group_theory Herstein Sol.pdf

7/22/2019 group_theory Herstein Sol.pdf

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Solutions to

TOPICS IN ALGEBRA

I.N. HERSTEIN

Part II: Group Theory

Department of Mathematics

Indian Institute of Science, Bangalore


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No rights reserved.

Any part of this work can be reproduced or transmitted in any form or by anymeans.

Version: 1.1Release: Jan 2013Author: Rakesh Balhara


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Preface

These solutions are meant to facilitate the deeper understanding of the book,Topics in Algebra, 2nd edition. We have tried to stick with the notations devel-oped in the book as far as possible. But some notations are extremely ambigu-ous, so to avoid confusion, we resorted to alternate commonly used notations.

The following notation changes will be found in the text:

1. a mappingToperating on an elementxis represented throughT(x) ratherthanxT.

2. subgroup generated bya is represented through a rather than (a)

Any suggestions or errors are invited and can be mailed to: [email protected]

3


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Problems (Page 35)

1. In the following determine whether the systems described are groups. If theyare not, point out which of the group axioms fail to hold.(a)G = set of all integers, a b a b.(b)G = set of all positive integers, a b= ab, usual product of integers.(c)G = a0, a1, , a6 where

ai aj =ai+j ifi +j 7).(d) G = set of all rational numbers with odd denominators, a b a +b, theusual addition of rational numbers.Solution:(a) Clearly the binary operation is well-defined. Also a b G a, b G. Sothe closure property also holds good. Now a (b c) = a (b c) = a (b c) =a b + c. On the other hand (a b) c= (a b) c= (a b) c= a b c. Soa (b c)= (a b) c. So the associativity does not hold good. Hence G is not agroup.

(b) Clearly the binary operation is well-defined. Also we can check the binaryoperation satisfies closure and associativity too. Again 1 is the required identity

as 1 a = a 1 = a for all positive integers a. But we can see inverse of anyelement except for 1 does not exist. So the existence of inverses fails. Hence Gis not a group.

(c) We can easily check G is a group with a0 as identity element and a7i asinverse element ofai.

(d) Again we can easily check G is a group with 0 as identity and mn

as inverseelement of m

n.

2. Prove that ifG is an abelian group, then for all a, b G and all integers n,(a b)n =an bn.Solution: We resort to induction to prove that the result holds for positiveintegers. For n= 1, we have (a b)1 =a b= a1 b1. So the result is valid forthe base case. Suppose result holds forn = k 1, i.e. (a b)k1 =ak1 bk1.


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We need to show result also holds good for n = k. We have

(a b)k = (a b)k1 (a b)

= (ak1 bk1) (a b)

= (ak1 bk1) (b a)

= (ak1 bk) a

=a (ak1 bk)

=ak bk

So the result holds for n = k too. Therefore, result holds for all n N. Nextsupposen Z. Ifn= 0, then (ab)0 =ewhereethe identity element. Therefore(a b)0 =e = e e= a0 b0. So the result is valid for n = 0 too. Next suppose nis a negative integer. So n = m, wherem is some positive integer. We have

(a b)n = (a b)m

= ((a b)1)m by definition of the notation

= (b1 a1)m

= ((a1) (b1))m

= (a1)m (b1)m as the result is valid for positive integers

= (am) (bm)

=an bn

So the result is valid for negative integers too. Hence the result that (a b)n =an bn holds in an abelian group for all n Z.

3. IfG is a group such that (a b)2 =a2 b2 for all a, b G, show thatG mustbe abelian.Solution: We have for all a, b G

(a b)2 =a2 b2

(a b) (a b) = a2 b2

a ((b a) b) = a ((a b) b)

(b a) b= (a b) bas a1 exists in G

b a= a b as b1 exists in G

Butb a= a b a, b G impliesG is an abelian group. Hence the result.

4. IfG is a group in which (a b)i = ai bi for three consecutive integers i forall a, b G, show that G is abelian.


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Solution: Letn, n + 1,n + 2 be some three consecutive integers. Therefore wehave

(a b)n =an bn (1)

(a b)n+1 =an+1 bn+1 (2)

(a b)n+2 =an+2 bn+2 (3)

Using (2) we have

(a b)n+1 =an+1 bn+1

(a b)n (a b) = an+1 (bn b)

(an bn) (a b) = (an+1 bn) b, Using (1)

((an bn) a) b= (an+1 bn) b

(a

n

b

n

) a= (a

n

a) b

n

an (bn a) = an (a bn)

bn a= a bn (4)

Again using (3), analogously we have

bn+1 a= a bn+1

b (bn a) = a bn+1

b (a bn) = a bn+1, Using (4)

(b a) bn = (a b) bn

b a= a b

So we have a b= b a a, b G. And hence G is abelian.

5. Show that the conclusion of the Problem 4 does not follow if we assume therelation (a b)i =ai bi for just two consecutive integers.Solution: Suppose (a b)i = ai bi for i = n and i = n + 1. We claim Gis abelian if and only if (a b)n+2 = an+2 bn+2. Clearly, from last Problemwe have (a b)n+2 = an+2 bn+2 G is abelian. Also if G is abelian, then(a b)i = ai bi i Z; in particular result holds for i = n+ 2. ThusG isabelian if and only if (a b)n+2 =an+2 bn+2. So the result of Problem 4 mightnot follow if we assume (a b)i =ai bi for just two consecutive integers.

