer homomorphism. We proved that k e r ( v ) is the subgroup A P ( G ) , which, by definition, is the smallest normal subgroup of G such that the corresponding factor group is an abelian p-group. One goal of transfer theory is to find a relatively smal l subgroup N , where P C N C G , and such that G / A * ( G ) = N / A P { N ) . (In other words, we want TV to control p-transfer i n G.) If this happens, it might allow us to prove that G is nonsimple by s tudying the smaller group N and showing that N has a nontr iv ia l abelian p-factor group. T h e n A P ( N ) < N , and thus by control of p-transfer, A P ( G ) < G. Assuming that G does not have order p, therefore, it cannot be simple.

A natural candidate for control of p-transfer is the Sylow normalizer TV = N G ( P ) , but this does not always work. For example, N G ( P ) fails to control p-transfer if G is the alternating group A s , w i t h p = 2. There N G ( P ) = P , and A p { P ) has index 4 in P , but G is simple.

If the Sylow subgroup P is abelian, we proved i n Chapter 5 that N G ( P ) actual ly does control p-transfer in G . Th i s was done i n two steps: first, L e m m a 5.12 shows that N = N G ( P ) controls G-fusion in P , which means that whenever two elements of P are conjugate in G , they are already conjugate in N , and then by Coro l la ry 5.22, we know that whenever F C J V C G and TV controls G-fusion in P , it also controls p-transfer in G .

If the Sylow p-subgroup P is nonabelian, the Sylow normalizer usually does not control fusion in P . Nevertheless, even in this case, there are situations where N G ( P ) can be proved to control p-transfer. The main result

295

296 1 0 . M o r e T r a n s f e r T h e o r y

of this section is a powerful theorem of this type due to T . Yoshida . It strengthens earlier results of P . H a l l and H . Wie landt , and it guarantees that the Sylow normalizer controls p-transfer if the Sylow subgroup P is not too far from being abelian. More precisely, N G ( P ) controls p-transfer if the nilpotence class of P is less than the prime p. (If p = 2, this corollary of Yoshida 's theorem tells us nothing new, although the theorem itself has content for a l l primes.) Our proof was inspired by Yoshida 's original character-theoretic argument, but it is more elementary than his, and it does not rely on character theory.

For a given prime p , the key to Yoshida 's theorem is the wreath product W = C p l C p , where Cp is the cyclic group of order p . The group W can be defined as a semidirect product A x U , where A is the direct product of p copies of Cp and U ^ Cp. Of course, to construct the semidirect product W, we need a specific action of U on A , and to describe it we write

A = ( a i ) x (a2> x ••• x (a p ) ,

and U = ( u ) , and we let U act on A by setting ( a i ) u = a i + l for 1 < i < p , and ( a p ) u = a r . Fol lowing the usual custom, we view A and U as subgroups of W = Cp I Cp.

Before stating Yoshida 's theorem, we make a few routine observations about the group W = Cp I Cp. F i r s t , we see that the subgroup A of W satisfies \ W : A \ = p and | A | = j p . A l so , A is elementary abelian, which means that it is abelian and that x p = 1 for a l l elements x € A . Furthermore, A is generated by the subset { a u a 2 , • • • , a p } , which consists of p elements that are conjugate in W, and whose product is not the identity.

In general, if P is an arbitrary nonabelian group having an abelian subgroup A of prime index p, then Z ( P ) C A . (Otherwise A Z ( P ) > A and thus P = A Z { P ) is abelian.) In this si tuation, if a € A - Z ( P ) , then C P { a ) = A , and hence the conjugacy class of a in P has size p . A p p l y i n g this to the group W, we conclude that the generating elements <n form a full conjugacy class of W.

1 0 . 1 . T h e o r e m (Yoshida) . L e t P € S y l p ( G ) a n d N = N G ( P ) , w h e r e G i s a finite g r o u p . T h e n N c o n t r o l s p - t r a n s f e r i n G unless P has a h o m o m o r p h i c i m a g e i s o m o r p h i c t o C p \ C p .

10 .2 . C o r o l l a r y . L e t P € S y l p ( G ) a n d N = N G ( P ) , w h e r e G i s a finite g r o u p . If t h e n i l p o t e n c e class of P i s less t h a n p , t h e n N c o n t r o l s p - t r a n s f e r m G .

P r o o f . B y L e m m a 10.3, below, the group Cp \ Cp has nilpotence class p, so it cannot be a homomorphic image of a p-group w i t h class less than p. The result is now immediate from Theorem 10.1. •

1 0 A 297

O f course, we can also conclude that N G ( P ) controls p-transfer in G if the Sylow p-subgroup P satisfies one of the several other conditions on a p-group that are sufficient to guarantee that Cp } Cp is not a homomorphic image. For example, if P has exponent p, then Cp I Cp cannot be a homomorphic image because Cp I Cp contains an element of order p 2 , namely u a i . More generally, if for every two elements x,y € P , there exists an element c in the derived subgroup of ( x , y ) such that { x y ) P = x P y P C

p , then Cp i Cp cannot be a homomorphic image. ( A p-group that satisfies this cond i t ion is said to be r e g u l a r . There is a fairly extensive theory of regular p-groups, which we w i l l not discuss here, but we mention that the theorem of H a l l and Wie landt that is generalized by Yoshida 's theorem asserts that the normalizer of a regular Sylow p-subgroup controls p-transfer.)

To apply Yoshida 's theorem, one must show that Cp I Cp is not a homomorphic image of some given p-group, but to prove the theorem, we need conditions on a p-group sufficient to guarantee that Cp I Cp actually is a homomorphic image. Tha t is the goal of the next few results.

Since the group W = C p l C p satisfies the hypotheses of the following lemma w i t h t = p, the lemma implies that \ Z ( W ) \ = p, that W is elementary abelian of order p P ~ l , and that the nilpotence class of W is p.

1 0 . 3 . L e m m a . L e t P be a p - g r o u p , a n d l e t A 2. Suppose t h a t A i s g e n e r a t e d by t h e elements of some c o n j u g a c y class of P . T h e f o l l o w i n g t h e n h o l d .

(a) \ Z ( P ) \ = p .

(b) P ' i s e l e m e n t a r y a b e l i a n of o r d e r p 4 " 1 .

(c) T h e n i l p o t e n c e class of P i s t.

P r o o f . Let AT be a conjugacy class of P that generates A . Since \ A \ = p l > p and A is elementary abelian, we see that A is not cyclic, and thus the class K contains more than one element. It follows that P is nonabelian, and hence Z ( P ) C A .

N o w P ' C A because P / A has order p, and so if a € AT, we have ( a )P ' C A . Since ( a )P ' contains P ' , it is a normal subgroup of P , and hence it contains the entire conjugacy class K . B u t K generates A , and thus we have ( a )P ' = A , and since \ {a) \ = p, it follows that \ A : P ' \ | A | / | P ' | = | Z ( P ) | > p, where the equality follows .by L e m m a 4.6, and the final inequali ty holds since P is a p-group. Thus | Z ( P ) | = p = \ A : P ' | , and (a) and (b) are proved. A l so P has nilpotence class t by Theorem 4.7, and thus (c) holds. •

B y wr i t ing the group operation in an elementary abelian p-group as addi t ion, the group can be viewed as a vector space over the field Z / p Z of order p, and this enables us to use linear-algebra techniques. In particular, since the subgroups of an elementary abelian group are exactly the sub-spaces, and since every subspace of a finite dimensional vector space is a direct summand, it follows that i f X is an arbi trary subgroup of an elementary abelian group E , then E = X x Y for some subgroup Y C E . Th is observation and some other standard facts from linear algebra w i l l be used in the proof of the next result.

Reca l l that the elements a { of the group W = Cp I Cp, form a conju-gacy class of W contained i n the elementary abelian subgroup A , and that the product of these elements is not the identity. Together w i th the fact that \ Z { W ) \ = p established i n L e m m a 10.3, this provides a useful way to recognize Cp I Cp, and this is the content of the following.

10 .4 . T h e o r e m . L e t P be a p - g r o u p , a n d l e t A < P , w h e r e A i s e l e m e n t a r y a b e l i a n a n d \ P : A \ = p . Suppose t h a t \ Z ( P ) \ = p , a n d assume t h a t t h e p r o d u c t of t h e elements of some n o n c e n t r a l c o n j u g a c y class of P c o n t a i n e d i n A i s n o t t h e i d e n t i t y . T h e n P ^ C p l C p .

P r o o f . Since P has a noncentral conjugacy class, it is nonabelian, and thus Z ( P ) C A . B u t A is elementary abelian, and so we can write A = Z ( P ) x B , for some subgroup B C A . T h e n \ A : B \ = \ Z ( P ) \ = p , and we have \ P : B \ = p 2 . A l so , since B n Z ( P ) = 1 and every nonidentity normal subgroup of P meets Z ( P ) nontrivially, it follows that c o r e F ( P ) = 1, and thus the r ight-mult ipl icat ion action of P on the p 2 right cosets of B in P is faithful. We conclude that P is isomorphic to a subgroup of the symmetric group S p 2 .

Let a G AT, where A" is a noncentral class of P such that AT C A and the product of the elements of AT is not the identity. Choose u <E P - A , and l e t T : A ^ A b e the map x x u . T h e n Tp is the identity map on A since up € A and A is abelian. A l so , AT = { a T l | 0 < i < p } .

Now view A as a vector space over the field F of order p . T h e n T is a linear operator on A and Tp = I , the identity operator on A . Since F has characteristic p , we have { T - I f = Tp - 1 = 0, and we let k be the smallest positive integer such that (T - I ) k = 0. T h e n k A ( T - I ) > A ( T - I ) 2 > • • • > A ( T - > A ( T - i f = 0 ,

and it follows that d i m ( A ) > k.

Next , we compute w i t h polynomials over the field F to obtain

( X - l ) p - 1 = ^ ~ 1 ) P = X P ~ 1 = 1 + X 4- X 2 + + X p ~ l

1 1 X - l X - l

1 0 A 299

If fc < p , then

0 = (T - I f ' 1 = I + T + T2 + ••• + T p _ 1 ,

and thus

a + a T + aT2 + ••• + aTp~l = 0 .

Since K = {aT* | 0 p and | A | > p p , and we have | P | > f + l .

B y the result of the first paragraph, P is isomorphic to a subgroup of the symmetric group S p 2 . Since the full p-part of { p 2 ) \ is p P + l and we have shown that \ P \ > p P + 1 , i t follows that i n fact, P is isomorphic to a Sylow p -subgroup of S p 2 . B u t a l l Sylow p-subgroups of this group are isomorphic, and so P is uniquely determined up to isomorphism. The group Cp I Cp satisfies the hypotheses of the theorem, however, and it follows that P Cp I Cp, as required. •

We mention that the proof of Theorem 10.4 yields another way to th ink about the group C p \ C p : it is a Sylow p-subgroup of the symmetric group S p 2 . We shall not need this observation, however.

10 .5 . C o r o l l a r y . L e t P be a p - g r o u p , a n d l e t A < P , w h e r e A i s a n element a r y a b e l i a n s u b g r o u p of i n d e x p . L e t a € A - Z ( P ) , a n d assume t h a t t h e p r o d u c t of t h e p elements i n t h e c o n j u g a c y class of a i n P i s n o t t h e i d e n t i t y . T h e n Cp \ C p i s a h o m o m o r p h i c i m a g e of P .

P r o o f . Let {a» | 1 < i < p } be the conjugacy class of a, and write b = a i a 2 - - - a p , so that b £ 1 by assumption, and (b) C Z ( P ) . If | Z ( P ) | = p , then P ^ Cp I Cp by Theorem 10.4, and there is nothing more to prove. We can thus assume that Z ( P ) > ( b ) , and we let z € Z ( P ) - ( b ) . T h e n z £ A , and thus Z = ( z ) has order p , and b £ Z .

Now consider the group P = P / Z . Since b £ Z , we see that b is nontriv¬ial , and it is the product of the conjugate elements a l for 1 < i < p . If these factors were al l equal, then since (a i )* = 1, we would have 6 = ( a x ) P = 1, which is not the case.

Since { a l } is a conjugacy class of P contained i n A , and its cardinal i ty exceeds 1, it follows that this class has size p . The elements a~ are thus distinct, and the product of these elements is the nonidentity element b. Thus P satisfies the hypotheses of the lemma, and working by mduct ion on P , we can conclude that Cp I Cp is a homomorphic image of P . It is thus also a homomorphic image of P , as required. •

We also need to consider p-groups for which we are not given an elementary abelian subgroup of index p . Th i s is where transfer becomes relevant, and so perhaps a brief review of some basic transfer theory would be useful.

Let H C G have finite index, and let T be a right transversal for H in G. (Recal l that this means that exactly one member of T lies in each right coset of H in G.) If t G T and g G G, we write t>g to denote the unique element of T ly ing in the coset H t g . Th is defines an action of G on T, and we see that tg[t-g)-1 lies i n H . A pretransfer map V : G -* H is defined by the formula

V(g) = Htg(t-g)-\ t e r

where the factors are mul t ip l ied according to some fixed, but arbi trary l in ear order on the elements of T. In general, the map V depends on the transversal T and the part icular order on the elements of T, but the composi t ion of V w i t h the canonical homomorphism H — H / H ' is the uniquely defined transfer homomorphism v : G — H / H ' , which is independent of the transversal and the linear order. In particular, since the transfer v is a homomorphism from G into the abelian group H / H ' , we see that G' C ker(v), and thus if x , y G G and x = y mod G', it follows that v ( x ) = v ( y ) , and thus V ( x ) = V { y ) m o d H ' .

A t one point i n our proof of Yoshida 's theorem, we w i l l need to consider the degenerate case where H = G. In that si tuation, we can take T = {1}, and we see that the corresponding pretransfer is the identi ty map. It follows that i f V : G - > G is any pretransfer, then V { g ) = g mod G'.

10 .6 . L e m m a . L e t M M be a p r e t r a n s f e r m a p . C h o o s e a t r a n s v e r s a l T f o r M i n P . T h e n f o r x G P , w e h a v e

(a) I f x e M , t h e n V ( x ) = T\ x l m o d M ' . t e r

(b) I f x & M , t h e n V ( x ) = x p mod M ' .

P r o o f . F i r s t , observe that modulo M ' , the pretransfer V is uniquely determined, and so we can compute it using any convenient transversal. A l so , since M < P , it follows that the set S = { r 1 | t G T } is a transversal for M in P . (One way to see this is to observe that the coset M t ~ l is the inverse of M t in the factor group P / M , which is closed under taking inverses.)

U x e M , we use the transversal S to compute V ( x ) . We have s x s " 1 G M , and thus M s x = M s , and it follows that s-x = s. T h e n s x ( s - x ) - 1 = sxs~l = x l , where t G T and s = t~l. Thus

V ( x ) = JJ s x s - 1 = JJ x* mod M ' ,

1 0 A 301

and this proves (a).

Now let x £ P - M . Then the set P = { x i | 0 < i < p ) is a transversal for M i n P , and this is the transversal we use to compute V ( x ) . To determine u-x for u G P , consider first the case u = x { for 0 < i < p - 1. Clear ly u-x = x i + 1 , and the corresponding factor i n the computat ion of V ( x ) is u x { u < x ) - 1 = 1. The remaining case is u = x ^ " 1 , where we have u-x = 1. The corresponding factor in this case is u x ( u - x ) " 1 = a*. M u l t i p l y i n g the factors for u £ U, we get V ( x ) = x? m o d M ' , as wanted. •

Before we state our next result, we recall that if M is a p-group, then the F ra t t i n i factor group M / $ ( M ) is elementary abelian, and thus i n part icular, M ' C $ ( M ) . In fact, we can say more: $ ( A f ) / M ' is exactly the set of p th powers of elements of the abelian group M / M ' .

10 .7 . L e m m a . L e t M < P , w h e r e P i s a p - g r o u p a n d \ P : M | = p . L e t V : P ^ M be a p r e t r a n s f e r m a p , a n d assume t h a t V ( M ) % $ ( M ) . T h e n Cp i C p i s a h o m o m o r p h i c i m a g e of P .

P r o o f . Wr i t e P = P / * ( M ) , and observe that since M ' C $ ( M ) , the element F ( x ) G M is uniquely determined for a l l elements x G P . B y hypothesis, V ( M ) ^ $ ( M ) , and_thus we can choose an element x G M such that V ( x ) ^ $ ( M ) . T h e n V ( x ) + 1, and by L e m m a 10.6(a), we have

t e T

where T is a transversal for M in P .

Since ( X ) P = 1 a n d _ ^ | = p , it follows that the e l e m e n t e d are not a l l equal, and thus since M is elementary abelian and \ P : M | = p, we see that the elements x* are distinct, and they form a full conjugacy class i n P . The product of the elements i n this class is not the identity, and thus Corol la ry 10.5 guarantees that Cp l C p i s & homomorphic image of P . It is therefore also a homomorphic image of P , as wanted. •

Next , we present a general transfer theory result.

10 .8 . T h e o r e m (Transi t ivi ty of transfer). L e t G be a g r o u p , a n d suppose t h a t H C K C G , w h e r e \ G : P | < oo. L e t

U : G ^ K ,

W : K -> P and

V : G - * H

be p r e t r a n s f e r maps. T h e n f o r a l l g e G , w e h a v e V { g ) = W ( U ( g ) ) m o d H ' .


P r o o f . Let T be the right transversal for K in G that was used to construct the pretransfer map U. Thus

U ( g ) = I I k t .

where Jfet = t g ^ - g ) - 1 £ A", and the factors A* are mul t ip l ied i n some definite but unspecified order. Let w : K -> Tf/Tf ' be the transfer, so that is the coset of f f ' i n f f that contains W ( k ) for fc G K . Since «; is a homomorphism, we have

w{U{g)) = Y [ w ( k t ) , t e r

and thus W(U(g)) = l [ W { k t ) mod f f ' .

t e r Now let 5 be the right transversal for f f i n K that was used to construct the pretransfer map W. T h e n

w(kt)=n h t , s ,

where / i M = s k t ( s - k t ) - 1 G H , and where for each element t e T, the mul t ip l ica t ion is carried out i n some definite but unspecified order. Work ing modulo H ' , the order of these factors is irrelevant, and we have

W ( U ( g ) ) = l [ h t , s m o d / P .

We w i l l show that the set ST = { s t | s e S, t G T } is a right transversal for H i n G, and that ( s i ) 5 ( ( s i ) ^ ) _ 1 = h t , a . It w i l l follow that V ( g ) is congruent modulo H ' to the product (in any part icular order) of the elements h t , s for s e S and t e T , and this w i l l complete the proof.