6. InS3 give an example of two elements x, y such that (x y)2 =x2 y2.Solution: We assume, as described in Example 2.2.3,S3 = {e,,

2, , , } with 3 =e, 2 = e and i = i. We choosex= and y =.We have

( )2 = ( ) ( ) = ( ) = 1 =e


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Whereas

2 2 =2 e= 2

Thus ( )2 =2 2.

7. InS3 show that there are four elements satisfying x2 =e and three elements

satisfyingy 3 =e.Solution: Again, as in Problem 6, we assume S3 = {e,,

2, , , }with3 =e,2 =e. We have (e)2 =e; ()2 =2; (2)2 =; ()2 =e; ( )2 =e;( )2 = e. Thuse,, , are the elements with their square equal toidentity. Also we have (e)3 =e; ()3 =e; (2)3 =e; ()3 =; ( )3 = ;( )3 = . Thus we havee, ,2 with their square equal to identity. Hencethe result.

8. IfG is a finite group, show that there exists a positive integer N such thataN =e for all a G.Solution: Since G is finite, we assume G = {g1, g1, , gm} for some positiveinteger m. For somegi G, consider the sequence gi, g2i , g

3i , . Since G is

finite and closed under binary operation, so there must be repetition in the se-quence, i.e. gji = g

ki for some positive integers j and k with j > k. But that

meansgjki =e, wheree is the identity element. Letj k= ni. Thus we havenicorresponding to everygi such thatg

nii =e. LetN=n1 n2 nm. But

thengNi =e for all i, showing the existence of required positive integer N.

9. (a) If the group G has three elements, show it must be abelian.(b) Do part (a) ifG has four elements.(c) Do part (a) ifG has five elements.Solution:(a) LetG be a group of order 3 and some a, b G with a =b.Case 1 , Either of a or b equals to the identity element: Supposea = e thena b= e b= b = b e= b a. Similarly, ifb = e, we have a b= b a. Thus ifeithera or b equals toe, we have a b= b a.Case 2, Neither ofa or b is identity element: Consider a b. We have a b=a,otherwise it would mean b = e. Similarlya b = b as a = e. AlsoG has onlythree elements, so a b has not option but to be equal to the identity element.Thereforea b= e. A similar argument will show thatb a= e. Thusa b= b ain this case too.

So we have a b= b a a, b G. Hence G is abelian for o(G) = 3.

(b) Again letG be the group of order 4 and let some a, b G.Case 1 , Either ofa, b equals to e: In this case, clearly a b= b a.Case 2, Neither ofa, bequals toe. Considera b. Clearly,a b=a and a b=b.But sinceG has four elements, letc =e be the fourth element. So a bhas two


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options, either equals toe or equals to c.

Ifa b= e, thena = b1

b a= b b1

b a= e. Thusa b= b a= e.

Ifa b= c, then consider b a. Clearly, b a=a and b a=b. So b a hasonly two options, eitherb a= e or b a= c. But ifb a= e, then it wouldimplya b= e, which is not true. So b a= c too. Thus a b= b a= c.

Thus we have a b= b afor all a, b G. HenceG is abelian for o(G) = 4.

(c) For this part, we have to make use of the material not presented till nowin the book. Since the order ofG is prime, therefore it is cyclic. But a cyclicgroup is abelian, so G must be abelian for order 5.

10. Show that if every element of the groupG has its own inverse, then G isabelian.Solution: Let some a, b G. So we have a1 =a and b1 =b. Also a b G,thereforea b= (a b)1 =b1 a1 =b a. So we have a b= b a, showingGis abelian.

11. IfG is a group of even order, prove it has an element a =e satisfyinga2 =e.Solution: We prove the result by contradiction. Note that G is a finite group.Suppose there is no element x satisfying x2 =e except for x = e. Thus if someg = e belongs to G, then g2 = e, i.e. g = g1. It means every non-identityelement g has another element g1 associated with it. So the non-identity ele-ments can be paired into mutually disjoint subsets of order 2. We can assume

the count of these subsets equals to some positive integer n as G is a finitegroup. But then counting the number of elements ofG, we haveo(G) = 2n + 1,where 1 is added for the identity element. So G is a group of odd order, whichis not true. Hence there must exist an element a =e such thata2 =e for G isa group of even order.

12. LetG be a nonempty set closed under the an associative product, which inaddition satisfies:(a) There exists an e G such that a e= a for all a G.(b) Givea G, there exists an elementy(a) G such that a y(a) = e.Prove thatG must be a group under this product.Solution: In order to show G is a group, we need to show

1. e a= a a G, and

2. Ifa y(a) = e, theny (a) a= e.

Suppose some a G, therefore there exists y(a) G such that a y(a) = e.Again y(a) G implies there exists y(y(a)) G such that y(a) y(y(a)) = e.


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So we have

y(a) a= (y(a) a) e= (y(a) a) (y(a) y(y(a)))

= ((y(a) a) y(a)) y(y(a))

= (y(a) (a y(a)) y(y(a))

= (y(a) e) y(y(a))

=y(a) y(y(a))

=e

Thusy(a) a= e (1)

Now using (1), we have e a= (a y(a)) a= a (y(a) a) = a e= a. Thus

e a= a (2)

Form (1) and (2), we conclude G is a group.

13. Prove, by an example, that the conclusion of Problem 12 is false if we as-sume instead:(a) There exists an e G such that a e= a for all a G.(b) Given a G, there exists y(a) G such that y(a) a= e

Solution: Consider G =

a ab b

|a, b Q+

with usual matrix multiplica-

tion as binary operation , where Q+ denotes the positive rational numbers.

We define e = 1 10 0. Also forM = a ab b, we define y(M) = 1a+be. Onecan easily see that M e = M and y(M) M = e. AlsoG is not a group asinverses do not exist.

14. Suppose a finite setG is closed under an associative product and that bothcancellation laws hold inG. Prove that G must be a group.Solution: Since G is a finite group, so we can assume G = {g1, g2, . . . , gn}, forsome positive integern. Let somea G. ConsiderS= {a g1, a g2, , a gn}.We assert each element ofS is distinct as ifa gi = a gj with gi = gj , thenleft-cancellation implies gi = gj . Therefore,o(S) = o(G), combining with thefact that S G, we conclude S = G. So ifa G, therefore a S. But that

means a = a gk, for some gk G. We claim gk is the right-identity. Forestablishing our claim, we consider S = {g1 a, g2 a, , gn a}. Again sinceright-cancellation too holds good, proceeding in an analogous way, we can seeS = G. Let some x G. So x S, therefore x = gi a for some gi G.Now x gk = (gi a) gk = gi (a gk) = gi a = x. Thus we have shown,x gk = x x G. So gk is the right-identity. Now, sincegk G, therefore


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gk S. So gk =a gl, for some gl G. But that shows the existence of right-inversegl, for an arbitrarily chosen element a G. Thus right-inverse exists for

each element in G. With the existence of right-identity and right-inverses, weconcluded that G is group.

15. (a) Using the result of Problem 14, prove that the nonzero integers modulop, p a prime number, form a group under multiplication mod p.(b) Do part (a) for the nonzero integers relatively prime to n under multiplica-tion mod n.Solution:(a) Let G be the set consists of non-zero integers modulo p. We noticed thatG is a finite set with multiplication mod p as well-defined binary operation.Using Problem 14, G would be a group if we show that the multiplication mod

p is associative and the both cancellation laws hold good. One can easily seethat a (b c) = (a b) c = abcmod p. So associativity is not an issue.Next suppose a b = a c, we need to show that it would imply b = c. Buta b = a c ab = ac mod p a(b c) = 0 mod p p | a(b c). But pbeing prime, so either p | a or p | b c. Since p|a, so p | (b c). Also p|b and

p|c, so p | (b c) implies b c= 0, or b = c. Thus left-cancellation law holdsgood. Similarly we can see right-cancellation also holds good. Using previousproblem, we conclude G is a group.(b) LetG be the set consists of non-zero integers relatively prime to n. ClearlyG is a finite set. Also multiplication mod n is well-defined binary operationover G. To show G is a group under multiplication mod n, we need to showthat associativity and both cancellation laws hold good. It is easy to see thata (b c) = (a b) c= abc modn. So associativity holds good. Next suppose

ab= ac. But that means ab= ac mod n a(bc) = 0 mod n n | a(bc).But gcd(a, n) = 1, thereforen | a(bc) impliesn | (bc). Also gcd(b, n) = gcd(c,n) = 1, therefore n | (b c) implies b = c. Thus we see left-cancellation holdsgood. Similarly, we can check right-cancellation too holds good. And so G is agroup.

16. In Problem 14 show by an example that if one just assumed one of thecancellation laws, then the conclusion need not follow.Solution: We construct a set G with n 2 elements, equipped with a binaryoperation : G GG such thatx y= y. Clearly the binary operation iswell-defined. Next, we seex (y z) = y z =z . Also (x y) z = z . Thus thebinary operation is associative. Next we check left-cancellation. Suppose we

havex y= x z. Butx y= y and x z = z. Thusx y= x z y = z, showingleft-cancellation holds good. On the other hand, if we have x y = z y, wecannot conclude x= z as x y = z y = y x, z G. Thus right-cancellationdoes not hold good. Finally, we proveG is not a group. SupposeG is group,therefore it must have the identity element. Let e be the identity element. Butthen x e= e x G, showing all elements are identity elements. Thus we get


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G = {e}, but we have assumed G has n elements with n 2, hence a contra-diction. ThusG has no identity element. Hence G is not a group. So if a finite

set with well-defined binary operation is satisfying associativity and one sidedcancellation law, it need not to be a group under that binary operation.