Let x G G be an arbi trary element. T h e n K x = K t for some element t G T , and we have x t ' 1 G K . T h e n P ^ i " 1 ) = P s for some element s G S, and thus / f x = H s t . Th i s shows that the coset H x contains st, and we must show that st is the unique element of ST contained in H x . Suppose then that H s t = H s ' t ' , where s' G S and t' G T. Since f f s and f f s ' are contained in K , it follows that K t = K t ' , and thus t = t! because T is a right transversal for K i n G. N o w f f s i = f f s Y = H s ' t , and hence f f s = H s ' , and we conclude that s = s' because 5 is a right transversal for f f i n f i . Th is shows that ST is indeed a right transversal for f f in G, as claimed.

Wr i t e ( s t ) - g = s't', where s ' e S and £' G T . Then H s t g = f f s Y , and since f f s and f f s ' are contained in K , we have K t g = K t ' , and thus t - g = t' and kt = t g ( t - g ) - 1 = t g ( t ) - 1 . A l so ,

f f sfc£ = H s t g { t ' ) - 1 = H s ' t ' { t ' ) - 1 = H s ' ,

303

and thus s-kt = s' and h t , s = s k t ( s - k t ) - 1 = s k t ( s ' ) - 1 . Us ing the equations

kt = t g { t ' y l and h t : S = s k t { s ' ) - \

we obtain

( s t g ^ i s t y g ) - 1 = s t g ( s ' t ' ) - 1

= s t g ( t ' ) - 1 ( S ' ) - 1

= s k t ( s ' ) - 1

= h t , s •

N o w mul t ip l ica t ion over t G T and s G 5 yields Y(#) = VK(t/(#)) m o d i f ' , as wanted. •

F ina l ly , we state a fairly technical sufficient condi t ion for a p-group to have a homomorphic image isomorphic to C p \ C p . It is this result that we w i l l use i n the proof of Yoshida 's theorem. Reca l l that we write o { x ) to denote the order of an element x .

10 .9 . T h e o r e m . L e t S S be a p r e t r a n s f e r . F i x a n element x G P , a n d w r i t e R = (s e S \ o { s ) < o { x ) ) . If V ( x ) & R $ ( S ) , t h e n Cp l C p i s a h o m o m o r p h i c i m a g e of P .

P r o o f . Since S S be pretransfers. If U { x ) 6 $ ( M ) , then f/(x) is a p th power modulo M ' , and we can write U ( x ) = y p

mod M ' for some element y € M . Work ing m o d S', the pretransfer W yields the transfer homomorphism w : M - c 5 / 5 ' , and M ' C ker(w) since to maps to an abelian group. We thus have

V ( x ) = W ( U ( x ) ) = W ( y p ) = W { y ) p m o d S',

where the first congruence holds by Theorem 10.8; the second holds because U ( x ) = y p mod M ' and M ' C ker(«;) , and the th i rd congruence holds because w is a homomorphism. B u t W ( y ) G S, so W { y ) p G $ ( 5 ) , and thus V ( x ) G $ ( 5 ) since S" C $ ( 5 ) . Th is is a contradict ion since we assumed that V ( x ) # R $ ( S ) , and we conclude that U ( x ) £ $ ( M ) . If x G M , the result follows by L e m m a 10.7.

We can assume now that x & M , and thus x ^ 1 and o ( x p ) < o ( x ) . We work to derive a contradict ion i n this case. L e m m a 10.6(b) yields x p = U ( x ) mod M ' , and thus V ( x ) = W ( U ( x ) ) = W ( x p ) m o d S'. (Note that we cannot carry this one step farther and write W { x ) p because W is defined only on M , but x £ M . )

B y the transfer-evaluation lemma (Lemma 5.5), we know that W ( x p ) is congruent modulo S' to a product of conjugates of x p ly ing i n S. These conjugates are elements of S having smaller order than x , and thus they lie

i n R C i ? $ ( 5 ) . Since also S' C J £ * ( S ) , we conclude that £ i ? $ ( S ) , contrary to hypothesis. •

A s we mentioned, when Yosh ida proved his result, he used character theory, and in part icular, he appealed to a representation-theoretic result known as Mackey 's theorem. We replace this w i t h the next result, which is a transfer analog of Mackey 's theorem. To state the Mackey transfer theorem, we consider the decomposition of a group G into ( H , AT)-double cosets, where H and K are arbitrary subgroups of G. These double cosets are the subsets of G of the form H g K , where g G G. Clearly, H g K is the orbit containing g of the action of the external direct product H x AT on G defined by g - ( h , k ) = h~lgk. Since the sets H g K are orbits i n an appropriate action, we see that distinct ( H , AT)-double cosets are disjoint, and they par t i t ion G.

We say that a subset A C G is a set of representatives for the ( H , A O -double cosets i n G if each of these double cosets contains exactly one element of X . If X is such a set of representatives, then the distinct ( H , AT)-double cosets in G are exactly the double cosets H x K for x G A , and i n particular, | A | is equal to the number of these double cosets.

Note that the double coset H g K is invariant under left mul t ip l ica t ion by elements of H , and thus it is a union of right cosets of H . Since G acts by right mul t ip l ica t ion on the right cosets of H , the subgroup K acts by right mul t ip l ica t ion on these cosets, and it is easy to see that the right cosets of H that comprise the double coset H g K are exactly the members of the AT-orbit containing H g in this action.

Now assume that \ G : H \ < oo, so that H has just finitely many right cosets i n G. T h e n each double coset H g K is the disjoint union of the finitely many right cosets of H that it contains. These cosets of H form the AT-orbit containing H g , and so we can use the fundamental counting principle to determine their number. The stabilizer of H g i n G is H 9 , and it follows that the size of the AT-orbit containing H g is |AT : K n H 9 \ , and so this is the number of right cosets of H in the double coset H g K . In particular, this number is finite, and so we have |AT : AT n H 9 \ < oo. Furthermore, since the total number of right cosets of H in G is finite and each ( H , AT)-double coset is a union of some of these, it follows from the fact that the distinct double cosets are disjoint that there are only finitely many of them, and thus if X is a set of representatives for the ( H , AT)-double cosets in G, we have | X \ < oo.

10 .10 . T h e o r e m (Mackey transfer). L e t X be a set of r e p r e s e n t a t i v e s f o r t h e ( H , K ) - d o u b l e cosets m a g r o u p G, w h e r e H a n d AT a r e s u b g r o u p s , a n d \ G : H | < oo. L e t V : G - c H be a p r e t r a n s f e r m a p , a n d f o r each element x £ X , letWx : AT -+ AT n H x be a p r e t r a n s f e r m a p . T h e n f o r k G AT, w e

1 0 A 305

h a v e

V { k ) = JJ x W x ^ x - 1 m o d H ' . xex

P r o o f . F i r s t , observe that W x { k ) £ K n P x C H x , and that W^fc) is uniquely determined modulo (AT n H x ) ' C ( i i 1 ) ' = ( P ' ) x - It follows that x W ^ A ^ x - 1 G P , and this element is uniquely determined modulo P ' . Thus the product on the right is an element of H that is uniquely determined modulo H ' , and the order i n which the factors is mul t ip l ied is irrelevant.

For each element x £ X , let Sx be the right transversal for K n H x in K used to construct the pretransfer Wx. T h e n 1^1 = \ K : AT n P 9 | , which is the number of right cosets of P i n H x K . In fact, these right cosets form the AT-orbit of H x under right mul t ip l ica t ion . Since A T n P x is the stabilizer of H x in AT, it follows that the elements of a right transversal for K D H X in A" carry the coset P x to a l l of the different members of this P - o r b i t . Th is shows that the cosets H x s for s G Sx are a l l of the right cosets of H in H x K . A s x runs over A and s runs over S*, therefore, we account for each right coset of H i n G exactly once, and thus the disjoint union

T = \ J xSx

is a right transversal for H i n G. It follows that

V ( k ) E [ ] J] ^ ( ( a s ) . * : ) - 1 mod H ' xex sesx

for a l l elements k G G . We w i l l complete the proof by showing that if k G K , then for a l l elements x G A , we have

J] s s fc i ixs ) -* ; ) - 1 = xWsCAOaT 1 mod H ' . sesx

We need to compute ( x s ) - k , where x G X , s G and k G AT. Th i s element of T lies i n the coset H x s k , which is contained i n the double coset H x K because s G 5 X C K and fc G K . It follows that (xs)-fc lies i n xSx, and so we can write ( x s ) - k = xs', where s' G 5 X . T h e n H x s k = H x s ' , and if we left-mult iply by x " \ we get ( H x ) s k = ( H x ) s ' , and so s k i s ' ) ' 1 £ Tf*. B u t also s k i s ' ) ' 1 G AT since s,s' £ Sx C K and jfc G K , and thus we have s k i s ' ) ' 1 e K H H x . It follows that

i K C \ H x ) s k = i K r \ H x ) s ' ,

and thus s-k = s', where here, the dot denotes the action of K on the right transversal Sx for K n i P i n AT. T h e n

xsfc((xs).fc)- 1 = x s ^ x s T 1 = x s f c ^ ) " 1 ^ " 1 = x ^ s - A ; ) - 1 ) * - 1 ,

306 1 0 . M o r e Transfer T h e o r y

and this is an equality of elements of P . Now mul t ip ly over s G Sx in some fixed but arbi trary order, and work modulo H ' to get

J] x s k { { x s ) - k ) - x =x(H skis-ky^x-1 = x W x { k ) x ~ l m o d H ' , seSx seSx

as wanted. •

Yoshida 's theorem asserts something about a Sylow p-subgroup P of G under the assumption that N G ( P ) does not control p-transfer. The following lemma w i l l allow us to exploit the failure of transfer control in the proof of Yoshida 's theorem.

1 0 . 1 1 . L e m m a . L e t P C N C G, w h e r e P i s a S y l o w p - s u b g r o u p of t h e finite g r o u p G a n d N i s a s u b g r o u p t h a t does n o t c o n t r o l p - t r a n s f e r i n G. T h e n t h e r e exists a s u b g r o u p M < N , w i t h \ N : M \ = p , a n d such t h a t U { G ) C M f o r a l l p r e t r a n s f e r maps U : G ^ N .

Proof. Let V : G - > P and W : N -* P be pretransfer maps, and let v and w be the corresponding transfer homomorphisms obtained by composing V and W w i t h the canonical homomorphism P - > P / P ' . If U : G ^ N is a pretransfer, it follows by t ransi t iv i ty of transfer (Theorem 10.8) that v ( G ) = w ( U ( G ) ) C w ( N ) . N o w G / A ? { G ) ^ v ( G ) and N / A P ( N ) = w ( N ) , and since we are assuming that G / A P { G ) and N / A P ( N ) are not isomorphic, it follows that v ( G ) ^ w ( N ) , and thus v ( G ) < w ( N ) .

Since w ( N ) is a p-group and v ( G ) is a proper subgroup, there exists a subgroup L such that v ( G ) C L < w { N ) and \ w ( N ) : L \ = p . B y the correspondence theorem applied to the homomorphism w : N P / P ' , the inverse image of L in iV is a subgroup M of N such that M < N , and \ N : M \ = \ w { N ) : L \ = p . A l so , since w ( U ( G ) ) = v { G ) C L for an arbitrary pretransfer map U : G - > N , it follows that C/(G) C M , as wanted. •

F ina l ly , we are ready to prove Yoshida 's theorem.

P r o o f of T h e o r e m 1 0 . 1 . Assume that AT = N G ( P ) does not control p-transfer in G , and let V : G -> A^ be a pretransfer map. B y L e m m a 10.11, we can choose a subgroup M < N w i th |AT : M | = p, and such that V ( G ) C M , and we note that N ' C M .

Choose an element n G A" — M such that the order o(n) is as small as possible. If g is any prime divisor of o ( n ) , then o(n 9 ) < o(n), and so n« G M , and thus the image of n in N / M has order g. B u t \ N / M \ = p, and it follows that g = p, and thus o(n) is power of p. Since P is the unique Sylow p-subgroup of N , we have n G P .

Now let A be a set of representatives for the ( N , P)-double cosets in G , and for each element x G A , fix a pretransfer map : P -> P n A/" s .

P r o b l e m s 1 0 A 307

Since n G P , we can apply the Mackey transfer theorem (Theorem 10.10) to deduce that

v ( n ) = I I x W x { n ) x ~ l mod TV'. xex

Also , V ( n ) G V"(G) C M , and since also N ' C M , it follows that

Since P C N , one of the ( A , P) -double cosets is A/"IP = N , and so exactly one member of X lies in N . If we cal l this element y , we see that P n A/T = P and thus Wj, is a pretransfer from P to itself, and we have Wy(n) = n mod P ' . Since P ' C N ' C M and n G" M , it follows that W y ( n ) g M = M y , and thus y W y { n ) y - x g M .

The product of the elements x W x { n ) x - 1 lies i n M , and since the factor corresponding to x = y does not lie in M , it follows that some other factor must also fail to lie in M . There exists an element x G G - N , therefore, and a pretransfer map W : P -> P n N x , such that x W ^ x ' 1 £ M . We thus have W ( n ) 6 ( P n A ^ ) - ( P n M x ) .

Now wr i t ing 5 = P C\ N x and Q = P fl M x , we have Q C 5 C P and VK(n) e S - Q. B y the choice of the element n , we know that a l l elements of N w i t h order smaller than o ( n ) lie in M , and thus a l l elements of N x w i t h order smaller than o(n) lie i n M 1 . It follows that the subgroup R = (s G S | o(s) < o(n)) is contained in Q . A l so ,

\S : Q \ = \S : S n M x \ = \ S M X : M x | < \ N X : M x \ = p ,

and therefore, $ ( 5 ) C Q . Thus P $ ( 5 ) C Q, and since W ( n ) g Q , we have W ( n ) ^ i ? * ( 5 ) .

F ina l ly , recall that x £ TV = N G ( P ) , and thus P x j i P . B u t P x is the unique Sylow p-subgroup of G contained in N x , and thus P £NX, and we have S = P C \ N X < P . B y Theorem 10.9, therefore, CV\CV is a homomorphic image of P . •

P r o b l e m s 1 0 A

1 0 A . 1 . Show that a 2-group P is regular if and only if it is abelian.

H i n t . F i rs t , observe that factor groups of regular p-groups are regular. If P is a min ima l counterexample to the problem, show that | P ' | = 2.

10A.2. Let P be a regular p-group. Show that the set { x e P \ x p = 1} is a subgroup.

Hint . F i r s t , observe that subgroups of regular p-groups are regular. Suppose that P is a min ima l counterexample to the problem. Show that i f x,y € P , where x has order p, then some proper subgroup of P contains bo th x and x», and deduce that [ x , y f = 1. Conclude that the p th power of every element of P ' is t r iv i a l .

10A.3. Let P be ap-group w i t h | Z ( P ) | = p , and let A be abelian and have index p i n P . If Z ( P ) is a direct factor of A , deduce that P is isomorphic to a subgroup of Cp I Cp, and show that A is elementary abelian.

10A.4. Suppose that P e S y l 2 ( G ) , and that P is isomorphic to the direct product of the quaternion group Q8 w i t h the cyclic group C 2 . Show that G' < G.

Hint . Observe that C 2 l C 2 = D 8 . A l so , i f N = N G ( P ) , show that N / P has a nontr iv ia l fixed point in its natural action on P / $ ( P ) .

10A . 5 . Let Q C P , where P is a p-group and | P : Q\ = p 2 , and suppose that Q n Z ( P ) = 1. Show that P is isomorphic to a subgroup of C p l C p .

10A.6. Let G act transit ively on a set ft, and let P be a point stabilizer. Show that G is 2-transitive on ft if and only if G = P U H g H for some element g e G.

10A.7. Let A < P , where \ P : A \ = p and A is an elementary abelian p -group. Suppose that P - A contains an element of order p and also an element of order p 2 . Show that Cp l C p i s a homomorphic image of P .

10B

A group is metacyclic if it has a cyclic normal subgroup wi th a cyclic factor group. In this section, we use Yoshida 's theorem to prove the following result of B . Hupper t .

10.12. T h e o r e m (Huppert) . L e t P e S y l p ( G ) , w h e r e G i s a finite g r o u p a n d p > 2 , a n d suppose t h a t P i s n o n a b e l i a n a n d m e t a c y c l i c . T h e n p d i v i d e s \ G : G ' | .

We begin w i t h some easy observations.

10.13. L e m m a . L e t P be a m e t a c y c l i c g r o u p .

(a) I f U o P , t h e n P / U i s m e t a c y c l i c .

(b) I f V C P , t h e n V i s m e t a c y c l i c .

1 0 B 309

P r o o f I _ L e t _ C < P,_where C and P / C are cycl ic . For (a), wr i te_P = P / U . T h e n C < _ P , and C is a homomorphic image of C, and thus C is cyclic . A l so , P / C ^ J P J U C , and this is a homomorphic image of the cyclic group P / C . Thus P / C is cyclic, and hence P is metacyclic, as required. For (b), observe that V C \ C < V and V n C is cyclic since it is a subgroup of C. A l so ,

n C ) ^ F C / C , which is a subgroup of P / C , and so it is cyclic . It follows that V is metacyclic. •

10.14. L e m m a . Assume t h a t p > 2. T h e n t h e g r o u p W = C V \ C V i s n o t a h o m o m o r p h i c i m a g e of a m e t a c y c l i c g r o u p .

Proof. B y L e m m a 10.13(a), it suffices to show that W is not metacyclic. If W is metacyclic, however, then W is contained i n a cyclic normal subgroup, and so W is cyclic . Th is is not the case, however, since W is elementary abelian of order p P ~ l > p by L e m m a 10.3(b). •

Now assume the hypotheses of Theorem 10.12. A s we have just seen, C p l C p cannot be a homomorphic image of the metacyclic Sylow p-subgroup P , and hence Theorem 10.1 guarantees that N = N G ( P ) controls p-transfer i n G. If we can show that A P { N ) < N , therefore, it w i l l follow that A P ( G ) < G, and thus p divides \ G : G'\, as wanted. It suffices, therefore, to prove the following result.

10.15. Theorem. L e t P < N , w h e r e N i s a finite g r o u p a n d P e S y l p ( A 0 , a n d assume t h a t P i s n o n a b e l i a n a n d m e t a c y c l i c , a n d t h a t p > 2. T h e n p d i v i d e s \ N : N ' \ .

We need Maschke's theorem, which is usually viewed as belonging to representation theory, but which is useful when considering coprime actions on elementary abelian p-groups. We use the more-or-less standard proof of Maschke's theorem to establish Theorem 10.16, whose statement is somewhat more general than the usual statement of Maschke's theorem.

10.16. T h e o r e m (Maschke). L e t K be a g r o u p of finite o r d e r m, a n d s u p pose t h a t K acts v i a a u t o m o r p h i s m s o n a g r o u p V w i t h s u b g r o u p s U a n d W such t h a t V = U xW, w h e r e U i s a b e l i a n a n d K - i n v a r i a n t . Assume t h a t t h e m a p u ^ u m i s b o t h i n f e c t i v e a n d s u r j e c t i v e o n U. T h e n t h e r e exists a K - i n v a r i a n t s u b g r o u p N < V such t h a t V = U x N .

Observe that if U is finite and has order coprime to m , then the map u i-> u m is automatical ly both injective and surjective. Th i s is because it has an inverse, namely the map u i-> un, where the integer n is chosen so that m n = 1 m o d | P | .

In the standard version of Maschke's theorem, V is a vector space over some field P , and U is a given AT-invariant subspace. In that case, it is

automatic that U is a direct summand of V, and so no explici t mention of W appears i n the usual statement of the result. Here, of course, the group V is wr i t ten additively, and so the key requirement is that the map u ^ m-u should be injective and surjective, and this happens if the characteristic of the field F does not divide TO. ( A n d it fails otherwise.)

We need the following case of Maschke's theorem for the proof of Theorem 10.15.

10.17. Corol lary. L e t K be a finite g r o u p t h a t acts v i a a u t o m o r p h i s m s o n a n e l e m e n t a r y a b e l i a n p - g r o u p V, a n d assume t h a t p does n o t d i v i d e \ K \ . Suppose t h a t U C V i s a K - i n v a r i a n t s u b g r o u p . T h e n t h e r e exists a K - i n v a r i a n t s u b g r o u p N C V such t h a t V = U x N .

Proof. Wr i t e TO = \ K \ and observe that \ U \ is coprime to TO, SO that as we observed, the map u ^ u m is bo th injective and surjective on U, as required for Theorem 10.16. It suffices, therefore, to show that V = U x W for some subgroup W C V. A s we mentioned in the discussion preceding Theorem 10.4, this follows easily by viewing V as a vector space over the field Z / p Z . •

P r o o f of T h e o r e m 10.16. Since V = U x W, we can define the map 0 : V U by setting 9 { u w ) = u i o r u e U a n d w e W , and we observe that 6 is a homomorphism and that 6 { u ) = u for u E U. Now let x E V and k E K . T h e n 0 ( x k ) G U, and since U is AT-invariant, we see that 6 ( x k ) k ~ 1 E U. Since K is finite, we can define the map U by

^ ( x ) = J] e{xk)k~\ k e K

and we observe that this is well denned since each factor i n the product lies i n U and U is abelian. In fact, <p is a homomorphism since if x , y E V and k E K , w e have { x y ) k = x k y k , and hence

rtxv) = n ^ ( ^ r 1

k e K

= l[(e(xk)e(yk))k~1

= n e(xk)k~l n k e K k e K

where, of course, the th i rd equality holds because U is abelian.

1 0 B 311

N o w let N = k e v ( i p ) . To show that N is /^-invariant , l e t a e K and compute that

¥ ? ( ^ ) a " 1 = ( n e{xak)k~1)a~1= \ \ ^ ( x a f c ) ( a f c ) _ 1 = < p { x ) , k e K k e K

where the last equality holds because a k runs over a l l elements of K as k does. It follows that < p ( x a ) = t p ( x ) a , and hence i f <p{x) = 1, then also < p ( x a ) = 1 for a l l a e K . Th is shows that iV is AT-invariant, as claimed.

O f course, N = ker(v?) < V, so to complete the proof, we must show that U n N = 1 and U N = V. If u € P , then u f c € P for a l l keK, and thus 0 ( u f c ) f c _ 1 = ( t i f c ) f c _ 1 = u , and it follows that cp(u) = u m , where we recall that m = \ K \ . B y hypothesis, therefore, the restrict ion of tp to P is bo th injective and surjective, and in part icular, U n N = C/nker(yj) = 1. F ina l ly , we have < p ( U N ) = <p(U) = U, and thus also <p(V) = U. Since U N Z} N , it follows by the correspondence theorem that U N = V, as required. •

We can now prove Theorem 10.15, and as we have observed, Hupper t ' s theorem (Theorem 10.12) w i l l then follow.

P r o o f o f T h e o r e m 10 .15 . Reca l l that P < N , where P is a nonabelian metacyclic Sylow p-subgroup of N , and p > 2. O u r goal is to show that N has an abelian homomorphic image of order divisible by p , and we proceed by induct ion on \P\.

Since P is metacyclic, P ' is cyclic, and of course P ' < N . A l so P ' > 1, and thus P ' has a unique subgroup of order p , which is necessarily normal i n N . N o w let Y be an arbi trary normal subgroup of order p i n N . T h e n P / Y is a normal Sylow p-subgroup of N / Y , and it is metacyclic by L e m m a 10.13(a). If P / Y is nonabelian, it follows by the inductive hypothesis that N / Y has an abelian homomorphic image of order divisible by p , and thus N has such a homomorphic image and we are done. We can assume, therefore, that P / Y is abelian, and thus P ' C Y. It follows that Y = P ' , and thus \P'\ - p , and P ' is the unique normal subgroup of order p i n N .

Since \P'\ = p , it follows that P ' is central in P , and thus P has ni lpo-tence class 2. Because p > 2, Theorem 4.8(a) guarantees that the set V = { x € P | = 1} is a subgroup of P , and of course, V < N . A l so V is metacyclic by L e m m a 10.13(b), and since every cyclic subgroup and cyclic factor group of V has orderiat most p , it follows that \V\ 2, it follows by Theorem 6.11 that P has more than one subgroup of order p , and we deduce that \V\ - p 2 , and V is an elementary abelian p-group.

Now write Z = Z ( P ) . l i V C Z , then P is in the kernel of the conjugat ion action of N on V, and so N / P acts on V. Th i s is a coprime action, and

since P ' is an invariant subgroup of order p, it follows by Maschke's theorem (or more precisely, by Coro l la ry 10.17) that we can write V = P ' x W, where W is (W/P)- invar ian t , and thus W < N . B u t \W\ = p , and this is a contradict ion since P ' is the unique normal subgroup of order p in N . We conclude that V % Z .

Since P 1 C Z , the group P / Z is abelian. A l so , it follows that for each element x G P , the map [ - , x ] is a homomorphism. If y G P , therefore, we have [ y p , x ] = [ y , x } P = l , where the second equality holds because | P ' | = p . Thus x centralizes y P , and since x G P is arbitrary, we see that y p £ Z for al l y G P , and we conclude that P / Z is elementary abelian.

Now AT acts on P / Z , and P is contained i n the kernel of this action because P / Z is abelian. It follows that N / P acts on P / Z , and of course, this is a coprime action. A l so V Z / Z is an invariant subgroup, and hence by Maschke's theorem, we can write P / Z = ( V Z / Z ) x ( H / Z ) , where H / Z is (AVP)- inva r i an t , and thus P < N . A l so V D H C V Z D H = Z , and since V £ Z , it follows that V £ P , and thus \ V f \ H \ < p 2 . We conclude that H has a unique subgroup of order p , and thus P is cyclic.

Now V is abelian and Z is central, and thus V Z is abelian. T h e n V Z / P , so Y Z / Z < P / Z , and thus P / Z is nontr ivia l , and P is not central in P . Now write C = C N ( H ) , so that P % C, and thus A T / C has order divisible by p . A l so , N / C is isomorphically embedded i n A u t ( P ) , and since P is cycl ic , A u t ( P ) is abelian, and thus N / C is an abelian homomorphic image of W w i t h order divisible by p . It follows that p divides |JV : JV' | , as required. •

P r o b l e m s 1 0 B

1 0 B . 1 . G iven a prime p and an integer n > 0, let C = ( x ) be a cyclic group of order p n . Let a G A u t ( G ) w i t h x a = X P + 1 , and let P = CX (CT ) . Show that P is a metacyclic p-group w i t h nilpotence class n .

H i n t . Use Theorem 4.7.

N o t e . Th i s problem shows that metacyclic p-groups can have arbi t rar i ly large nilpotence class, and we know that if p > 2, such a group cannot have C p l C p as a homomorphic image. It follows that Yoshida 's theorem is s t r ic t ly stronger than the assertion that a Sylow normalizer controls p-transfer if the Sylow p-subgroup has nilpotence class less than p.

1 0 B . 2 . Let E < G, where E is an elementary abelian p-group, and suppose that | G : C G ( P ) | is not divisible by p. Show that E C Soc(G) .

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I O C

If H is an arbi t rary subgroup of a finite group G, then of course, the derived subgroup G' is contained i n the kernel of the transfer homomorphism G - > H / H ' . In this section we discuss another connection between G' and transfer; we consider what happens if G' plays the role of H .

10 .18 . T h e o r e m . L e t G be a finite g r o u p . T h e n t h e t r a n s f e r h o m o m o r p h i s m G - > G ' / G " i s t h e t r i v i a l m a p .

For applications w i th in group theory, we generally seek results (such as Yoshida 's theorem) that allow us to prove i n certain cases that the transfer homomorphism is nontr ivia l . B u t Theorem 10.18, does just the opposite: it shows that the transfer map is t r i v i a l when the target is the derived subgroup. There do not seem to be many group-theoretic applications of this result, but it does have applications to algebraic number theory, and i n part icular to class field theory. (In that context, Theorem 10.18, which was first proved by P. Furtwangler, is usually called the "principal ideal theorem".) We have decided to present this result because it is s t r iking and easy to state, and yet it seems quite difficult to prove. T h e proof we are about to present is essentially due to G . Hochschi ld . It uses some elementary r ing theory, and so it is quite different i n flavor from everything else in this book, but we hope that readers w i l l find it to be accessible.

Le t G be a finite group and let R be any ring. (Recal l that by definition, rings contain uni ty elements.) The g r o u p r i n g R [ G ] is the set of formal sums of the form £ r g g , where the sum runs over g G G , and the coefficients r g lie i n R. Elements of the group r ing are added i n the obvious way:

and this makes R [ G ] into an additive group. We view G as a subset of R [ G ] by identifying the group element g G G w i t h the sum r x x , where r g = 1 and r x — 0 for x ^ g . F r o m this point of view, an arbi t rary element f > 9 5

in R [ G ] is no longer just a formal sum; it is an actual linear combinat ion of the elements of G w i t h coefficients i n R.

To define mul t ip l ica t ion in R [ G ) , we use the mul t ip l ica t ion in G together w i t h the mul t ip l ica t ion in R and the distr ibutive law. Thus

Y r 9 9 + Y s 9 9 = Y ( r 9 + s 9 ) 9 > geG geG geG

where

h = Y x , y e G x y = g


It is routine to check that these definitions of addi t ion and mul t ip l ica t ion make R [ G ] into a r ing i n which the identity element of G is the uni ty element.

We mention that the group r ing R [ G ] can be defined even if G is an infinite group, but i n that case, the elements of R [ G ] are just those formal sums r g g in which a l l but finitely many of the coefficients r g are zero. Some of our results hold in this more general setting, and in fact, some of our proofs go through essentially unchanged. Nevertheless, we w i l l generally assume that G is finite in what follows.

We shall need only the integer group r ing Z[G] , where Z is the ordinary r ing of integers. In this case, where R = Z , there is a natural connection between the group r ing and actions of G on abelian groups, and we w i l l exploit this i n the proof of Theorem 10.18.

Suppose that M is an arbi t rary addit ively wri t ten abelian group, and let G be a finite group that acts v i a automorphisms on M . If x G M and g G G , we would usually write x» to denote the element of M resulting from the act ion of g on x , but in this setting, where M is wr i t ten additively, it is more convenient to write x g in place of x 9 . In this notation, we have x l = x and ( x g ) h = x ( g h ) for x G M and g , h e G . A l so , since the given action is assumed to be v i a automorphisms, we have ( x + y ) g = x g + y g for x , y G M and g G G . If n G Z and x G M , then of course, x n is just the element that would be wri t ten x n in the standard mult ipl icat ive notat ion, and since the action is v i a automorphisms, we have ( x n ) g = [ x g ) n for g G G , n G Z and x G M .

N o w let £ egg G Z[G] , where, of course, eg G Z for g G G . If x G M , we can define a "right action" of the r ing Z[G] on M by setting

It is routine to check for a , ¡3 G Z[G] and x , y G M that

(a) x { a + (3) = x a + x 0 ,

(b) ( x a ) p = x ( a ( 3 ) ,

(c) (x + y ) a = x a + y a and, of course

(d) x l = x , where 1 is the uni ty of Z[G] .

In other words, M is a right Z[G]-module. O f course, if we had started wi th a "left action" v i a automorphisms of G on M , so that g ( h x ) = [ g h ) x for g , h G G and x G M , the analogous construction would make M into a left Z[G] , module.

Temporar i ly put t ing modules aside, we define the a u g m e n t a t i o n map 6 : Z[G] -> Z by setting

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so that 8 ( a ) is s imply the sum of the coefficients appearing i n a G Z [ G ] . It is routine to check that 8 is a r ing homomorphism, and so its kernel, denoted A ( G ) , is an ideal of Z[G]; it is called the augmentation ideal.

10.19. L e m m a . L e t G be a finite g r o u p . T h e n A ( G ) i s t h e a d d i t i v e s u b g r o u p of Z [ G ] g e n e r a t e d by t h e elements g - 1 f o r 1 / g G G, a n d i n f a c t , these g e n e r a t o r s f o r m a Z-basis f o r A ( G ) .

We should explain what we mean by a "Z-basis". Let M be an addi-t ively wr i t ten finitely generated abelian group, and let { x u x 2 , . . . , x n } be a generating set. T h e n the subset of M consisting of elements of the form J2 e i X r for e% G Z is clearly a subgroup of M that contains a l l of the generators x % . It is thus a l l of M , and hence every element of M is a linear combinat ion of the f o r m ^ e ^ . The given generating set is a Z-basis for M if the only way to get = 0 is for a l l coefficients e% = 0. In this case, each element of M is u n i q u e l y of the form e%x%. For example, if G is a finite group, then G is a Z-basis for the addit ive group of Z \ G \ .

O f course, not every finitely generated abelian group has a Z-basis; those that do are said to be free abelian groups. A nont r iv ia l finite abelian group, for example, cannot have a Z-basis, and so it is definitely not free abelian.

In fact, one can define infinite Z-bases analogously, and the conclusion of L e m m a 10.19 holds for infinite groups too.

P r o o f of L e m m a 10.19. Certainly, 8 ( g - 1) = 0 for g G G , where 8 is the augmentation map, and thus g - 1 G ke r ( i ) = A ( G ) . Now let a be an arbi t rary element of A ( G ) , and write a = £ egg, w i t h eg G Z. T h e n

and since £ eg = 8 ( a ) = 0, we see that a is a Z-linear combinat ion of the elements g - 1 for 1 / g G G , and hence these elements form a generating set for A ( G ) .

To check that these generators form a Z-basis, suppose that there exist coefficients f g G Z for 1 + g G G such that f g ( g - 1) = 0. T h e n

and since G is a Z-basis for Z[G], it follows that the coefficient f g of g is zero for a l l nonidenti ty elements g e G . •

Suppose that { x u x 2 , • • •, x n } is a Z-basis for some addit ively wr i t ten abelian group M , and let A be an arbi t rary (mult ipl icative) abelian group.

Choose arbi t rary elements a* € A for 1 < i < n , and define the map 9 : M - • A by setting

9 ( J 2 e l X i ) = J]K)ei-Note that 9 is well denned since every element of M is uniquely of the form £ e i X l w i t h e{ G Z, and it is t r i v i a l to check that 9 is a group homomorphism. T h i s shows that i f M is a free abelian group and A is any abelian group, then there is homomorphism 9 : M -• A such that 9 carries the members of a Z-basis for M to arbi t rar i ly selected elements of A .

N o w let M be a left ^ -module , and l e t X C R and Y C M be additive subgroups. (The following remarks also apply i n the important case where M = R, and where X and Y are additive subgroups of R . ) Reca l l that by definition, XY is the set of a l l (finite) sums and differences of elements of the form xy i n M , where x G X and y G Y. (Equivalently, XY is the additive subgroup of M generated by a l l products of the form x y . ) Observe that i f X and Y are finitely generated (as additive groups) by subsets { x r } and { y 3 } respectively, then XY is finitely generated by the products xlyj.

In part icular , i n Z[G], the augmentation ideal A ( G ) is an additive subgroup generated by the elements g - 1 for g G G , and thus A ( G ) 2 is the additive subgroup generated by elements of the form (g - l ) ( h - 1) for g , h G G . Note that A(G)Z[G] C A ( G ) since A ( G ) is an ideal, and thus A ( G ) 2 C A ( G ) .

10 .20 . Theorem. L e t G be a finite g r o u p . Then G/G' i s i s o m o r p h i c t o the a d d i t i v e g r o u p A ( G ) / A ( G ) 2 v i a a n i s o m o r p h i s m c a r r y i n g the coset G'g t o the coset ( g - 1) + A ( G ) 2 .

Proof. Define <p : G A ( G ) / A ( G ) 2 by setting <p(g) = (g - 1) + A ( G ) 2 . For elements x, y G G , we have xy- 1 = ( x - 1) + (y-1) + (x - l ) ( y - 1), and since ( x - l ) ( y - 1) G A ( G ) 2 , it follows that c p ( x y ) = <p(x) + <p(y), and hence ip is a homomorphism. Al so , since the additive group A ( G ) is generated by elements of the form x - 1 w i t h x G G , we see that A ( G ) / A ( G ) 2 is generated by the cosets (x - 1) + A ( G ) 2 , and thus <p(G) is a generating set for A ( G ) / A ( G ) 2 . B u t <p(G) is a subgroup, and thus it must be the whole group A ( G ) / A ( G ) 2 , and hence ). To complete the proof, we show that i n fact, G' = ker(</?), and thus <p defines an isomorphism from G/G' onto A ( G ) / A ( G ) 2 .

Next , we construct a map i n the opposite direction. B y L e m m a 10.19, the set { g - l \ 1 ^ g G G } is a Z-basis for A ( G ) , and since G/G' is an abelian group, there is a homomorphism 9 : A ( G ) G/G' such that 9 ( g - 1) = G'g for 1 ^ g G G . (Of course, this formula also holds when g = 1 since 9 is a homomorphism, and thus 9 ( 0 ) is the identity element of G / G ' . )

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Since 0 is a homomorphism from the addit ive group A ( G ) to G / G ' , and % - 1) = G ' g for a l l elements g E G, we can apply 0 to both sides of the identity xy - 1 = ( x - 1) + (y - 1) + (x - l ) ( y - 1) to get G ' (xy ) = ( G ' x ) ( G ' y ) 6 ( ( x - l ) ( y - l ) ) for elements x , y G G . B u t G ' ( x y ) = ( G ' x ) ( G ' y ) , and it follows that 0((x - l ) ( y - 1)) is the identity, and thus (x - l ) ( y - 1) G ker(0). T h e elements (x - l ) ( y - 1) generate A ( G ) 2 as an abelian group, however, and thus A ( G ) 2 C ker(0).