17. Prove that in Problem 14 infinite examples exist, satisfying the conditions,which are not groups.Solution: We define An = {in | i Z+}, where n is a positive integer greaterthan 1. ClearlyAn with usual multiplication as binary operation satisfies allconditions of Problem 14, but is not a group as inverses do not exist for all ele-ments. Also since there are infinite positive integers greater than 1, so we haveinfinite examples satisfying the conditions of Problem 14 but are not groups.

18. For anyn > 2, construct a non-abelian group of order 2n. (Hint: imitatethe relation in S3)Solution: LetG be the group that we are going to construct. Let denotes itsbinary operation and let e be its identity element. Thus we have constructedan element ofG, which is e. Next we construct an elementa = e with ordern 3. Thus we have constructedn element, which are e,a,a2, , an1. Fi-nally we construct an element other than already constructed, b, with order 2,i.e. b2 = e. We interconnect a, b with the rule: b a b1 = a1. We claimwe have got 2n elements, i.e. G = {e,a,a2, , an1, b , b b, , b an1}. Toestablish our claim, we first notice that b ai b= ai. With this rule at hand,one can easily check any expression resulting from the product ofai andbj willbelongs to those 2n elements. To give readers more insight, suppose we have

some expressiona

i

b

j

a

k

. Now eitherj = 0 orj = 1. Ifj = 0, then expressionequals to ai+k, and thus belong to G. Whereas ifj = 1, then letx = ai b ak.We have b x= (b ai b) ak =ai ak =aki. Left-multiplying by b, we getx= b aki. Thus x G. Also a b= b a1; sincen = 2, thereforea b=b a.Thus we have gotG, a non-abelian group of order 2n.

19. IfSis a set closed under an associative operation, prove that no matter howyou bracket a1a2 an, retaining the order of the elements, you get the sameelement inS(e.g., (a1 a2) (a3 a4) = a1 (a2 (a3 a4)); use induction on n)Solution: Let Xn denote a expression we get by bracketing the n elements,keeping the order of elements same. We need to show all expressions Xn aresame for alln. We will use induction over n. Forn = 1 andn = 2, we have only

one expression possible, so trivially all expressions are same. For n = 3, we havetwo ways of bracketinga1a2a3, i.ea1 (a2 a3) and (a1 a2) a3. Since the binaryoperation is satisfying associativity, therefore both expression are same. Nextsuppose the result is true forn i1. We need to show that the result is equallytrue for n = i. Let Xi be some expression. Then it must be product of sometwo shorter expressions, i.e Xi = Y Zi, where < i. But all expressions


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with number of elements less than i are same. SoY = a1 Y

1, where Y

1

is some expression consists of 1 elements. Similarly, Zi = a+1 Z

i1.

Thus we have

Xi= Y Zi

= (a1 Y

1) (a+1 Z

i1)

=a1 X

i1,

where Xi1 is some expression of i 1 terms. But all expressions containing

i 1 terms are same (order of elements is assumed to remain same). But thenall expression having i elements turns out to be equal to a1 X

i1. Thus allexpressions havingi elements are same. The induction hypothesis implies resultis valid for all n. Hence the result.

20. Let G be the set of all real 2 2 matrices

a bc d

, where ad bc= 0 is a

rational number. Prove that the G forms a group under matrix multiplication.Solution: It is easy to check thatG is a group under matrix multiplication,

with

1 00 1

as identity element and 1(adbc)

d bc a

as inverse element of

a bc d

. Note that ad bc = 0 is given and of the inverses exist for all ele-

ments.

21. Let G be the set of all real 2 2 matricesa b

0 d

where ad = 0. Prove

thatG forms a group under matrix multiplication. Is G abelian?

Solution: We have G closed under multiplication as

a b0 d

a b

0 d

=

aa ab + bd

0 dd

G. Now sincead = 0, therefore G forms a group under ma-

trix multiplication with

1 00 1

as identity element and

1/a b/ad

0 1/d

as in-

verse element of

a b0 d

. Finally we have

a b0 d

a b

0 d

=

a b

0 d

a b0 d

implies ab +bd =ab+bd. Since ab +bd =ab+bd for all values ofa,b,d,

a

, b

, d

, soG is not an abelian.

22. Let G be the set of all real 2 2 matrices

a 00 a1

where a = 0. Prove

thatG is an abelian group under matrix multiplication.


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Solution: Easy to check G is a group under multiplication. Also we have

a 00 a1

a 00 a1

= a 00 a1

a 00 a1

= aa 00 a1a1

HenceG is abelian too.

23. Construct in theG of Problem 21 a subgroup of order 4.

Solution: Leta =

1 00 1

, b =

1 00 1

. Clearlya2 =Iandb2 =I, where

I =

1 00 1

. Also ab = ba =

1 00 1

. Thus H = {I,a,b,ab} forms a

subgroup ofG.

24. Let G be the set of all 2 2 matrices

a bc d

where a,b,c,d are integers

modulo 2, such that ad bc= 0. Using matrix multiplication as the operationin G, prove that G is a group of order 6.Solution: Its trivial to check that

G=

1 00 1

,

0 11 0

,

0 11 1

,

1 01 1

,

1 10 1

,

1 11 0

Henceo(G) = 6

25. (a) LetGbe the group of all 2 2 matrices a b

c d wheread bc= 0 anda,b,c,dare integers modulo 3, relative to matrix multiplication. Showo(G) = 48.(b) If we modify the example ofG in part (a) by insisting that ad bc= 1, thenwhat is o(G)?Solution: See Problem 26, which is general case of this problem.