We have seen that ker(v?) D G' and that it suffices to prove equality here. To establish the reverse containment, let k E ker(p) . Since <p(k) is the 0 element of A ( G ) / A ( G ) 2 , and by definition, i p ( k ) = ( k - 1) + A ( G ) 2 , it follows that k - l E A ( G ) 2 C ker(0), and thus G'k = 6 ( k - 1) is the identity of G / G ' , and hence k G G ' , as required. •

Let K C G be a subgroup, and view Z [ K ] C Z[G]. Let T be a right transversal for K i n G , so that each element g G G is uniquely of the form k t , where k f] K and t ET. If a G Z[G] is arbitrary, therefore, we can write

where e f c t G Z and a t = £ f c efciyfc G Z [ K ] . We refer to a t as the i - c o m p o n e n t of a w i t h respect to T, and we observe that for each element t G T, the map f t : a ^ a t from Z[G] to Z [ K ] is a well defined homomorphism of additive groups. We shall need the following technical result.

1 0 . 2 1 . L e m m a . L e t K C G, w h e r e G i s a finite g r o u p , a n d l e t T be a right t r a n s v e r s a l f o r K i n G such t h a t 1 G T. W o r k i n g i n Z [ G ] , l e t a G A ( K ) A ( G ) , a n d f o r t G T, w r i t e a t t o d e n o t e t h e t - c o m p o n e n t of a w i t h r e s p e c t t o T. T h e n a t G A ( K ) a n d J 2 t

a t E A ( K ) 2 .

P r o o f . Let f t be the map carrying an element of Z[G] to its i-component w i t h respect to T, and write / = Y \ f t . T h e n / and al l of the maps f t are additive homomorphisms, and our goal is to show that f t ( A ( K ) A ( G ) ) C A ( K ) for a l l t E T, and that / ( A ( K ) A ( G ) ) C A ( K ) 2 . Since the elements ( k - l ) ( g - 1) for k E K and g E G generate the additive group A ( K ) A ( G ) , it suffices to prove the lemma i n the case a = ( k - l ) ( g - 1).

Firs t , suppose that g E K . T h e n a = ( k — l ) ( g — 1) G Z [ K ] , and thus the 1-component of a is a , and a l l other components are zero. Since a E A ( K ) 2 C A ( K ) i n this case, there is nothing further to prove.

We can now assume that g £ K , and we write g = hs, where h E K and s E T — {1}. T h e n

a = ( k - l ) ( g - 1) = ( k - l ) g - ( k - 1) = ( k - l ) h s - ( k - 1)1 ,


and thus a s = ( k — l ) h G A ( K ) and a x = - ( k — 1) G A ( K ) . A l l other components of a are zero, and thus a l l components lie i n A(AT), as required. A l so , f ( a ) = ( k — l ) h - (fc - 1) = (fc - l ) ( h — 1) G A(AT)2, and the proof is complete. •

10.22. Corol lary. L e t K C G, w h e r e G i s a finite g r o u p . T h e n i n Z[G], w e h a v e A ( K ) 2 = A(AT)A(G) n A { K ) = A(AT)A(G) n Z [ K ] .

Proof. It is clear that A(AT)2 C A(AT)A(G) n A(AT) C A ( K ) A ( G ) n Z [ K ] , so it is sufficient to show that A ( K ) A ( G ) n Z [ K ] C A(AT)2. Let T be as in L e m m a 10.21, and observe that if a G A ( K ) A ( G ) n Z [ K ] then the 1-component of a is a itself, and al l other components are zero. Thus a is the sum of its components, and so a G A(AT)2 by L e m m a 10.21. •

If K C G, then since A(G) is an ideal of Z[G], we have A(AT)A(G) C A(G). O f course A(AT)A(G) is a normal subgroup of the additive group A(G), and i n the next several results, we study the factor group A(G) = A(G)/A(AT)A(G). We use the standard "bar convention", so that if A is any additive subgroup of A(G), then A is the image of A under the canonical homomorphism A(G) A(G). In part icular, since A { K ) C A(G), we see that A (AT) is a subgroup of A(G).

10.23. Corol lary. L e t K C G, w h e r e G i s a finite g r o u p . W o r k i n g i n Z [ G ] , l e t

A(G) — A(G)

A(AT)A(G) ' T h e n A(AT) = AT/AT', w h e r e k — 1 maps t o K ' k f o r elements k G AT. Proof. The map k - l ^ k - l defines a homomorphism from A (AT) onto A ( K ) w i t h kernel equal to A(AT) n A(G)A(AT). B y Corol la ry 10.22, this kernel is equal to A(AT)2, and thus A ( K ) ^ A(AT)/A(A/)2 v i a the map k ~ ^ ~ l H-> (fc - 1) + A (AT)2. B y Theorem 10.20 applied to the group AT, we know that A(AT)/A(AT)2 ^ AT/AT' v i a the map (fc - 1) + A(AT)2 ^ AT'fc. The map i n the statement of the corollary is the composi t ion of these two isomorphisms. •

We are interested in the transfer homomorphism v : G - > K / K ' , where K Ç G. Since K / K ' g A (AT) in the notat ion of Corol la ry 10.23, we can ask which subgroup of A(AT) corresponds to v ( G ) under this isomorphism. If V : G K is a pretransfer map, then v ( G ) is the image of V ( G ) i n K / K ' . The subgroup of A (A/) corresponding to v ( G ) , therefore, is exactly the set of elements of the form V ( g ) - 1 for g G G. We shall be able to give a useful description of this subgroup i n the case where K is normal in G.

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Assume now that AT < G. If g e G, then conjugation by g defines an automorphism of AT, and hence g - l A ( K ) g = A (AT). T h e n A ( K ) g = g A { K ) , and since g A ( G ) = A ( G ) , we have g A ( K ) A ( G ) = A ( K ) g A ( G ) = A ( A / ) A ( G ) , and thus the additive group A ( A T ) A ( G ) is invariant under left mul t ip l ica t ion by elements of G . T h i s subgroup of Z[G] , therefore, is a left ideal. (It is also a right ideal, but we shall not need that fact.) O f course, A ( G ) is also a left ideal, and thus A ( G ) = A ( G ) / A ( A / ) A ( G ) is a left Z[G]-module. Clearly, A ( G ) is annihilated by left mul t ip l ica t ion by elements of A(AT), and thus ( k - l ) a = 0 for a e A ( G ) and k € K . T h e n k a = a , and since the action of K is t r iv ia l , we have a left act ion of G / K on A ( G ) .

Now let T be a transversal for K i n G , and note that since we are assuming that AT is normal , there is no dis t inc t ion between left and right transversals. Let a e Z[G] be the sum of the elements of T, and observe that left mul t ip l ica t ion by a defines a homomorphism of additive groups S : A ( G ) -> A ( G ) . A l so , since left mul t ip l ica t ion by elements of AT fixes al l elements of A ( G ) , the map 3 is independent of the transversal T.

10 .24 . T h e o r e m . L e t v . G ^ K / K ' be t h e t r a n s f e r h o m o m o r p h i s m , w h e r e G i s a finite g r o u p a n d K < G. W o r k i n g i n Z [G] , w r i t e

X 7 c T _ A ( G ) 1 ; A ( A ~ ) A ( G ) '

a n d l e t E : A ( G ) -> A ( G ) be t h e m a p d e f i n e d by left m u l t i p l i c a t i o n by t h e sum of t h e elements of a t r a n s v e r s a l f o r K i n G. T h e n v { G ) 9* S ( A ( G ) ) .

P r o o f . Let V : G -> AT be a pretransfer map, so that

V ( g ) = l [ t g ( t - g ) - 1 , t e r

where T is some transversal for AT in G , and the product is taken in some fixed but unspecified order. T h e n V ( g ) e K , and s o V ( j ) - l € A (AT). Under the isomorphism between A(AT) and K / K ' i n Coro l la ry 10.23, this element corresponds to the image of V ( g ) i n K / K ' , which is v { g ) .

We argue next that

V ( g ) - l = E ( g - l ) .

To see this, wri te t g = k t ( t - g ) for each element t e T, where kt € AT. T h e n = ]"]>*, and since by Coro l la ry 10.23, the map fc ^ fc^T is a homomorphism from K to the additive group A (AT), we have

V { g ) - l = Y d

k t ~ 1 -ter


A l s o , recall that 3 ( a ) = era = era, where a = Y \ t, and thus

3 ( ^ = 1 ) = £ i f o - l ) .

We have

g*c - 1 ) = 5 3 * ^ - 5 3 *

k t { t - g ) ~ Y , t - 9 t e T t e r

= J 2 ( k t - l ) { t - g ) t e T

= Y , ( k t - l ) mod A ( K ) A ( G ) ,

where the final congruence holds because ( k - l ) x = ( k - l ) mod A ( A T ) A ( G ) for a l l x € G and A; € AT. Thus

3 ( ^ 1 ) = 5 3 ^ 1 = ^ ) - ! .

as claimed.

N o w let X be the image of the map g ^ V ( g ) - 1 from G to A(AT) . T h e n A corresponds to v ( G ) under the isomorphism of L e m m a 10.23 between A ( K ) and AT/AT', and it follows that A is a subgroup. The map 3 is a homomorphism from A ( G ) into itself, and we have shown that 3 carries the generators of A ( G ) onto the subgroup X . It follows that 3 ( A ( G ) ) = A = v ( G ) , as wanted. •

Reca l l that the assertion of Theorem 10.18 is that the transfer homomorphism v : G -> G ' / G " is always t r iv ia l . Ac tua l ly , something a l i t t le more general is true.

10 .25 . T h e o r e m . L e t G be a f i n i t e g r o u p , a n d l e t v : G A T / A " be t h e t r a n s f e r h o m o m o r p h i s m , w h e r e G' C AT C G . T/ien w(# ) l K : G ' l = 1 / o r aiZ elements g e G .

O f course, if AT = G ' , then Theorem 10.25 asserts that v ( g ) = 1 for al l g € G , and so Theorem 10.18 is an immediate consequence. To prove Theorem 10.25, we need the following general result about modules for commutative rings.

10 .26 . T h e o r e m . L e t A be a left R - m o d u l e , w h e r e R i s a c o m m u t a t i v e r i n g , a n d l e t U C R be a n i d e a l . Assume t h a t A a n d U a r e finitely g e n e r a t e d as

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a d d i t i v e g r o u p s , a n d suppose t h a t U A has finite i n d e x m i n A . T h e n t h e r e exists a n element r G R such t h a t r A = O a n d r = m - l m o d U.

We review a bit of mat r ix theory before we begin the proof. If S is an n x n mat r ix over the commutat ive r ing R, the c l a s s i c a l a d j o i n t T of S is the n x n mat r ix whose ( i , j ) -en t ry is plus-or-minus the determinant of the ( j , i ) -minor of S, where the sign is . T h e fact about the classical adjoint that we need is that TS = d l , where d = det(S) and I is the n x n identi ty matr ix .

P r o o f o f T h e o r e m 10 .26 . B y the fundamental theorem of abelian groups, the finite abelian group A / U A of order m is a direct product of cyclic subgroups d for 1 < i < k. Wr i t e m , = \ C X \ , and observe that m = \ A / U A \ = W r r n . N o w choose elements a { G A such that a { generates d modulo U A , and let B = fa | 1 < * < k ) , so that A = B + U A .

Next , we expand the set {a, | 1 < i < k } to obtain an appropriate generating set for A . To do this, we observe that U A is finitely generated as an additive group since both U and A are finitely generated, and so we can choose a finite generating set X for U A . Choose notat ion so that X = { a i | k < i < n } , where n = k + \ X \ , and observe that since A = B + U A , the set { a i | 1 < i < n } generates A . Let = 1 for k < i < n and note that n ? = i m = m and r m a i G U A for 1 < i < n .

Since mm G U A for 1 < i < n and A is generated by the elements a j

for 1 < j < n , we can write

where s i d G £/, and we let S be the n x n mat r ix over R w i t h ( i , j ) -en t ry equal to ' S i J . A l so , let M be the diagonal n x n ma t r ix w i t h (t, i )-entry equal t o m r l , and finally, let v be the n x 1 co lumn vector w i t h entry i equal to en. We thus have M v = Sv, and thus ( M - S)v = 0.

Now let T be the classical adjoint of the n x n mat r ix M - 5 over i?. T h e n T ( M - S) = r l , where r = d e t ( M - 5) and / is the n x n identi ty mat r ix over R, and we have

and thus r a { = 0 for 1 < i < n . Since the elements a { generate A , we have r A = 0, as required. A l so , the entries of S lie i n U, so we have

3=1

0 = T ( M - S)v = r l v = r v

r = d e t ( M - S) = d e t ( M ) = f[ m i • ! = m - l mod U

and the proof is complete. •

We return now to group rings of finite groups. Recal l that i i K < G then A ( G ) = A { G ) / A ( K ) A ( G ) is a left Z[G]-module.

10.27. L e m m a . L e t K < G, w h e r e G i s a finite g r o u p . I n 1\G\, l e t

A ( G ) = A ( K ) A ( G ) '

a n d l e t E : A ( G ) -> A ( G ) be t h e map i n d u c e d by left m u l t i p l i c a t i o n by t h e sum of t h e elements of a t r a n s v e r s a l f o r K i n G. Suppose t h a t e e Z[G] s a t i s f i e s e A ( G ) = 0, a n d l e t m = 5 { e ) , w h e r e 5 i s t h e a u g m e n t a t i o n m a p . Then \ G : K \ d i v i d e s m, a n d ( m / \ G : K \ ) > E ( A ( G ) ) = 0.

P r o o f . Let T be a transversal for K i n G w i t h 1 6 T , and let a = Y \ t, so that S ( a ) =<ra = ra for a g A ( G ) . G i v e n t 6 T, we know that the lef t-mult ipl icat ion maps on A ( G ) by a l l elements of the coset K t are equal, and thus it is no loss to assume that

e = J2 e t t

t e r

for some integers et w i t h V \ et = m. If g € G , then since e A ( G ) = 0, we have

( J 2 e t t ) ( g - l ) = 0 mod A ( K ) A { G ) . t e T

N o w for t 6 T, write t g = k t { t - g ) , where kt G A". T h e n

(E a t ) ( g - l ) = J2 e t h ( t - g ) - J2 e t t

t e T t e T t e T

t e T t e T

= ^ { e t k t - e t . g l ) { t > g ) , t e T

and by the previous paragraph, this element lies i n A ( A - ) A ( G ) . Now the (i-o)-component of this element w i t h respect to T is etkt - et .pl , and by L e m m a 10.21, this component must lie i n A ( K ) . It follows that et - et.g = 0 for a l l t € T and g € G.

Since the dot action of G is transitive on T , the numbers et must a l l be equal, say to e, and thus e = ea. A l so , m = V > i = e\T\ = e\G : K \ , and thus | G : K \ divides m , as wanted. F ina l ly , i f a 6 A ( G ) , we have

( m / \ G : K \ ) - E ( a ) = eaa = ea = 0,

and the proof is complete. •

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P r o o f o f T h e o r e m 10 .25 . A s usual, wri te A ( G ) = A { G ) / A ( K ) A ( G ) , and observe that AT < G since G' C AT. Let E : A ( G ) —> A ( G ) be the (uniquely determined) map induced by left mul t ip l ica t ion by the sum in Z[G] of the elements of an arbi t rary transversal for AT i n G. B y Theorem 10.24, the transfer image v ( G ) is isomorphic to E ( A ( G ) ) , and so to show that v { g ) \ K - G ' \ = 1 for al l g i n G, it suffices to show that the map | A : G ' \ E is identically zero on A ( G ) .

Wr i t e A - A ( G ) , and recall that AT acts t r iv i a l ly (by left mult ipl icat ion) on A , and thus the abelian group G / K acts on A by left mul t ip l ica t ion , and so A is natural ly a left module for the commutat ive r ing R = Z [ G / K ] . We want to apply Theorem 10.26 w i t h U = A ( G / K ) , and so we must compute the index \ A : U A \ .

We argue that \ A : U A \ = \ G : G'\. To see this, observe that the act ion of an element K g e G / K on A is just the action of g on A , and thus

U A = A ( G / K ) A = E ( ( K g ) - 1 ) A = J 2 ( 9 - 1 ) A = & ( G ) A . K g e G / K g e G

Also ,

A ( G ) A = A ( G ) A ( G ) = A { G Y ,

and thus

\ A : U A \ = \ A j G j : A ( G p | = | A ( G ) : A ( G ) 2 | = \ G : G'\,

where the second equality holds because A ( A ) A ( G ) C A ( G ) 2 , and the th i rd holds by Theorem 10.20.

Theorem 10.26 applies since A and A ( G / K ) are finitely generated, and it follows that there exists an element 7 6 Z[G/AT] such that 7 A = 0 and 7 = \ G : G ' l - l mod A ( G / K ) . Observe that this congruence modulo A ( G / K ) implies that the augmentation of 7 i n the group r ing Z [ G / K ] is equal to | G : G ' | .

Since the element K g of G / K acts on A as g does, we can use 7 to construct an element e G Z[G] that acts on A as 7 does, and for which the augmentation is equal to the augmentation of 7, namely \ G : G'\. Thus eA = 0 and 5(e) = \ G : G ' | , and therefore L e m m a 10.27 guarantees that ( | G : G ' | / | G : AT|)S is identically zero on A ( G ) . Since | G : G ' | / | G : K \ = \ K : G'\, we are done. •

We close w i t h an applicat ion of Theorem 10.18, original ly proved by J . A l p e r i n and T . K u o w i t h a different proof.

10 .28 . C o r o l l a r y . L e t G be a finite g r o u p , a n d l e t A = G' n Z ( G ) . T h e n

g\G:A\ _ 1 jQr an elemenfs g ^ G .

P r o o f . Let V : G - > A , U : G ->• G' and W : G' -> A be pretransfer maps, and let g e G. Since A is abelian, V is actually the transfer homomorphism from G to A and is the transfer homomorphism from G' to A . Now A C Z ( G ) , and thus = by Theorem 5.6. B y Theorem 10.8, which is the t ransi t iv i ty of transfer theorem, and again using the fact that A is abelian, we have V { g ) = W { U { g ) ) . N o w Theorem 10.18 tells us that U { g ) G G", and thus U ( g ) G k e r ( W ) . We now have g \ a A \ = V ( g ) = W ( U ( g ) ) = 1, as wanted. •

Problems IOC

1 0 C . 1 . Let H be a group homomorphism. Show that ip can be extended to a r ing homomorphism 9 : Z [ G ] Z [ H ] , and show that A ( N ) Z [ G ] = k e r ( 9 ) = Z [ G } A { N ) , where N = k e v { < p ) .

1 0 C . 2 . Let A be an addit ively wri t ten finitely generated free abelian group. Show that a l l Z-bases for A have the same finite size.