26. (a) LetG be the group of all 2 2 matrices

a bc d

where a, b, c, d are the

integers modulo p, p being a prime number, such that adbc = 0. G formsgroup relative to matrix multiplication. What is o(G)?(b) LetHbe the subgroup of the G of part (a) defined by

H=a b

c d

G | ad bc= 1

.

What is o(H)?Solution:(a) We will first count the number of ways in which ad bc= 0. We separate


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this counting into two cases. In first case we count the ways whenad = bc= 0.In second case, we will count the number of ways in which ad = bc = 0.

Case 1 : ad= bc = 0: Whena = 0, we can choosed in p ways; and whend = 0,we can choose a in p ways. But in this a = d = 0 has been counted twice.Therefore there are 2p 1 ways in which ad = 0. Similarly, bc = 0 in 2p 1ways. Thus there are (2p 1)2 ways in which ad = bc= 0.Case 2: ad = bc = 0: Since ad = 0, therefore a = 0 and d = 0. Similarly,b = 0 and c = 0. We chose some value ofa = 0 in p 1 ways; some value ofd= 0 in p 1 ways; some value ofb in p 1 ways; then find the value ofc withthese chosen values ofa, d,b. Since ad= bc modp has unique solution in c fornon-zero values ofa, d,b; thus we get unique value ofc. Thus ad = bc = 0 canbe chosen in (p 1)3 ways.Finally, the number ways of choosing a,b,c,d with ad bc= 0 equals numberways of choosing a,b,c,d without any restriction minus the number of ways ofchoosinga, b, c, dwithad bc= 0. Thus the number of ways of choosing a, b, c, dwithad bc= 0 is p4 (p 1)3 (2p 1)2. On simplification, we get

o(G) = p4 p3 p2 +p

(b) We separate our counting ofa, b, c, dfor whichad bc= 1 in three separatecases.Case 1 : ad = 0: This restrictbc= 1. The number ways of choosing a, d forwhich ad = 0 is: 2p 1. On the other hand, the number of ways of choosingb, c for which bc= 1 is: p 1. Thus when ad = 0, we can choose a,b,c,d in(2p 1)(p 1)Case 2: bc= 0: Analogous to previous case, we get number of ways of choosinga,b,c,d as (2p 1)(p 1).

Case 3: ad = 0 and bc = 0: In this case we get number of ways of choosinga,b,c,d as (p 1)(p 1)Thus total number of ways of choosing a, b, c, d for whichad bc= 1 is: 2(2p 1)(p 1) +

p2

(p 1)2. On simplifying we get

o(G) = p3 p


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Problems (Page 46)

1. If H and K are subgroups of G, show that HK is a subgroup of G.(Can you see that the same proof shows that the intersection of any number ofsubgroups ofG, finite of infinite, is again a subgroup ofG?)Solution: Let some x, y H K. Thereforex H andy H. But H beinga subgroup, so xy1 H. Again x K andy K. K being subgroup impliesxy1 K. Butxy1 Handxy1 K impliesxy1 H K. Thus we havefor all x, y H K, xy1 H K. HenceH Kis a subgroup ofG.


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Problems (Page 70)

1. Are the following mappings automorphisms of their respective groups?(a)G group of integers under addition, T :x x.(b)G group of positive reals under multiplication, T :x x2.(c)G cyclic group of order 12, T :x x3.(d)G is the group S3, T :x x

1

Solution:(a) Yes, as G is an abelian group.

(b) No, as mapping is not a homomorphism.

(c) No, as ifG= a, then o(a) = 12, while o(T(a)) = 4, showing T is not anautomorphism. Note that an isomorphism preserve the order of elements.

(d) No, asG is a non-abelian group, so the mapping T :x x

1 fails to be ahomomorphism.

2. Let G be a group, H a subgroup of G, T an automorphism of G. LetT(H) = {T(h)| h H}. ProveT(H) is a subgroup ofG.Solution: We need to show T(H) is a subgroup of G. Let x, y T(H),therefore x = T(h1) and y = T(h2) for some h1, h2 H. But then xy1 =T(h1)(T(h2))

1 =T(h1)T(h12 ) =T(h1h

12 ) =T(h3) for some h3 H as H is

a subgroup ofG. Soxy1 T(H) x, y H. This showsT(H) is a subgroupofG.

3. Let G be a group, Tan automorphism of G, N a normal subgroup ofG.Prove thatT(N) is a normal subgroup ofG.Solution: We need to showT(N) is a normal subgroup ofG. Clearly by lastProblem,T(N) is a subgroup ofG. Let someg G, thereforeg = T(g) for someg G as T is an onto mapping. But thengT(N)g1 = T(g)T(N)T(g)1 =T(g)T(N)T(g1) =T(gN g1) =T(N) as N is normal in G. Thus we havegT(N)g1 =T(N) g G. HenceT(N) is normal in G.

4. ForG = S3 prove that G I(G).Solution: We have, from Lemma 2.8.2

G

Z I

(G) (1)

Also we claim forSn withn 3, we have Z= {I}, whereIis the identity map-ping. We assumeA = {x1, x2, , xn}is the set of elements on which mappingsfrom Sn operate. Suppose some T Sn with T =I, therefore there is some xisuch that T(xi)= xi. Let T(xi) = xj , where i= j . Since n 3, therefore we


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can have xk with k =i and k =j . Define T :A A such that T(xi) =xi;T(xj) = xk; T

(xk) = xj ; and T(xl) = xl for the rest ofxl A. But then

T T(xi) = T(T(xi)) = T(xi) = xj ; and TT(xi) = T(T(xi)) = T(xj) = xk,showing T T = TT. So if someT = I, then T T = TT for some T Sn.Thus T / Z T = I; implying Z = {I}. Finally, using (1), we haveSn I(Sn) n 3

5. For any groupG prove that I(G) is a normal subgroup ofA(G) (the groupA(G)/I(G) is called the group of outer automorphismsofG).Solution: We have some change of notations, so we prefer to give detailedsolution of this problem. We define Tg : G G such that Tg(x) = gxg1

(Note that this is different from what is given in the book). Clearly Tg A(G)for all g G. Next, we define I(G) = {Tg | g G}. Let some some Tg1 ,Tg2 I(G). But then for all x G, we have Tg1Tg2(x) = Tg1(Tg2(x)) =Tg1(g2xg

12 ) = g1(g2xg

12 )g

11 = (g1g2)x(g1g2)

1 = Tg1g2(x). Thus Tg1Tg2 =Tg1g2 . ButTg1g2 I(G), thereforeTg1Tg2 I(G). Again let someTg I(G),therefore for all x G, we have Tg(x) = gxg1. So we have, for all x G

Tg(g1xg) = g(g1xg)g1

Tg(g1xg) = x

T1g (Tg(g1xg)) = T1g (x)

g1xg= T1g (x)

T1g (x) = (g1)x(g1)1

T1g (x) = Tg1(x)

Thus T1g = Tg1 . But Tg1 I(G), therefore T1g I(G). Hence I(G) is

a subgroup ofA(G). Finally, suppose someT A(G) and some Tg I(G),then we have for all x G

T TgT1(x) = T(Tg(T

1(x)))

=T(g(T1(x))g1)

=T(g)T(T1(x))T(g1)

=T(g)x(T(g))1

=TT(g)(x)

Thus T TgT1

= TT(g). But TT(g) I(G). Thus T TgT1

I(G) T A(G). Hence I(G) is a normal subgroup in A(G).

6. Let G be a group of order 4, G = {e,a,b,ab}, a2 = b2 = e, ab = ba.Determine A(G).


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Solution: The possible proper subgroups ofGare {e, a}, {e, b}, {e,ab}. We aimat finding a mapping Twhich is an automorphism. Since T is homomorphism,

thereforeT(e) = e. Also once we have found T(a) andT(b), the value ofT(ab)get decided by itself as T(ab) = T(a)T(b). Now what could be the possiblevalues ofT(a) so that T is an automorphism. Sinceo(a) = 2, so o(T(a)) mustbe 2. The elements with order 2 area,b,ab, so T(a) has three choices namelya,b,ab. OnceT(a) is decided, what could be the possible values forT(b). Againorder of b is 2, so the order ofT(b) must be 2. Again we have three possiblecandidates, a,b,ab, out of which one has already been fixed to T(a). So T(b)has two choices. Thus we have 3 2 = 6 possible automorphisms. Thus

A(G) =

e a b abe a b ab

,

e a b abe a ab b

,

e a b abe b a ab

,

e a b abe b ab a

,

e a b abe ab a b

,

e a b abe ab b a

7. (a) A subgroupCofGis said to be acharacteristic subgroupofGifT(C) Cfor all automorphisms T ofG. Prove a characteristic subgroup ofG must be anormal subgroup ofG.(b) Prove that the converse of (a) is false.Solution:(a) We, for some g G, define Tg : G G such that Tg(x) = gxg1. It isroutine to checkTg is an automorphism ofG for all g G. But it is given thatT(C) C for all automorphisms T. So Tg(C) C g G. But that meansgCg1 C, orgcg1 C g G & c C. Thus Cis normal in G.

(b) We simply give an example to show that converse of part(a) need not betrue. For G = {e,a,b,ab} with a2 = e, b2 = e and ab = ba; and C = {e, a};

andT =

e a b abe b a ab

, we haveCis a normal subgroup ofG but T(C) C.

Note thatTdefined above is an automorphism ofG.

8. For any group G, prove that the commutator subgroupG is a characteristicsubgroup ofG. (See Problem 5, Section 2.7)Solution: Let U ={xyx1y1 |x, y G}. So G =U. Also if some u U,then u1 U too. So if somea G, then a =

i xiyix

1i y

1i , where is


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some finite index set. But then

T(a) = T(i

xiyix1i y1i )

=i

T(xiyix1i y

1i )

=i

T(xi)T(yi)T(x1i )T(y

1i )

=i

T(xi)T(yi)(T(xi))1(T(yi))

1

=i

xiy

ix1i y

1i

So T(a) G a G. ThusT(G) G. HenceG is a characteristic sub-group ofG.