H i n t . Let p be a prime number and let B = p A . Show that A / B is a finite dimensional vector space over the field F of order p .

1 0 C . 3 . Let i f be a subgroup of the group U of invertible elements in Z [ G ] , and suppose that I f is a Z-basis for the additive group of Z[G] . Show that there is a subgroup K C U such that K ^ H and K is also a Z-basis for Z [G] , and such that 8 { k ) = 1 for a l l elements k € K , where 8 is the augmentation map on Z[G] .

N o t e . It had been conjectured that in this si tuation, H and G were necessarily isomorphic, but this is now known to be false,

1 0 C . 4 . A s i n the previous problem, let H be a Z-basis for Z [ G ] , where H is a subgroup of the group U of invertible elements i n Z [ G } . Show that G I G ' ^ H / H ' .

1 0 C . 5 . Le t | G | = n and suppose that the elements of G are numbered, so that G = {01, g 2 , . . . , g n } - G iven an element a € Z [G] , let M { a ) be the n x n matr ix over Z whose ( i , j ) -entry is equal to the coefficient of g 5 in 9 i a . Show that the trace t r ( M ( a ) ) is ne, where e G Z is the coefficient of 1 i n a .

1 0 C . 6 . In the si tuation of the previous problem, suppose that a m = 1 for some integer m . Show that a l l eigenvalues of the mat r ix M { a ) are roots of unity, and deduce that e G { - 1 , 0 , 1 } .

H i n t . Show that M ( a ) m is the identity matr ix , and use the fact that the trace of an n x n mat r ix is equal to the sum of its n eigenvalues, counting mult ipl ic i t ies .

Appendix: The Basics

In this appendix, we provide a quick review of some of the basic facts of elementary group theory that are assumed i n the foregoing chapters. The presentation here is neither detailed nor exhaustive, however, and readers may wish to consult the appropriate sections of a general abstract-algebra text for a more comprehensive treatment of some of this material . (Parts of our book A l g e b r a : A G r a d u a t e C o u r s e , for example, would be a good supplementary source for the theory needed here.)

A group is a set G together w i t h an associative binary operation on G that satisfies two further conditions: there must be a (two-sided) identi ty element in G, and each element of G must have a (two-sided) inverse w i t h respect to that identity. We generally write our groups mult ipl icat ively, denoting the identi ty by the symbol " 1 " , and wr i t ing " a T 1 " for the inverse of an element x 6 G. (It makes sense to fix this nota t ion because, as is easy to check, a group G has a unique identity, and each element x e G has a unique inverse.)

A group that plays a fundamental role i n the theory is the symmetric group on a nonempty set X . Th i s group, denoted S y m ( A ) , is the set of al l bijections from X onto itself, and the binary "mul t ip l ica t ion" in S y m ( X ) is function composit ion. (The elements of S y m ( A ) are also referred to as permutations of A . ) The group identity i n S y m ( A ) is the identity map on A , and this explains why the identi ty of a group is called the "identity" element. (Note that i n this book, we compose functions from left to right, so that i f a , P € S y m ( A ) , then a ( 3 means "do a , then /?", and so we can write ( x ) ( a / 3 ) = { { x ) a ) ( 3 . ) In the case where A = { 1 , 2 , . . . , n } we write Sn

for S y m ( A ) , and we note that \ S n \ = n \ .

325

326 A p p e n d i x : T h e B a s i c s

A transposition i n S y m ( X ) is a permutat ion that interchanges two points, leaving al l other members of X fixed, and it is not hard to see that every permutat ion is a product of transpositions. Less t r iv ia l is the fact that i f a G S y m ( X ) is wr i t ten as a product of r transpositions, then al though the integer r is not uniquely determined by a , its pari ty is. In other words, either every representation of a as a product of transpositions involves an even number of transpositions, or else every such representation involves an odd number of transpositions. E x a c t l y half of the permutations on a set X containing at least two points are products of an even number of transpositions, and these even permutations form a group called the alternating group on X , denoted A l t ( X ) . If X = {1, 2 , . . . , n} , we write A n for A l t ( X ) , and we note that \ A n \ = (n!)/2.

N o w let x G G, where G is an arbi trary group. If n is an integer, we define x n as follows. If n > 0, then x n = xx • • • x is the product of n copies of x ; i f n < 0, then, of course, - n is positive, and we define x n = ( x - 1 ) " " , and finally, we set x ° = 1. W i t h these definitions, it is easy to check that x m x n = yjn-\-n gj oj ^ x m ) n — x m n for al l integers m, n G Z .

In group theory, the word "order" has two different, but not entirely unrelated meanings. The order of a group G is its cardinality, denoted \ G \ , and the order of an element x G G, denoted o(x) , is the smallest positive integer n such that x n = 1. (If there is no such positive integer, we say that x has infinite order.) If o { x ) = n , then for integers a and b, we have x a = x b i f and only if a = b m o d n , and thus there are precisely n different elements of G that are powers of x , and it follows that \ G \ > o(x) . If x has infinite order, then the only way to get x a = x h is for a = b, and so i n this case, a l l of the elements x a w i th a G Z are distinct. In part icular, if x has infinite order, there are infinitely many different powers of x , and thus G must have infinite order.

A subgroup of a group G is a subset H C G such that H is a group i n its own right, using the mul t ip l ica t ion i n G. If H is a subgroup, it is easy to check that its identity must be the identi ty of G, and the inverse i n H of an element x G H is the inverse of x i n G . A subgroup of G, therefore, is a nonempty subset that is closed under mul t ip l ica t ion and inverses, and it is easy to see that conversely, every such subset is a subgroup. If x G G has finite order n > 1, then x ~ l = x n ~ x is a power of x , and so if G is finite, it follows that it suffices to check that a nonempty subset is closed under mul t ip l ica t ion i n order to establish that it is a subgroup; closure under inverses then follows automatically.

The p r imary objects of interest i n group theory are the subgroups of a given group G, so it is almost always the case that when we mention some

A p p e n d i x : T h e B a s i c s 327

subset of G , we are th inking of a subgroup. If we write "let H C G " , therefore, then although we may neglect to say so, we generally intend that H should be a subgroup, and we do not use any special notat ion to denote subgroups. (For clarity, however, especially in statements of theorems, we w i l l often not rely on this default, and we w i l l say expl ic i ty that H is subgroup.) In situations where we want to consider subsets that may not be subgroups, we indicate that by wr i t ing something like "let X C G be a subset".

O f course, a set that we c o n s t r u c t should not automatical ly be assumed to be a subgroup. For example, given subsets X , Y C G , the subset X Y is defined by X Y = { x y \ x G X , y£Y}. B u t even i f X and Y are subgroups, i t is not generally true that X Y is a subgroup.

X . l . L e m m a . L e t H a n d K be s u b g r o u p s of a g r o u p G. T h e n H K i s a s u b g r o u p if a n d o n l y if H K = K H .

P r o o f . If H K = K H , then ( H K ) ( H K ) = H ( K H ) K = H ( H K ) K = ( H H ) ( K K ) = H K , and thus H K is closed under mul t ip l ica t ion. A l so , i f h G H and k G K , then ( h k ) ' 1 = k ' ^ h ' 1 G K H = H K , and hence H K is closed under inverses. Since 1 G H K , we see that H K is nonempty, and thus it is a subgroup. Conversely, suppose that H K is a subgroup. T h e n K C H K and H C H K , and since H K is closed under mul t ip l ica t ion , we have K H C H K . To prove the reverse containment, let x G H K . Since H K is closed under inverses, we have x ~ l G H K , and thus we can write x ~ l = hk, w i th h G H and k G AT. T h e n x = (/ifc)" 1 = A T 1 / * - 1 G K H , and this completes the proof. •

There is a simple formula for the cardinal i ty of H K . (We do not say "order" since H K may not be a group.)

X . 2 . L e m m a . L e t H a n d K be finite s u b g r o u p s of a g r o u p G. T h e n \ H K \ = \ H \ K \ / \ H n K \ .

P r o o f . Wr i t e D = H n K . The set H x K of ordered pairs ( h , k ) w i t h he H and k e K has cardinali ty | / f | |AT | , and we have a map 6 : H x K ^ H K defined by 0((/i,fc)) = /ifc. To complete the proof, it suffices to show that each element x G H K has exactly | D | preimages w i t h respect to 9. To see this, wri te x = h k w i t h h G H and jfc G K . If d £ D , then also ( T 1 G D since both 7f and K are closed under inverses, and thus ( h d , d ~ l k ) <E H x K and 9 { { h d , d - l k ) ) = h k = x . The pairs { h d , d ~ l k ) are dist inct for distinct elements d G D , and this yields | D | preimages for x . To see that we have al l possible preimages, suppose that 9 ( ( h ' , jfc')) = x w i th h ' £ H and jfc' G AT. T h e n h k = x = ft'jfc', and thus jfc(jfc')"1 = h ^ t i , and this element, which we cal l d, lies in H n AT = £>. We have / i _ 1 f t ' = d, so h ' = hd, and also,


d = k ( k ' ) - \ so k' = d~lk. In other words, ( h ' , k ' ) is one of the preimages that we have already constructed, and the proof is complete. •

The following is another useful fact about products of two subgroups.

X . 3 . L e m m a (Dedekind). L e t H a n d K be s u b g r o u p s of a g r o u p G, a n d l e t H C U C G, w h e r e U i s a l s o a s u b g r o u p . T h e n H K n U = H ( K n U ) .

P r o o f . It is clear that H ( K n t / ) C H K , and since H C U, we also have H { K n U ) C U. Thus H ( K n U ) C FAT n U, and i t suffices to prove the reverse containment. Let u G H K n t/, and write u = hk, w i th ft e 7f and k e K . T h e n k = h~lu G £/ since H C\ U and u £ U. Thus k E U, and we have A; G AT n £/, and so u = h k G H ( K n t/), and the proof is complete. •

The following shows how Dedekind's lemma can be used. Before stating the result, however, we mention that intersections of arbi trary collections of subgroups are always subgroups. (This is clear since intersections of sugroups are closed under mul t ip l ica t ion and inverses, and they contain the identity, so they cannot be empty.)

X . 4 . C o r o l l a r y . Suppose t h a t H a n d K a r e s u b g r o u p s of a g r o u p G, a n d assume t h a t H K i s a l s o a s u b g r o u p . L e t D = H n K , a n d l e t X a n d fV be t h e c o l l e c t i o n s of s u b g r o u p s d e f i n e d by

X = { X | H C A C H K } and y = { Y \ D C Y C K } .

D e f i n e t h e m a p 9 : X -> y by i n t e r s e c t i o n w i t h K , so t h a t 9 { X ) = X n K . T h e n 9 i s i n j e c t i v e , a n d i t s i m a g e i s t h e set W = { W G fV | W H = H W ) .

H K

D

In the diagram, the parallelogram formed by K , H K , H and D is a direct diamond, which means that the lower node corresponds to the intersection of the subgroups represented by the side nodes, and the upper node corresponds to their product. The members of X lie along the upper edge, jo in ing H and H K , and the members of y lie along the parallel lower edge. The dashed lines represent the map 9, so that, for example, 9 { X ) = Y. In fact, each dashed line divides the or iginal figure into two new direct diamonds. For example, the upper parallelogram formed by the dashed line jo in ing X to Y is a direct d iamond because X n K = Y and (as should be clear) X K = H K . The paralellogram formed by Y, X , H and D is also a direct d iamond because H n Y = D and H Y = X , where as we shall see i n the proof, the latter equality is a consequence of Dedekind 's lemma. The node on the lower edge that is not joined to a corresponding node on the upper edge is intended to represent the fact that the map 9 need not be surjective.

P r o o f of Corol lary X . 4 . It should be clear that intersection wi th K actual ly does define a map from X to fV. Now let X G X . T h e n H C X C H K , and therefore X = X n H K = H ( X n K ) = H 9 ( X ) , where the second equality follows by Dedekind's lemma, which applies since H C X . Th i s shows that X is determined by its image 9 { X ) , and thus 9 is injective. A l so , since H 9 ( X ) = X is a subgroup, it follows by L e m m a X . l that 9 ( X ) H = X 9 ( U ) , and thus 9 { X ) G W . F ina l ly , i f W G W , then W H is a group by L e m m a X . l , and since W C K , we have W H G X . N o w 9 { W H ) = W t f n i f = W ( 7 f n K ) = W, where the second equality follows by Dedekind's l emma and the th i rd equality holds because H n K C W . •

Let G be an arbi trary group. If g G G , the subset G = | n G Z } is easily seen to be a subgroup, and clearly, G is contained i n every subgroup of G that contains g . We can thus describe G as the unique smallest subgroup of G that contains g , where "smallest" should be understood in the sense of containment. In fact, if X is an any subset of G , there is a unique smallest subgroup that contains X , namely the intersection of a l l of the subgroups of G that contain X . (This makes sense because there certainly is at least one subgroup of G containing A , namely G itself, and i n general, intersections of subgroups are subgroups.) The unique smallest subgroup of G that contains X is said to be the subgroup generated by A , and it is denoted ( A ) .

A maximal subgroup of a group G is a proper subgroup M such that there is no subgroup H w i t h M < H < G. (It is clear that every proper subgroup of a finite group is contained in at least one max ima l subgroup,

330 A p p e n d i x : The Basics

but some infinite groups, for example the additive group of the real numbers, have no max ima l subgroups at all.) The intersection of a l l of the max ima l subgroups of a finite group G is the Fratt ini subgroup, denoted $ ( G ) . A n interesting property of $ ( G ) is that it is exactly the set of elements of G that cannot contribute to the construction of a generating set for G.

X . 5 . L e m m a . L e t G be a finite g r o u p . If X C G i s a subset such t h a t ( A ) < G, then a l s o ( I U $ ( G ) } < G. Conversely, if u G G has the p r o p e r t y t h a t whenever ( A ) < G, a l s o ( X U { u } ) < G, then u € $ ( G ) .

Proof. Le t A C G be a subset such that ( A ) < G . T h e n ( A ) is contained i n some max ima l subgroup M of G , and since also $ ( G ) C M , we see that A U $ ( G ) C M , and thus ( A U $ ( G ) ) C M < G.

Now suppose that u € G is an element such that ( X U { u } ) < G for al l subsets A of G such that ( X ) < G. To prove that n e $ ( G ) , w e must show that u G M for a l l max ima l subgroups M of G , so suppose u 0 M , where M is a max ima l subgroup. T h e n ( M ) = M < G , and hence ( M u { u j ) < G , by hypothesis. We have M < ( M U {«}) < G , and this is a contradiction since M is a max ima l subgroup. •

We have seen that a subgroup G = (#) generated by a single element g G G is exactly the set of powers of g i n G . (A group generated by a single element is said to be cyclic.) If o { g ) is finite, the elements g l are distinct for 0 < i < o ( g ) , and since these are easily seen to be a l l of the elements of G , it follows that \ C \ = o { g ) i n this case. O n the other hand, if g has infinite order, it is clear that G has infinite order.

If G is a cyclic group, so that G = (g) for some element g G G , it is easy to describe a l l of the subgroups of G .

X . 6 . L e m m a . L e t G be a c y c l i c g r o u p w i t h G = ( g ) , a n d l e t H C G be a n o n i d e n t i t y s u b g r o u p . Then g m G H f o r some p o s i t i v e i n t e g e r m, a n d if m i s the smallest such i n t e g e r , then m d i v i d e s every i n t e g e r n such t h a t g n G H . A l s o , H = ( g m ) , a n d i n p a r t i c u l a r , a l l subgroups of G a r e c y c l i c .

Proof. Since al l elements of G are powers of g and H contains some non-identi ty element, it follows that g a G H for some nonzero integer a . If a < 0, then g ~ a = ( 5

a ) " 1 G H , and thus i n any case, H contains an element of the form g m w i t h m > 0, and we can choose m to be as small as possible.

Suppose now that g n G H for some integer n. B y the "division algor i thm" we can write n = q m + r , where the "remainder" r satisfies 0 < r < m. We have g r = g n { g m ) - q , and this element lies i n H since bo th g n e H and g m G H . Since r < m, it follows by the min imal i ty of m that we cannot have r > 0, and thus r = 0, and m divides n , as required.

To show that H = ( g m ) , we must show that an arbi t rary element h G H is a power of g m . B u t G = ( g ) , so we can certainly write h = g n for some integer n , and thus since g n G H , we know that n/m is an integer. T h e n h = g n = { q m ) n l m , and this completes the proof. •

X . 7 . C o r o l l a r y . L e t G = (g) be a finite c y c l i c g r o u p of o r d e r n , a n d f o r each p o s i t i v e d i v i s o r d o f n , l e t Gd = \ g n / d ) . T h e n Gd i s t h e u n i q u e s u b g r o u p o f G h a v i n g o r d e r d, a n d these s u b g r o u p s Gd a r e a l l of t h e s u b g r o u p s o f G .

P r o o f . Observe that o ( g ) = \ G \ = n and \ G d \ = o ( g n ' d ) . Now { g n l d ) d = g n = 1, and if 0 < e < d, then ( n / d ) e < n , and so { g n ' d ) e / 1. It follows that o { g n / d ) = d, and thus \ G d \ = d.

We show next that an arbi trary subgroup H of G is one of the subgroups Gd. If H is t r i v i a l , it is G j , and so we can assume that H is nontr ivia l , and we let m be the smallest positive integer such that g m G H . Since g n = l e H , L e m m a X . 6 guarantees that m divides n and that H = { g m } . It follows that H = Gd, where d = n / m , and this completes the proof. •

We have just seen that if G is a finite cyclic group and H is a subgroup, then \ H \ divides | G | . In fact, this is true for a l l finite groups G, even if they are not cyclic. To establish this theorem of Lagrange, we consider cosets.

G i v e n a subgroup H C G and an element x G G , where G is an arbi t rary group, the set H x — { h x \ h G H } is a r i g h t cose t of H i n G . The number of these right cosets, which may be infinite, is the i n d e x of H i n G , which is denoted | G : H \ . (Of course, the index is the number of d i f f e r e n t right cosets of H in G . If H x = H y , then this coset w i t h two names should be counted only once.)

X . 8 . T h e o r e m (Lagrange). L e t H be a s u b g r o u p of a n a r b i t r a r y g r o u p G. T h e f o l l o w i n g t h e n h o l d .

(a) L e t y G H x , w h e r e x G G . T h e n H y = H x .

(b) D i s t i n c t r i g h t cosets of H i n G a r e d i s j o i n t .

(c) A l l r i g h t cosets of H i n G h a v e c a r d i n a l i t y e q u a l t o \ H \ .

(d) I f \ G \ i s finite, t h e n \ H \ d i v i d e s \ G \ a n d \ G \ / \ H \ = \ G : H \ .