9. IfG is a group, Na normal subgroup ofG, M a characteristic subgroup ofN, prove that Mis a normal subgroup ofG.Solution: Let some g G. We defineTg : G G such that Tg(x) =gxg

1.Clearly Tg is an automorphism of G. Also Tg(N) = N as N is given to benormal in G. Now consider Tg : N N. Since Tg(N) = N, one can eas-ily see Tg is an automorphism of N too. But then Tg(M) M as M isgiven to be a characteristic subgroup of N. So gM g1 M g G, orgmg1 M g G & m M. HenceM is normal inG.

10. Let G be a finite group, Tan automorphism ofG with the property thatT(x) = x for x G if and only if x = e. Prove that every g G can berepresented as g = x1T(x) for somex G.Solution: First note that G is given to be a finite group. We define mapping : G G such that (x) = x1T(x). Clearly mapping so defined is well-defined. Also

(a) = (b) a1T(a) = b1T(b)

T(a)(T(b))1 =ab1

T(ab1) = ab1

But T(x) =x implies x = e, so (a) =(b) implies ab1 =e, i.e. a= b. Thusmapping is one-to-one. But since G is finite, therefore being one-to-oneimplies is onto too. But onto implies that if some g G, then it has itspre-image in G, i.e. g = x1T(x) for some x G. Hence every element g ofGcan be represented as x1T(x) for somex G.


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11. Let G be a finite group, Tan automorphism ofG with the property thatT(x) = x if and only ifx = e. Suppose further thatT2 =I. Prove thatG mustbe abelian.Solution: Using previous problem, if some g G, then g = x1T(x) for somex G. So we haveT(g) =T(x1T(x)) =T(x1)T(T(x)) = (T(x))1T2(x) =(T(x))1x = (x1T(x))1 = g1. Thus T(g) = g1 g G. Now leta, b G. So we have

T(ab) = (ab)1 =b1a1 (1)

AlsoT(ab) = T(a)T(b) = a1b1 (2)

Using (1) and (2), we have

b1a1 =a1b1 ab = ba

So we have ab = ba a, b G. Hence G is an abelian group.

12. Let G be a finite group and suppose the automorphism Tsends more thanthree-quarters of the elements ofG onto their inverses. Prove that T(x) = x1

for all a G and that G is abelian.Solution: Define a setH = {x G | T(x) = x1}. So we haveo(H) > 34 n.Let some h H. Consider the set Hh. Clearly o(Hh) > 34 n. Thereforeo(H Hh) > n2 . Let somex H Hh. Thereforex = h1 and x = h2h, forsome h1, h2 H. Also sincex H Hh H, therefore T(x) = x

1. So

T(x) = x1 =h11 (1)

T(x) = x1 = (h2h)1 =h1h12 (2)

T(x) = T(h2h) = T(h2)T(h) = h12 h

1 (3)

Using (2) and (3), we get h1h21 =h21h1, orhh2 = h2h. Alsoh2 = xh1,

sohh2 = h2h hxh1 =xh1h hx = xh. Thus we have

hx= xh x H Hh (4)

Now consider N(h), i.e. normalizer subgroup ofh. The equation (4) impliesH Hh N(h). Soo(N(h)) > n2 . But N(h) being a subgroup ofG, there-fore o(N(h)) | o(G), forcing N(h) = G. But that means hg = gh g G.

So h Z h H, where Z is the center subgroup of G. So o(Z) > 3

4 n.Again Zbeing a subgroup ofG, so o(Z) | o(G), forcing Z = G. But Z = Gimplies G is abelian. Also if G is abelian, then if some x, y H, we haveT(xy1) = T(x)T(y1) = x1(y1)1 = x1y = (y1x)1 = (xy1)1, show-ing xy1 too belongs to H. Hence withG abelian, Hbecomes a subgroup ofG. But then o(H) | o(G), therefore H= G as o(H)> 34 n. But H=G implies


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T(x) = x1 x G, what we need to show.

13. In Problem 12, can you find an example of a finite group which is non-abelian and which has an automorphism which maps exactly three-quarters ofthe elements ofG onto their inverses?Solution: Consider the Quaternion group Q = {1, i, j, k}. Let T =

1 1 i i j j k k1 1 i i j j k k

. We left it to the reader to check T is

an automorphism of Q which transfer exactly 34 elements of Q into their in-verses.

14. Prove that every finite group having more than two elements has a non-

trivial automorphism.Solution: Let G be some group with order greater than 2. Now eitherG isnon-abelian, orG is abelian. IfG is non-abelian, then we haveZ=G, whereZis the center subgroup ofG. So there is some g G such that g /Z. We definefor some g /Z, mappingT :G G such that T(x) =gxg1. We claim T sodefined is an automorphism which is not equal to identity mapping. Tis an au-tomorphism ofG is easy to check. Also ifT=I, thenT(x) = x x G. Butthat means g xg1 =x gx = xg x G, i.e. g Zwhich is not the case.SoT =I. So for non-abelian groups, we have found a non-trivial automorphism.