P r o o f . We show first that i f h G H , then H h = H . (Note that this is the special case of (a) where x = 1 and y = h . ) Since h ' 1 G H and H is closed under mul t ip l ica t ion , we have H h ~ l C H . R igh t mul t ip l ica t ion by h yields H C H h , and the reverse containment is another consequence of the fact that H is closed under mul t ip l ica t ion . Asser t ion (a) follows since we can wri te y = hx, where h G H , and thus H y = H h x = H x . Now (b) is immediate since i f z G H x n H y , then H x = H z = H y by two applications of (a).

For (c), we construct a bijection 9 : H - > H x , where Ex is an arbi trary right coset of i f i n G . Le t 9 ( h ) = hx and observe that 9 is automatical ly surjective since by definition, every element of H x has the form hx for some element he H . A l so , 9 is injective since i f hx = kx for elements h , k G H , then right mul t ip l ica t ion by x ~ l yields h = k.

Fina l ly , since # G i f p for a l l g G G , we see that G is the union of the | G : i f | right cosets of H i n G . These cosets are disjoint by (b), and each of them has cardinal i ty | f f | by (c), and thus | G | = | G : / f | | / f | , proving (d). •

X . 9 . Corollary. L e t g G G, w h e r e G i s a finite g r o u p . T h e n o ( g ) d i v i d e s \ G \ .

Proof. We have o { g ) = \ ( g ) \ , which divides | G | by Lagrange's theorem. •

The exponent of a finite group is the least common mult iple of the orders of its elements. B y Corol la ry X . 9 , therefore, the exponent of G divides \ G \ for finite groups G .

Every th ing we d id w i t h right cosets works as well w i t h left cosets of the form x H = { x h \ h G i f } . In part icular, the "mirror image" of Theorem X.8(d) shows that if i f C G , where G is finite, then the number of left cosets of H in G is also equal to | G | / | f f |, and thus the sets { H x \ x G G } and { x H | x G G } have equal cardinal i ty (although in general, they are different sets). B u t even if G is infinite, the cardinalities of the sets of left and right cosets of i f in G are equal. To see this, consider the bijection from the set of a l l subsets of G to itself defined by A ^ { x ~ l \ x G A } , and check that it carries each left coset of i f i n G to a right coset and v i c e v e r s a .

X . 1 0 . Corol lary. L e t H C AT C G , w h e r e G i s a finite g r o u p a n d H a n d K a r e s u b g r o u p s . T h e n \ G : i f | = | G : K \ \ K : H \ .

Proof. Since | G : K \ = \ G \ / \ K \ and \ K : H | = \ K \ / \ H | , we conclude that | G : K \ \ K : H \ = \ G \ / \ H | = | G : i f |. •

X . l l . Corollary. L e t i f a n d K be s u b g r o u p s of a finite g r o u p G. T h e n \ K : H n K \ < \ G : H \ , a n d e q u a l i t y h o l d s if a n d o n l y if H K = G.

Proof. Since H K is a subset of G , L e m m a X . 2 yields

IG| > \ H K \ = l ^ l ^ l 1 1 - 1 1 | f f n K \ '

w i t h equality i f and only if i f AT = G . T h e n

again w i t h equality if and only if i f K = G. •

Another useful fact i n this vein is the following.

X.12 . Corol lary. L e t H a n d K be s u b g r o u p s of a finite g r o u p G, a n d s u p pose t h a t \ G : H \ a n d \ G : K \ a r e r e l a t i v e l y p r i m e . T h e n H K = G.

Proof. Since i i i l i f C i i C G , it follows by Corol la ry X . 1 0 that \ G : H \ divides \ G : H n K \ , and similarly, \ G : K \ divides \ G : H D K \ . Since the indices \ G : H \ and \ G : K \ are coprime, their product must divide \ G : H n K \ , and thus

T h i s yields

\ G : H \ = ^ \ < , , = IA/ : P D P I , ' ' | P | ~ \ H H K \ 1 1 '

and thus by Corol la ry X . l l , equality holds and H K = G. •

Versions of these corollaries hold even for infinite groups but we w i l l not give proofs, or even precise statements.

ana Gr 2 are groups, a Dijection y : b i -> Gr 2 is said to be an i somorpmsm it » respects mul t ip l ica t ion , by wnicn we mean tnat ( ^ y j -V K x m v ) » r an elements x , y e L a . A l so it tnere is an isomorphism i r o m

r ~ r rrt I t i T w h i T o r a r e * s o m o r p m c ' a i ? . w e ™ , ~ J ; 1 8 " * ' ^ is an isomorpnism^then

, f " ' «Z 1 8 ^ ! S O m ° r p m S m ' . S O 1 1 U l ~ t n e n a l S O ^2 - O i , ana m iact, is an equivalence relation.j

A group theorist would say that a group is nothing but a certain mul t i pl icat ion rule on a set, and that what counts is the mul t ip l ica t ion rule, and not the set. F r o m this point of view, we see that i f the underlying set of a group is changed, but the mul t ip l ica t ion is somehow preserved, this w i l l not fundamentally alter the group. To a group theorist, therefore, isomorphic groups are essentially identical, and so if Gx =i G2, then G2 enjoys a l l of the "group theoretic" properties of Gx. For example, if Gx has exactly eight elements of order 3, then the same is true of G2, and i f G\ has an abelian subgroup of order 10, then so too does G2. ( A group G is abelian if xy = yx for a l l x,y <G G.)

O f course, isomorphic groups may look very different to someone who is not a group theorist. For example, the additive group of the real numbers is isomorphic to the mult ipl icat ive group of the positive real numbers v i a the map x ^ ex. More surprisingly, the additive group of the real numbers is isomorphic to the additive group of the complex numbers. (It requires an appeal to the ax iom of choice to establish the existence of such an isomorphism, however, and so it is not possible to describe an explici t map.)

The following lemma shows that for each positive integer n , there is essentially only one cyclic group of order n .

X . 1 3 . L e m m a . L e t B = (b) a n d C = (c) be c y c l i c g r o u p s of t h e same f i n i t e o r d e r n , a n d l e t 9 : B - > C be d e f i n e d by 9 ( b l ) = c% f o r a l l i n t e g e r s i. T h e n 9 i s a w e l l d e f i n e d i s o m o r p h i s m f r o m B o n t o C.

P r o o f . F i r s t , observe that o ( b ) = n = o(c). If bl = V, then i = j mod n , and thus & = c\ and it follows that B is unambiguously defined. Also , 9 is surjective since every element of C has the form c? for some integer j , and since \ B \ = \ C \ < oo, it follows that 9 is also injective. F ina l ly , i f x,y E B , write x = br and y = bs for integers r and s. T h e n

9 ( b r b s ) = 0 ( b r + s ) = c r + c = crcs = 9 ( b r ) 0 ( b s ) ,

and so 9 ( x y ) = 6 ( x ) 6 ( y ) , and 0 is an isomorphism. •

If 9 : G i -> G 2 is an isomorphism, it should be clear that 9 carries each subgroup of G i to a subgroup of G2, and in fact, the map H ^ 0 ( H ) is a bijection from the set of subgroups of Gx onto the set of subgroups of G2. Furthermore, i f H C Gx is a subgroup that satisfies a certain property in Gx, then its image 0 ( H ) satisfies the corresponding property in G2.

G i v e n a group G , consider, for example, the set of elements z G G that commute w i t h a l l elements of G . Th i s set, denoted Z ( G ) , is the c e n t e r of G , and it is easy to check that it is a subgroup. If 0 : Gx - > G2 is an isomorphism, it should be clear that 9 carries the center of Gx to the center of G 2 . More generally, if H is any subgroup of Gx that can be described wi th the definite article "the", then 9 ( H ) is the corresponding subgroup of G2. For example, if H is the Fra t t in i subgroup of Gx or the derived subgroup of G i , then the image 9 ( H ) is respectively the Fra t t in i subgroup or the derived subgroup of G 2 . (It is not necessary to know the definition of the derived subgroup to see that this must be true.)

A n isomorphism from a group G to itself is called an a u t o m o r p h i s m of G , and the set of a l l automorphisms of G is denoted-Aut(G) . Of course, A u t ( G ) is a subset of S y m ( G ) , and it is easy to see that in fact, A u t ( G ) is a subgroup of S y m ( G ) .

X . 1 4 . L e m m a . L e t G be a c y c l i c g r o u p . T h e n A u t ( G ) i s a b e l i a n , a n d if G i s finite, t h e n | A u t ( G ) | = y>(|G|), w h e r e ip i s E u l e r ' s t o t i e n t f u n c t i o n .

P r o o f . Wr i t e G = ( g ) . Observe first that if a , (5 € A u t ( G ) and a ( g ) = ( 5 ( g ) , then a ( g n ) = a ( g ) n = ( 5 ( g ) n = f 3 ( g n ) for a l l integers n . Since every element of G has the form g n for some integer n , it follows that a and (5 agree on a l l of G , and thus a = (3.

N o w let o~, r G A u t ( G ) . T h e n <r(g) = g s and r ( g ) = g t for some integers s and t, and it is easy to check that <JT and T O bo th carry g to g s t . Since CTT and T O agree on we conclude that these automorphisms are equal, and hence A u t ( G ) is abelian.

N o w suppose \ G \ = n , so that o ( g ) = n . Since automorphisms preserve orders of elements, every automorphism of G maps g to an element of the set X = { x G G | o(x) = n}, and by the first paragraph of the proof, distinct automorphisms of G carry g to dist inct elements of X . If x G A is arbitrary, then G = ( x ) , and hence by L e m m a X . 1 3 , there is an isomorphism 9 from G = (g) to G = (x) such that % ) = x. Thus 9 is an automorphism of G that carries # to x, and it follows that | A u t ( G ) | = | A | .

To complete the proof, we argue that | A | = <p(n). Since every element of G is uniquely of the form g r w i t h 0 < r < n , it suffices to show that g r G A if and only if r and n are coprime. To see this, let i f = ( g r ) , so that \ H \ = o ( g r ) , and let m be the unique smallest positive integer such that g m G i f . It follows by L e m m a X . 6 , that m divides both r and n, and thus if r and n are coprime, we have m = 1 and g G i f , and thus H = G. T h e n 0 ( 5

r ) = | i f | = n , and # r G X .

Conversely, suppose that g r G A and let m be the greatest common divisor of r and n. Wr i te d = n / m , so that n = md, which divides r d , and thus ( 5

r ) d = 1 and d > o ( g r ) = n . T h e n d = n and m = 1, as wanted. •

A subgroup H C G is characteristic if every automorphism of G maps i f onto i f . Since an isomorphism from one group to another carries the center to the center and the F ra t t i n i subgroup to the F ra t t in i subgroup, it follows that an automorphism of a group G carries Z ( G ) to Z ( G ) and $ ( G ) to $ ( G ) . M o r e generally, every subgroup of a group G that can be described unambiguously using the word "the" must be mapped to itself by al l automorphisms of G , and hence such a subgroup is characteristic. In part icular, if G is a finite cyclic group, then every subgroup of G is characteristic because if i f C G and | i f | = d, then by Corol la ry X . 7 , we can say that " i f is t h e subgroup of G having order d".

G i v e n elements x and g in a group G , we write x° to denote the element g ~ l x g , and we say that x 9 is conjugate to x i n G . (Note that x 9 = x if and only i f x g = g x . ) It is easy to see that conjugation by g is an automorphism of G . (This map is bo th injective and surjective because it has an inverse, namely conjugation by g ~ \ and it is t r iv i a l to check that {xy)<> = x 9 y 9 as required.) The automorphism x ^ x 9 is said to be an inner automorphism of G , and the set of a l l inner automorphisms of G is denoted Inn(G) . Since ( x 9 ) h = x 9 h , for a l l x , g , h G G , it is easy to see that Inn(G) is a subgroup of the full automorphism group A u t ( G ) .

If X C G is a subset and g € G, then the image of X under conjugation by g is the subset X 9 = { x 9 \ x € G } . Since the conjugation maps are automorphisms, it follows that if H C G is a subgroup, then a l l of its conjugates H 9 for g G G are subgroups too. N o w recall that a characteristic subgroup of G is a subgroup that is mapped to itself by a l l automorphisms of G . We can weaken this condi t ion and consider the normal subgroups of G , which are those subgroups that are mapped to themselves by all i n n e r automorphisms of G . A subgroup H C G is normal , therefore, precisely when H 9 = H for a l l elements g G G . We write H < G in this case. O f course, characteristic subgroups are automatical ly normal .

In general, it is not true that normal subgroups of normal subgroups of a group G are normal in G , but the following substitute is often useful.

X.15 . L e m m a . L e t N < G, w h e r e G i s a g r o u p , a n d l e t C be c h a r a c t e r i s t i c i n N . T h e n C < G.

Proof. Let g G G and let a be the inner automorphism of G defined by conjugation by g . T h e n a is injective, and it maps N onto N , and since a respects mul t ip l ica t ion , the restriction of a to A is an automorphism of N . A s C is characteristic i n N , we have C9 = a { C ) = C, and thus C < G. U

G i v e n a subgroup H C G , the set {# G G | H 9 = H } , is the normalizer of i n G , which is denoted N G ( t f ) . It is easy to see that N G { H ) is always a subgroup of G , and that it contains the subgroup H . It follows that N G (J f ) is the unique largest subgroup of G that contains H as a normal subgroup.

Since conjugation by an element g G G is an injective map, \ H 9 \ = \ H \ for subgroups H of G . Thus if \ H \ is finite and Jf» C H , then Jf» = H , and 5 G N G ( J f ) . In general, however, it can happen that H 9 is proper in H . B u t if H 9 C H for elements g G G , then also i f 3 " 1 C P . Conjugating both sides by g , we get H C and thus in fact, tf5 = i f . To show that H < G , therefore, it suffices to check that H 9 <Z H for a l l elements g G G .

X.16 . L e m m a . A e i i f k a s u b g r o u p of a g r o u p G. T h e n t h e f o l l o w i n g a r e e q u i v a l e n t .

(1) H < G.

(2) H x = x H f o r a l l x G G .

(3) E v e n / h</fti cose* of H i n G i s a left coset of H .

(4) E v e n / ie/t coset of H i n G i s a r i g h t coset of H .

(5) ( H x ) ( H y ) = H x y f o r a l l x , y e G .

(6) T h e set of r i g h t cosets of H i n G i s c l o s e d u n d e r m u l t i p l i c a t i o n .

Proof. It is clear that (2) implies (4). Conversely, i f (4) holds and i e G , then the left coset x E is a right coset of H containing x , and so it must be the coset Ex by Theorem X.8(a) , and this proves (2). Thus (2) and (4) are equivalent, and since (2) is left-right symmetr ic , it must also be equivalent to (3), which is the mirror image of (4). A l s o , (5) and (6) are easily seen to be equivalent. Tha t (5) implies (6) is obvious, and conversely, assuming that E x H y is a right coset of H , then since it contains the element xy, it must be the right coset Exy by Theorem X.8(a ) .

Now assume (1), so that E < G, and let x G G . T h e n E = H x = x ~ x E x , and left mul t ip l ica t ion by x yields x H = H x , and thus (2) holds. N o w assume (2), and let x,y G G. T h e n H x H y = H ( x E ) y = E ( H x ) y = Exy since E E = E , and this proves (5). To complete the proof, it suffices to show that (5) implies (1). To see this , assume (5) and let x G G. T h e n

E x = x ~ l H x C E x ' 1 Ex = E x ~ l x = E ,

and since this holds for al l x G G, we have H < G. •

A useful observation about a normal subgroup N < G is that i f H C G is an arbi trary subgroup, then N E is a subgroup. To see why this is so, observe that

N E = [j N h = y h N = E N , h e H h e H

and apply L e m m a X . l .

Now let A be a normal subgroup of G, where G is an arbi t rary group, and write G / N to denote the set of right cosets of N i n G. (Of course, L e m m a X . 1 6 implies that the word "right" i n the previous sentence is superfluous because for normal subgroups, there is no dis t inct ion between left cosets and right cosets.) B y condi t ion (5) of L e m m a X .16 , the set G / N is closed under mul t ip l ica t ion , and we observe that i n fact it is a group since the coset N l = N acts as an identity and ( N x ) ( N X - 1 ) = N l , so that A ^ x " 1

is the inverse of N x i n G/N. The group G / N is called the quotient group or factor group of G w i t h respect to the normal subgroup N . (We stress that N must be normal; otherwise by L e m m a X . 1 6 , there are right cosets whose product is not a right coset.)

If G and H are arbi trary groups, a map 9 : G - > H is called a homo-morphism if 9 ( x y ) = 9 ( x ) 9 ( y ) for a l l x,y G G. (Thus an isomorphism is a homomorphism that happens to be bo th injective and surjective.)

We consider some examples of homomorphisms. G i v e n an arbi t rary group G , let 9 { x ) G A u t ( G ) be conjugation by x . It is t r iv i a l to check that gxy = ( g x ) y for elements g , x , y G G , and thus 9 { x y ) = 9 ( x ) 9 ( y ) and 9 : G -> A u t ( G ) is a homomorphism. (Remember that mul t ip l ica t ion i n A u t ( G ) is defined to be left-to-right.) The image of G under 9, of course,

is Inn(G) . For another example, let Z be the additive group of the integers and let H be an arbi trary group. F i x an element h G H and consider the map n i-> h n from Z to H . Since h m + n = h m h n for m, n G Z , this map is a homomorphism, and the image of Z i n J i is clearly the subgroup ( h ) .

Perhaps the most important example is the c a n o n i c a l homomorphism 7T : G -> G / A , where N < G and 7r(s) = A ^ . Observe that TT really is a homomorphism since vr(xy) = W x y = ( N x ) ( N y ) = 7r(x)7r(y) for elements x , y G G . The canonical homomorphism TT is always surjective, but since al l elements of N map to the identity of G/N, it is never injective unless N = 1, the t r i v i a l subgroup.

G i v e n an arbi trary group homomorphism 0 : G -> Jf, the k e r n e l of 0, denoted ker(0) is the set { x G G | 0(x) = 1}. For example, the kernel of the homomorphism 0 : G -> A u t ( G ) , where 0(x) is conjugation by x, is the set of elements x G G such that g x = g for al l g G G . Since # x = 5 i f and only if gx = x g , we see that ker(0) = Z ( G ) , the center of G . For another example, consider the canonical homomorphism TT : G -> G / W , where N < G. Since the identity of G / A is the coset N , we see that g lies i n ker(vr) precisely when N g = N , and thus ker(vr) = A".

X . 1 7 . L e m m a . L e t 9 : G -• H be a g r o u p h o m o m o r p h i s m , a n d l e t N = ker(0). T h e f o l l o w i n g t h e n h o l d .

(a) 9 ( 1 ) = 1. (b) 9 ( x - l ) = 9 ( x ) - x f o r x e G .

(c) N i s a n o r m a l s u b g r o u p o f G .

(d) F o r x , y G G, w e h a v e 9 ( x ) = 9 ( y ) if a n d o n l y if N x = Ny.

(e) 9 i s i n j e c t i v e if a n d o n l y if N = 1.

P r o o f . We have 0(1)0(1) = 0(1-1) = 0(1), and left mul t ip l ica t ion by 0(1)~1

i n H yields 0(1) = 1, proving (a). Also , if x G G , we have 1 = 0(1) = 0(xx" : ) = 0(x)0(x"1), and (b) follows by left mul t ip l ica t ion by 9 ( x ) ~ x .

Now N is nonempty since 1 G AT by (a). Also , if x, y G N , then 9 ( x y ) = 9 ( x ) 9 ( y ) = 1-1 = 1, and thus xy G N , and N is closed under mul t ip l ica t ion. Furthermore, if x G N , then 0(x" : ) = 9 ( x ) ~ l = l " 1 = 1, and thus x " 1 G N , and this proves that A is a subgroup. To complete the proof of (c), let n G N and g G G . T h e n 9(n°) = 9 ( g - l n g ) = 9 ( g ) - 1 1 9 ( g ) = 1, and hence n 9 G N . Th i s shows that N g C N for a l l g G G , and thus N < G, proving (c).

For (d), observe that 0(x)0(y)"1 = 0(xy" 1 ) . The left side is equal to 1

i f and only if 0(x) = 9 ( y ) , and the right side is 1 if and only if x y " 1 G N , or equivalents , x G Ny, and this, we know, is equivalent to N x = Ny. Th i s proves (d), and (e) is an immediate consequence. •

A p p e n d i x : The Basics 339

Thus kernels of homomorphisms denned on a group G are normal subgroups. Since we saw previously that every normal subgroup N < G is the kernel of the canonical homomorphism G —> G/N, it follows that the normal subgroups of G are exactly the kernels of the homomorphisms defined on G.

The following homomorphism theorem asserts that an arbi trary surjec-tive homorphism 6 is essentially the canonical homomorphism from G onto G/N, where N = ker(0).

X.18 . T h e o r e m (Homomorphism). L e t 6 : G - > H be a s u r j e c t i v e g r o u p h o m o m o r p h i s m , a n d l e t N = ker(0). Then G / N ^ H . I n f a c t , t h e r e i s a n i s o m o r p h i s m r : G / N - > H such t h a t T ( N X ) = 9 { x ) f o r a l l x e G .

Proof. Let I G G . B y L e m m a X . 17(d), a l l elements of the coset TVx have the same image under 6, and this image is 6 { x ) . We can thus unambiguously define r on G / N by setting r ( N x ) = 9 { x ) , and we must show that r is an isomorphism. F i rs t ,

r { N x N y ) = r ( N x y ) = 6 { x y ) = 9 { x ) 6 { y ) = T ( N x ) r ( N y ) ,

and thus r is a homomorphism. Al so , since r ( N x ) = 9 ( x ) , we see that the image of r is the image of 6, and since this is a l l of H , it follows that r is surjective. To prove that r is injective, it suffices to check that ker(r) is t r iv i a l , and so we suppose that Nx e ker( r ) . T h e n 1 = T { N X ) = 9 ( x ) , and thus x € ker(0) = N , and so Nx = N , which is the identity of G/N. U

Actua l ly , Theorem X.18 even tells us something about homomorphisms that are not surjective. To explain this, suppose that 6 : G H is an arbi t rary homomorphism. If A C G is a subgroup, it is easy to check that the image 6 { X ) is a subgroup of H , and in part icular , 0 { G ) is a subgroup of H . We can thus view 9 as a surjective homomorphism from G onto 6 { G ) , and so we can apply the homomorphism theorem to deduce that 6 ( G ) ^ G /ker(0).

A s an example of how the homomorphism theorem can be used, we prove the following so-called " N / C - T h e o r e m " . To state it , we define the centralizer in G of a subset A C G to be the set of elements of G that commute w i t h a l l elements of A . It is easy to check that this set, denoted C G ( A ) , is actual ly a subgroup of G .

X .19 . Corol lary. L e t H be a s u b g r o u p of a n a r b i t r a r y g r o u p G, a n d w r i t e N = N G ( U ) a n d C = C G { H ) . Then C < N a n d N / C i s i s o m o r p h i c t o a s u b g r o u p o f A u t ( H ) .

Proof, l i x € N , then H x = H , and we write r x : H - > H to denote the map h ^ h x . Since conjugation by x is an automorphism of G , i t is easy to see that TX is an automorphism of H . A l so , if x , y G N and h e H , then ( h ) T x T y = ( h x ) y = h x y = [ h ) T X Y , and thus r x y = T X T V , and it follows that

the map 0 : N -> A u t ( i f ) denned by 0 ( x ) = TX is a homomorphism. A n element x £ H lies in ker(0) if and only i f r x is the identity map on i f , or equivalents , h x = h for a l l h e H . B u t / i x = h precisely when x h = hx, and thus k e v ( 0 ) = C G ( H ) = C. T h e n N / C 6 ( N ) C A u t f i f ) , and the proof is complete. •

If i f is a cyclic subgroup of G , for example, then N G ( H ) / C G ( H ) is an abelian group since it is isomorphic to a subgroup of A u t ( i f ) , which is abelian by L e m m a X .14 .

Another applicat ion of the homomorphism theorem is the following.

X . 2 0 . C o r o l l a r y . L e t N a n d H be s u b g r o u p s of a g r o u p G, a n d suppose t h a t N < G. T h e n H H N < H a n d N H / N H / ( H n N ) .

P r o o f . Reca l l that N H is a group, and observe that the right cosets of A" i n W i f a l l have the form N n h = N h , where n G N and h G i f . Now define 6 : i f -> N H / N by setting 0 ( h ) = N h , and observe that 6 is surjective, and that it is the restriction to i f of the canonical homomorphism TT : G -»• G / i V . It follows that 0 is a homomorphism, and k e r ( 6 ) = H D k e r ( n ) = H H N . Thus H n N < H and i f / ( i f n J V ) ^ A ^ i f / A by the homomorphism theorem. •

We have already mentioned that if 0 : G - > H is an arbi t rary group homomorphism and X C G is a subgroup, then 0 ( A ) is a subgroup of i f . The following correspondence theorem shows that if 0 is surjective, then every subgroup of H is the image of some subgroup of G . In fact, much more is true.

X . 2 1 . T h e o r e m (Correspondence). L e t 0 : G -> H be a s u r j e c t i v e g r o u p h o m o m o r p h i s m , a n d l e t N = k e x ( 0 ) . T h e n t h e m a p X H-» 0 ( A ) i s a b i j e c t i o n f r o m t h e set X of s u b g r o u p s of G t h a t c o n t a i n N o n t o t h e set y of a l l s u b g r o u p s of H . F u r t h e r m o r e , if X G X a n d Y G y a n d Y = 0 ( A ) , t h e n t h e f o l l o w i n g h o l d .

(a) A = { x G G | 0 ( x ) G Y } .

(b) T h e m a p a d e f i n e d by X g H-> 0 ( X g ) i s a b i j e c t i o n f r o m t h e set of a l l r i g h t cosets of X i n G o n t o t h e set of a l l r i g h t cosets of Y i n H , a n d i n p a r t i c u l a r , \ G : A | = | i f : Y\.

( c ) X < G if a n d o n l y i f Y < i f , a n d i n t h i s case, t h e m a p a of (b) i s a n i s o m o r p h i s m G / X 9* H / Y .

P r o o f . To prove that the map A H-» 0 ( A ) is a bijection from X onto y , it suffices to construct an inverse map from y to X . G iven a subgroup Y C i f , wri te 0~l(Y) = { x G G | 0 ( x ) G Y } , so that 0~l(Y) is the inverse image of

Y i n G . (Caut ion: we are not asserting that there is a map called 0 " 1 . ) It is easy to check that 9 ~ 1 ( Y ) is a subgroup of G, and since it clearly contains ker(0) = N , we have 9 ~ 1 { Y ) G X .

We proceed now to show that the map Y ^ 0 _ 1 ( Y ) from y to X is the inverse of the map X ^ 9 { X ) from X to y. (Assert ion (a) w i l l also: follow from this.) If Y G y, then clearly 9 ( 0 - 1 ( Y ) ) C Y. To prove equality, let y G Y and observe that since 9 is surjective, there exists an elenent x G G such that 0(x) = y. T h e n x G 0 " 1 ( * r ) , and thus y = 9 ( x ) lies in 0 ( 0 - ! ( Y ) ) , and we have 0 ( 0 _ 1 ( * r ) ) = Y, as needed.

If A G A ' , then clearly A C 0 " X ( 0 ( A ) ) . To prove equality, we consider an element g G 0 " 1 ( 0 ( A ) ) , and we show that g G A . Certainly, 0(#) G 0 ( A ) , and thus there exists x G A such that 9 ( g ) = 0(x) . It follows by L e m m a X.17(d) that N g = N x , and thus g G N g = N x C A , where the final containment holds because J V C I b y the definition of X . Thus A = 0 " X ( 0 ( A ) , and this shows that the maps A ^ 0 ( A ) and Y i-> 0 - : ( Y ) are inverse bijections between # and y, as required.

N o w l e t X E X and F e ^ , where y = 0 ( A ) . If g G G, then 0(A#) = 9 ( X ) 9 ( g ) = Y 9 { g ) , and so a really is a map from the set { X g \ g e G } to the set { Y h | ft G # } . Also , a is surjective since if h G H , then ft = 0(g) for some element g G G, and thus a ( A y ) = 9 { X g ) = Yf t . To prove that a is injective, suppose that a ( A a ) = a ( A 6 ) for elements a, 6 G G . T h e n Y 0 ( a ) = Y0(6) , and so 0(6) = y 0 { o ) for some element y G Y. Since Y = 0 ( A ) , we have y = 0(x) for some element x G A , and thus 0(6) = 0(x)0(a) = 0(xa). It follows that A^6 = A^xa, and since N C. X , this coset of N is contained i n bo th X a and A 6 , which, therefore, are not disjoint. It follows that X a = X b , and thus a is injectve, and the proof of (b) is complete.

Observe that 9 ( X 9 ) = 0 ( A ) ^ ) = Y9^) for a l l elements g G G. N o w suppose that A < G. Let ft G H , and write ft = 9 { g ) , where g G G . T h e n

= y«(s) = 9 ( X 9 ) = 0 ( A ) = Y , and thus Y < H . Conversely, suppose that Y < H . Let g G G , and observe that N = N<> C X*>, so that A 9 G Also , 0 (A9) = Y 0 ( f ) = Y = 0 ( A ) , and thus A 9 = A since the map defined on X by applying 0 is injective. It follows that A < G , and this shows that A < G if and only if Y • H / Y is an isomorphism. We know from (b) that a is bo th injective and surjective, and so it suffices to show that it is a homomorphism. Th i s is clear, however, since i f a, b G G , then 9 { X a X b ) = 0 ( A a ) 0 ( A 6 ) . •

If A G X i n the notat ion of the correspondence theorem, then clearly N G ( A ) G X , and we argue that 0 ( N G ( A ) ) = N H { 9 { X ) ) . To see why this is true, observe that since the normalizer of a subgroup is the largest

subgroup i n which it is normal , it suffices to show that i f U, V G X w i th U Q V , then U < V if and only i f 9 ( U ) < 6 { V ) . T h i s follows by applying the correspondence theorem to the surjective homomorphism from V to 9 ( V ) obtained by restricting 9 to V.

Next , we apply the correspondence theorem to the canonical homomorphism 7T : G —> G/N, where N < G. T h e n ker(7r) = N , so X is the set of a l l subgroups of G that contain N . If X G X , then T T (X) = X / N , and thus the correspondence theorem tells us that every subgroup of G / N is uniquely of the form X / N , where X is a subgroup satisfying TV C X C G. A l so , A < G if and only i f X / N < G/N, and in that case, G / X ( G / N ) / ( X / N ) .

We complete our review by discussing direct products. Suppose that we are given (not necessarily distinct) groups d w i t h 1 < i < r . Let P be the set of ordered r-tuples ( x x , x 2 , . . . , x n ) , and define mul t ip l ica t ion in P componentwise, so that

( x i , x 2 , • • - , x r ) ( y i , y 2 , . . . , y r ) = ( x x y x , x 2 y 2 , . . . , x r y r ) •

It is t r i v i a l to check that P is a group; it is called the external direct product of the groups G{, and we write P = Gx x G2 x • • • x Gr.

If 1 < i < r , let N i C P be the set of r-tuples of the form

( 1 , . . . , 1 , ^ , 1 , . . . , 1 ) ,

where the component i n posit ion i is an arbi trary element X i G G t and a l l other components are the identities of the respective groups. It is easy to see that N i is a subgroup of P , and that N i = G i . A l so , it should be clear that every element x G P is uniquely of the form x = n x n 2 - - - n r w i t h r n € N i t

and i n particular, we can write P = N X N 2 - • • N r .

The elements of N i and N j commute i i i ^ j , and so if 1 < k < r , we have N i C C P ( N k ) C N P ( A ^ f e ) for i ^ k. Since A^fe is also contained in Np(AT f e ) , which is a subgroup, we see that P = N X N 2 • • • N r C N P ( N k ) , and thus N k < P .

Given a group G and subgroups < G for 1 < i < r , we say that G is the internal direct product of the normal subgroups M» provided that every element g G G is uniquely of the form 5 = mxm2 • • • m r w i th m , G M , . Thus if P is the external direct product of the abstract groups G», then P is also the internal direct product of the normal subgroups N i that we defined above, w i t h N i ^ d .

Note that "external direct product" is a c o n s t r u c t i o n since given an arbi t rary finite list of groups, one can define their external direct product. B u t "internal direct product" is really a statement about a finite collection of normal subgroups of a given group; it may be true, but it is not automatical ly so. Despite these differences, and at the risk of confusion, we often use the

same notation, and we write G = M x x M 2 x • • • x M r if G is the internal direct product of its subgroups M x .

G i v e n a group G and subgroups M , < G for 1 < i < r , suppose that G = M \ M 2 • • • M r . The following result shows what addi t ional information is needed to prove that G = M x x M 2 x • • • x M r .

X . 2 2 . T h e o r e m . L e t G = M X M 2 • • • M r , w h e r e G i s a g r o u p a n d M % i s a n o r m a l s u b g r o u p f o r l < i < r . T h e n G = M x x M 2 x • • • x M r if a n d o n l y if

( M x M 2 - - - M k - X ) n M k = 1

f o r a l l s u b s c r i p t s k w i t h 1 < k < r .

P r o o f . Suppose first that G = M x x M 2 x • • • x M r . Let

x G ( M X M 2 • • • M k - X ) D M k

for some subscript k, and write x = n x n 2 • - - U k - i , w i t h n , G M t . We can then write x = mxm2 • • • m r , where r r n = n { for 1 k, and we can get a second decomposit ion of the element x by taking mt = 1 for a l l subscripts i different from k and mk = x . In both situations, mt G M , for 1 < i < r , and thus since we are working i n a direct product, the factors mt are uniquely determined. In the first case, we had mk = 1, and i n the second, mk = x , and thus x = 1.

Conversely, assume that { M X M 2 • • • M k _ x ) D M k = 1 for 1 < k < r , and suppose that mxm2 • • • m r = n x n 2 • • • n r , where m „ n { G for 1 1. We have

( n x n 2 - - - n k - 1 ) - 1 { m x m 2 - - - m k - x ) = { n k ) { m k ) - x ,

and this element lies in M X M 2 • • • M k _ x n M k since M X M 2 • • • M k _ x is a group. It follows that { n k ) { m k ) - 1 = 1, which is a contradict ion. •

In particular, if G = M x x M 2 x • • • x M r is an internal direct product, then M j D M f c = 1 i f j ^ fc. If r > 2, however, this necessary condi t ion is definitely not sufficient in the si tuat ion of Theorem X . 2 2 .

We need an easy general fact: if M and N are normal subgroups of a group G , and M n N = 1, then all elements of M commute wi th a l l elements of N . To see this, let m G M and n G N , and consider the c o m m u t a t o r [ m , n ] of m and n , which, by definition, is the element m ^ n ^ m n . T h e n [ m , n ] = m - l m n , which lies in M because M < G . Also , [m,n] = (n" 1 )""/! , which lies in A^ because N < G. Thus [m, n] = 1, and this yields m n = n m .

If G = M i x M 2 x • • • x M r is an internal direct product, we conclude that the elements of M 3 commute w i t h the elements of M k if j ^ k, and

it follows easily that G is the direct product of the subgroups M i in every possible ordering of these subgroups. G iven a finite collection X of normal subgroups of a group G , therefore, it is meaningful to say that G is the direct product of the members of X wi thout specifying an ordering.

We have already seen that the external direct product P of groups G{

is the internal direct product of certain subgroups N i < P , where N t ^ G{. Another connection between external and internal direct products is the following.

X . 2 3 . L e m m a . L e t G be a g r o u p , a n d suppose t h a t G i s t h e i n t e r n a l d i r e c t p r o d u c t of t h e n o r m a l s u b g r o u p s M t f o r i < i < r . T h e n G i s i s o m o r p h i c t o t h e e x t e r n a l d i r e c t p r o d u c t of t h e g r o u p s M i .

P r o o f . Let P be the external direct product of the M u and construct a map 9 from P to G by setting

9 ( ( m i , m 2 , • • • , m r ) ) = m\m2 • • • mr .

T h e n 9 is a bijection since every element of G is uniquely of the form m i m 2 - - - m r w i t h ra,eM, To show that 9 is an isomorphism, we need

( x i x 2 • • • x r ) ( y i y 2 • • • y r ) = x x y i x 2 y 2 • • • x r y r

for x i , y l e M l , and this is clear since X i and V j commute when i > j . U

For an example of how this can be used, suppose that G is an internal direct product of normal subgroups, each of which has a t r i v i a l center. To prove that Z ( G ) = 1, it suffices to prove that Z ( P ) = 1, where P is the isomorphic group constructed as the external direct product of the factors of G . It is clear, however, that an r- tuple ( m x , m 2 , . . . , m r ) is central in P i f and only i f each component is central in its respective group, and it follows that Z ( P ) = l .

Index

abelian Hall subgroup and 7r-length, 94 abelian normal subgroups all cyclic, 189 abelian Sylow subgroup intersection, 38 action, 1 action on cosets, 3, 148 action on subspaces of vector space, 228j¡f action on right transversal, 148 action via automorphisms, 68 action, Frobenius, 177ff action, semiregular, 177 actions fixing prime order elements, Ulff additive commutator in ring, 125 adjoint (classical) of matrix, 321 admit (of action), 131 Alperin, J., 55, 323 alternating group, 34 alternating group, definition, 326 alternating group, Schur multiplier of, 152 alternating group, simplicity of, 250 arrows in transitive action, 262 augmentation ideal, 315 augmentation map, 314 automorphism, definition, 334 automorphism of symmetric group, 33, 250 automorphism tower, 278 automorphism, order of, 70 axiom of choice, 333

Baer theorem, 55 Baer trick, 142 bar convention, 23 Bartels theorem, 290 base group of wreath product, 73 basis for free abelian group, 315 Bender, H., 31, 217, 271 Berkovich, Y., 86

binomial coefficient, 9 block (of action), 237 Bochert theorem, 246^ Brodkey theorem, 38, 71, 74 Burnside, book of, 181, 217 Burnside orbit count, 7 Burnside pV-theorem, 216# Burnside p-nilpotence theorem, 159 Burnside theorem, prime degree action, 240 Burnside, W., 30

Cameron, P., 249, 285 canonical homomorphism, 338 Carter subgroup, 91 Cauchy theorem, 8, 10 Cauchy-Frobenius orbit count, 7 center of group, 2, 334 central extension, 151 central prime order elements, 174 central series, 20, 115 central subgroup, transfer to, 154 centralizer, definition, 339 Cermak-Delgado subgroup, 43 chain stabilizer, 137 character theory, 53, 59, 147, 182, 216-217 character, permutation, 236 characteristic series, 7r-separable, 92 characteristic subgroup, 11 characteristic subgroup, definition, 335 characteristically simple group, 277 Chermak-Delgado theorem, 41 class 2 group, 122, 142 class (conjugacy), 4 class equation, 5 class (nilpotence), 22, 116, 119 class size, 6

345

346 I n d e x

class sizes, common divisor graph, 268 class sizes, number of, 128 class sizes, smallest, 128 classical adjoint of matrix, 321 collection of commutators, 120ff common divisor graph on class sizes, 268 common-divisor graph, components, 265ff commutator, 40, 48, 113jff commutator, additive, 125 commutator collection, I20ff commutator, multiple, 115 commutator subgroup, 80 see also derived

subgroup commutators in coprime action, 138 complement, 85 complement for normal subgroup, 65 complete group, 279 component of group, 273ff, 287 see also

layer components of common-divisor graph,

265# composition factor, 29 composition series, 29 composition series of solvable group, 84 conjugacy class, 4 see also class conjugacy of complements, 82 conjugacy of Sylow subgroups, 14 conjugates of subgroup, 6 conjugation action, 2 conjugation action on normal subgroup,

230 connected graph, 260 control of fusion, 158, 167 control of transfer, 167, 295# coprime action, 96# coprime action on abelian group, 140 coprime action, orbit sizes of, 102ff coprime action, commutators in, 138 coprime orbit sizes, 102-105 core of cyclic subgroup, 63 core, 3 corefree subgroup, 285 correspondence theorem, 340 coset, 331 coset action, 3 coset, double, 6, 304 crossed homomorphism, 76, 114 cyclic extension, 107 cyclic factor group, 118 cyclic group automorphism, 334 cyclic group, subgroups of, 330 cyclic subgroup, core of, 63 cyclic Sylow subgroups, 159#

Dedekind lemma, 328 degree, 225 depth, subnormal, 45

derived length, 82 derived length and nilpotence class, 128 derived length, large, 146 derived series, 80 derived subgroup, 80, 113 derived subgroup, finiteness, 155 derived subgroup, noncommutators in, 125 development, 15 Dietzmann theorem, 156 dihedral group, 55ff, 120, 189, 196 dihedral group, construction, 70 direct diamond, 329 direct product, 69# direct product, center of, 25 direct product, definition, 342 Doerk, K., 86 dot action on right transversal, 148 double coset, 6, 304 double transitivity, 225

E, 274 see also layer elementary abelian group, 27, 82, 202, 298,

310 Euler totient function, ip, 334 even permutation, 34, 326 existence of Sylow subgroup, 8, 9, 12 exponent, 116 extension, 66 external direct product, 342 extraspecial p-group, 123

F, 25 see also Fitting subgroup F*, 271 see also generalized Fitting

subgroup factor group, 337 faithful action, 2, 40, 133, 233 faithful action on normal subgroup, 145 faithful orbit (existence of), 74 Feit-Thompson theorem, 30, 75, 97 Feit, W., 30 finite index center, Schur theorem, 155 Fischer-Greiss monster, 31 Fitting subgroup, 25, 46, 53, 271 Fitting subgroup and subnormality, 47 Fitting subgroup of solvable group, 86 Fitting theorem, coprime action, 140 fixed points come from fixed points, 101 fixed-point subgroup, 96 focal subgroup theorem, 165 Frattini argument, 15 Frattini subgroup, 27, 95, 282, 330 Frattini subgroup of p-group, 27, 117 free abelian group, 315 Frobenius action, 177 , 232 Frobenius complement, 26, 179 Frobenius complement, nonsolvable, 181 Frobenius complement, odd order, 185, 193

I n d e x 347

Frobenius complement, properties, 186 , 193

Frobenius group, 181 Frobenius kernel, 179 Frobenius kernel, nilpotence of, 196#, 201 Frobenius kernel, nonabelian, 180 Frobenius kernel, solvable, 196 Frobenius p-nilpotence theorem, 171jff, 197 Frobenius theorem on kernel existence, 181,

186 Frobenius theorem, permutations, 183 Frobenius, G., 216 fundamental counting principle, 5 Furtwangler, P., 313 fused classes, 100, 158 fusion and p-nilpotence, 170 fusion, control of, 158

general linear group, 30, 203#, 228 general linear group, order of, 204 general linear gronp, Sylow subgroup of,

205 generalized Brodkey theorem, 39 generalized dihedral group, 57, 70 generalized Fitting subgroup, 271 ff, 276,

281 generalized quaternion group, 74, 110, 120,

153, 189, 196, 209 generating set of subgroup, 329 glasses, 96 GL, 30 see also general linear group Glauberman Z( J)-theorem, 217 Glauberman lemma, 97 Glauberman, G., 31, 217 global property, 59 Goldschmidt, D., 31, 55, 217 good theorem, 2 graph and transpositions, 241 graph, common divisor, 265ff graphs and primitivity, 260 graphs and orbitals, 259 greed, 46 group, definition, 325 group of order 120, 18, 256 group of order 21952, 17 group of order 24, 33 group of order 8, 190 group of order p 2 q 2 , 37 group of order p 2 q , 32 group of order p 3 g , 33 group of order p a q , 37 group of order p q , 32 group of order p g r , 38 group ring, 313 groups of small order, 38

half transitivity, 232jf

Hall C-theorem, 87 Hall D-theorem, 90 Hall E-theorem, 86 Hall E-theorem converse, 87 Hall 7r-subgroup, 86 Hall w-subgroups, nonisomorphic, 90 Hall subgroup, 12 Hall subgroup in solvable group, 86 Hall subgroup in 7r-separable group, 93 Hall subgroup, conjugacy, 87 Hall subgroup, normal, 75 Hall-Higman Lemma 1.2.3, 93 Hall-Wielandt transfer theorem, 167, 297 Hall-Witt identity, 125 Hall, P., 86, 120, 134 Hartley-Turull theorem, 102 Hawkes, T., 86 Higman, G., 196 Higman-Sims group, 257 Hochschild, G., 313 homomorphism, definition, 337 homomorphism theorem, 339 homomorphism, canonical, 338 Horosevskii theorem, 70 Huppert theorem on metacyclic Sylows, 308

imprimitive action, 237 induced action on factor group, 132 infinite p-group, 8, 19 inner automorphism (definition), 335 internal direct product, 342 intersection of abelian subgroups, 61 intersection of subnormal subgroups, 47 intersection of Sylow subgroups, 37-39 invariant classes, 100 invariant cosets, 101 invariant Sylow subgroups, 96 involution, 55 involutions, nonconjugate, 57 Ishikawa, K., 128 isomorphism, 333 Iwasawa lemma, 252

J , 198 see also Thompson subgroup Jacobi identity, 125 Janko group, 164 join of subgroups, 42 join of subnormal subgroups, 47ff Jordan set, 243 Jordan theorem, 211ff Jordan-Holder theorem, 29, 84 Jordan, C., 227

Kegel, O., 294 kernel of homomorphism, definition, 338 kernel of action, 2 kernel of crossed homomorphism, 76

348 I n d e x

kernel of transfer, 165 Klein group, 34, 46, 56 Kuo, T., 323

Lagrange theorem, 331 Lagrange theorem, converse, 9, 24 lattice, 42, 47 layer, 274ff, 282, 287 left transversal, 77 Lie ring, 126 linear representation, 147 local subgroup, 58 local subgroup and homomorphism, 59-60 local-to-global theorem, 59 lower central series, 116, 127, 281 lower triangular matrix, 205 Lucchini theorem, 63 Lyons, ft, 55

Mackey transfer theorem, 304 magic eyeglasses, 96 Mann subgroup, 128 Manning theorem and 3-transitivity, 264 Maschke theorem, 309 Mathieu group, 31, 226#, 256# matrix unit, 253 Matsuyama, H., 31, 55, 217 maximal class p-group, 120 maximal subgroup, 25, 27 maximal subgroup, definition, 329 maximal subgroup of solvable group, 84, 91 maximal subgroup, nilpotent, 168, 209 McKay, J., 8 measure, Chermak-Delgado, 41ff metacyclic group, 160, 308^ minimal-normal subgroup, 48, 54, 82, 90,

275 minimal-simple group, 61 monster simple group, 31 multiple commutator, 115 multiple transitivity, 225, 264 multiple transitivity and primitivity, 240

N-group, 61 n! theorem, 4 Navarro, G., 164 Neumann, P., 7 nilpotence and subnormality, 47 nilpotence class, 22, 116, 119, 296 nilpotence class and derived length, 128 nilpotence class of Mann subgroup, 129 nilpotent factor group, 27 nilpotent group, 20, 24 nilpotent injector, 91 nilpotent joins, Baer theorem, 55 nilpotent maximal subgroup, 168, 209 nilpotent normal subgroup, 26

nonabelian simple group, 29 normal p-complement, 159, 164 normal abelian subgroups cyclic, 189 normal closure, 51 normal Hall subgroup and splitting, 75 normal series, 20, 80 normal subgroup, definition, 336 normal subgroup of primitive group, 238 normal-J theorem, 210 normal-P theorem, 207 normalizer, 3, 336 normalizer of subnormal subgroup, 48 normalizers grow, 22

O'Brien, E., 19 odd permutation, 34, 326 odd-order theorem, 30 see also

Feit-Thompson theorem f i r of p-group, 120 Hi of class 2 p-group, 122 orbit, 4, 223 orbit size, 5 orbits, number of, 7 orbit sizes in coprime actions, 102 orbital, 257 see also suborbit orbits of subgroup, 236 order of automorphism, 70

P x Q-theorem, 139 P x Q-theorem, strong form, 144 p V theorem, 30, 217^ p-central element, 220 p-complement, 85, 88 p-complement, normal, 159 p-cycle in primitive group, 245 p-group, 8 p-group of class 2, 122 p-group of maximal class, 120 p-group, center of, 20 p-group, elementary abelian, 27 see also

elementary abelian p-group, infinite, 8 p-group, nilpotence of, 21 p-group, omega subgroup of, 120 p-group, subgroups of, 24 p-group with unique minimal subgroup, 189 p-groups, number of, 19 p-local subgroup, 58 p-local subgroup of p-solvable group, 140 p-local subgroups and p-nilpotence, 171 p-nilpotence and fusion, 170 p-nilpotent group, 159 see also normal

p-complement p-solvable group, 93, 95, 107, 140, 146, 207,

210 paired orbital, 257, 262 partitioned group, 186, 195, 232

I n d e x 349

Passman, D., 232, 240 path connected graph, 260 perfect group, 151 permutable subgroups, 6, 49 permutation character, 7, 236 permutation group, 223ff permutation isomorphism, 5, 103, 224 permutation representation, 2, 223 Pettet, M., 278

27 see also Frattini subgroup PGL, 205 7r-group, 12 7r-length, 94 7r-separable group, 91 7r-solvable group, 93 point stabilizer, 3 point stabilizer, primitive group, 239 pointwise stabilizer of set, 242 Praeger, C , 106, 249, 265, 285 pretransfer map, 149 prime degree action, Burnside theorem, 240 prime index subgroup, 6, 18, 85 prime order elements central, 174 prime subdegree, 269 primitive action, 237# primitive group containing p-cycle, 241,

242, 245 primitive group, simplicity criterion, 252 primitive solvable group, 248 primitive subgroup of symmetric group,

246# primitivity and multiple transitivity, 240 principal ideal theorem (of transfer), 313 pro-p-group, 18 product of subgroups, 6, 49, 327, 336 projective general linear group, 205 projective special linear group, 30 see also

PSL PSL, 30, 205 PSL, simplicity, 251ff

quasinormal subgroup, 50 quasiquaternion, 123 quasisimple group, 272 quaternion group, 74 see also generalized

quaternion group quotient group, 337

rank of abelian p-group, 203 regular p-group, 297, 307 regular action, 2, 225 regular orbit, existence, 71 regular subgroup of permutation group, 235 regular wreath product, 73 repeated commutator, 132^ representation, 147 representation group (Schur), 151, 272

representation, permutation, 2 right transversal, 77 right transversal, dot action on, 148

Saxl, J., 249, 285 Schenkman theorem, 283 Schenkman, E., 278 Schur multiplier, 123, 151, 153, 272 Schur representation group, 151, 272 Schur theorem, finite index center, 155 Schur-Zassenhaus theorem, 75# second center, 20 Seitz, G., 249, 285 self-centralizing subgroup, 129 self-normalizing Sylow subgroup, 164 semidihedral group, 74, 120, 189, 196 semidihedral group, Schur multiplier, 153 semidirect product, applications of, 70 semidirect product, existence of, 72 semiregular action, 177, 225 semisimple group, 274 series of subgroups, 20 setwise stabilizer of set, 242 sharp multiple transitivity, 226 sharp transitivity, 225 simple group, nonabelian, 29 simple group classification, 30 simple group, order of, 163 simple group, Sylow 2-subgroup of, 162 simple groups of small order, 35 Sims conjecture, 249, 285 SL, 30, 110 SL, order of, 204 socle, 48, 54 solvable group, 30, 80# solvable group, Fitting subbgroup, 86 solvable group, maximal subgroup, 84 solvable minimal normal subgroup, 82 solvable primitive group, 248 special linear group, 30 see also SL split extension, 65^ sporadic simple group, 30 stabilizer of chain, 137 stabilizer of point, 3 strongly conjugate subgroups, 54, 289ff strongly Jordan set, 243 subdegree, 259 subdegrees in imprimitive group, 265 subdegrees in primitive group, 26Iff subnormal 7r-subgroup, 53 subnormal closure, 289ff subnormal core, 294 subnormal depth, 45 subnormal nilpotent subgroup, 47 subnormal partition, 195 subnormal series, 7>separable, 92 subnormal subgroup, 45

350 I n d e x

subnormality, 45ff, 271ff suborbit, 257ff supersolvable group, 85 supersolvable group, Mann subgroup of,

131 Suzuki, M., 55 Suzuki group, 163 Sylow p-subgroup of symmetric group, 299 Sylow C-theorem, 14 Sylow counting theorem, 16 Sylow D-theorem, 15 Sylow E-theorem, 9 Sylow subgroup, 8 Sylow subgroup of Frobenius complement,

188, 193 Sylow subgroup of general linear group, 205 Sylow subgroup, cyclic, 159ff Sylow subgroups, intersection, 10, 16, 38 Sylow subgroups, number of, 15-16 Sylow subgroups, set of, 10, 14 Sylow system, 90 Sylow theorem, via Cauchy, 8, 12 symmetric group, 4, 325 symmetric group automorphism, 250 symmetric group, normal subgroup of, 251 symmetric group, primitive subgroup of,

246

Verlagerung, 149

weakly closed subgroup, 163 weight of commutator, 127 Weiss theorem on subdegrees, 262 Wielandt automorphism tower theorem,

278 Wielandt solvability theorem, 88 Wielandt subgroup, 54 Wielandt zipper lemma, 50ff wreath product, 73, 124, 296

Yoshida theorem, 296 Yoshida, T., 167

Zassenhaus, H., 181 Zenkov theorem, 61 zipper lemma, 50

Tate's theorem, 168 Taussky-Todd, O., 196 the, 11, 334 Thompson P x Q-theorem, 139 Thompson p-nilpotence theorem, 197, 201,

213 Thompson subgroup, 198, 20Iff Thompson theorem on Sims conjecture,

285ff Thompson, J., 30, 61, 169, 217 Thompson's thesis, 179, 196, 201, 214 three subgroups lemma, 126 totient function, 334 transfer evaluation lemma, 153ff transfer kernel, 165 transfer map, 149 transfer theory, 147ff, 295ff transfer to central subgroup, 154 transfer to derived subgroup, 313 transitive action, 7, 223 transitivity of transfer, 301 translate of subset in action, 237 transposition, 240, 326 transversal, 77 see also right transversal trivial block, 237

upper central series, 20 upper triangular matrix, 204

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F o r a comple t e l is t of t i t les i n th i s series, v i s i t the A M S B o o k s t o r e at www.ams.org/bookstore/.

http://www.ams.org/bookstore/

The text begins with a review of group actions and Sylow theory. It includes semidlrect products, the Schur-Zassenhaus theorem, the theory of commutators, coprime actions on groups, transfer theory. Frobenius groups, primitive and multiply transitive permutation groups, the simplicity of the PSL groups, the generalized Fitting subgroup and also Thompsons J-subgroup and his normal p-complement theorem.

Topics that seldom (or never) appear in books are also covered These include subnormality theory, a group-theoretic proof of Burnside's theorem about groups with order divisible by just two primes, the Wielandt automorphism tower theorem. Yoshida's transfer theorem, the "principal ideal theorem" of transfer theory and many smaller results that are not very well known.

Proofs often contain original ideas, and they are given in complete detail. In many cases they are simpler than can be found elsewhere.The book is largely based on the author's lectures, and consequently, the style is friendly and somewhat informal. Finally, the book includes a large collection of problems at disparate levels of difficulty.These should enable students to practice group theory and not just read about ft

Martin Isaacs is professor of mathematics at the University of Wisconsin. Madison. Over the years, he has received many teaching awards and is well known for his inspiring teaching and lecturing. He received the University of Wisconsin Distinguished Teaching Award in 1985. the Benjamin Smith Reynolds Teaching Award in 1989, and the Wisconsin Section MAA Teaching Award in 1993. to name only a few. He was also honored by being the selected MAA Pólya Lecturer in 2003-2005.

ISBN 9/8-0-8218 *3**-*

9II780821II8*)344<.

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Finite Group Theory-Martin Isaac

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