When G is abelian, either x2 = e for all x G or there is some x0 G suchthatx20 = e. If there is some x0 G such that x

20 =e, then we claim that the

mapping T : G G such that T(x) = x1 is a non-trivial automorphism of

G. SinceG is abelian, T surely is an automorphism. Also ifT = I, then wehave x1 = x x G, i.e x2 = e x G, which is not the case. SoT is anon-trivial automorphism ofG.

On the other hand, ifGis abelian, withx2 =e x G, then for some positiveintegerm

G Z2 Z2 Z2 m times

Note that here we have assumed G to be finite. But then there existm inde-pendent symbols, a1, a2, , am with o(ai) = 2 i. By independent we meanai=

j aj for any index set . So

G= a1 a2 am

Also o(G) 3 guarantees us the existence of atleast two such independentsymbols, saya1anda2. We define a mappingT :G Gsuch thatT(a1) = a2,T(a2) = a1and for the rest of independent symbolsai,T(ai) = ai. Once defined


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for independent symbols, we extend it for all elements ofG using:

T(j

aj) = j

T(aj)

Clearly extending Tthis way for all elements ofG makes T a homomorphism.Also we can check T is onto. Thus we have found an automorphism ofG whichis not equal to an identity mapping. Hence every finite group with order greaterthan 2 has a non-trivial automorphism.

15. Let G be a group of order 2n. Suppose that half of the elements ofG areof order 2, and the other half form a subgroup Hof order n. Prove thatH isof odd order and is an abelian subgroup ofG.Solution: Let Kdenotes the set of all elements with order 2. ThereforeG =

H K. Also the index ofH in G is 2, so H is a normal subgroup ofG (Seeproblem 2, page 53). Let some k K, therefore k1 = k. Also H beingnormal in G implies kH k1 = H. Let some h H, therefore h = kh1k

1 forsome h1 H. So we haveh = kh1k

1 hk = kh1; but hk K (Why), sohk = (hk)1. Thus kh1 = hk = (hk)

1 =k1h1 =kh1 h1 = h1. Thus

we have h = kh1k h H. Now let some x, y H, therefore x = k x1k,andy = ky1k. So

xy= (kx1k)(ky1k) = kx1kky1k = kx1y1k (1)

Alsoxy= k(xy)1k= ky1x1k (2)

Using (1) and (2), we have

kx1y1k= ky1x1k x1y1 =y1x1

yx = xy

Soxy = yx x, y H. Hence His an abelian subgroup ofG Also since thereis no element with order 2, so order ofHis odd (See Problem 11, page 35).

16. Let (n) be the Euler -function. If a > 1 is an integer, prove thatn | (an 1).Solution: Let G be a cyclic group of order an 1. The existence of suchG is guaranteed (Why). ThereforeG = x for some x G. Define mapping

T :G Gsuch thatT(y) = ya

. Now since gcd(a, an

1) = 1, thereforeTis anautomorphism ofG, i.e T A(G). AlsoT2(y) =T T(y) =T(ya) =ya

2

. So wehaveTn(y) = ya

n

. Butyan1 =e, whereeis the identity element ofG, therefore

yan

= y. So Tn(y) = y y G. ThereforeTn = I, where I is the identitymapping. Next suppose for some positive integer m < n, we have Tm = I.Then we have Tm(y) = y y G, in particular, we have Tm(x) = x. But


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Tm(x) = x xam

=x xam1 =e o(x)< an 1, which is not true. Thus

n is the smallest possible integer for which Tn = I. Thuso(T) = n in A(G).

Also G being finite cyclic group, thereforeo(A(G)) = (o(G)) = (an 1). Soby Lagrange Theorem, we get n | (an 1).

17. Let G be a group and Zthe center ofG. IfTis any automorphism ofG,prove that Z(T) Z.Solution: We need to show that Zis a characteristic subgroup ofG. For someautomorphism mapping T, we need to show T(Z) Z. Let somez T(Z),therefore there exist some x Z such that T(x) = z. But x Z im-plies xg = gx g G. Since mapping T is one-to-one, therefore we haveT(xg) = T(gx) g G. But T(xg) = T(gx) T(x)T(g) = T(g)T(x) zT(g) = T(g)z g G. But since T is onto, therefore z T(g) = T(g)z g G zg =g z g G, which says z Z. Therefore z T(Z) z Z. SoT(Z) Z.

18. LetG be a group andTan automorphism ofG. If, fora G,N(a) = {xG| xa = ax}, prove that N(T(a)) = T(N(a)).Solution: We need to showN(T(a)) = T(N(a)) a G & T A(G). Wehave N(T(a)) = {x G | xT(a) = T(a)x}. But sinceT is an onto mapping,therefore

N(T(a)) = {T(x)| x G & T(x)T(a) = T(a)T(x)}

={T(x)| x G & T(xa) = T(ax)}

={T(x)| x G & xa= ax}; since mapping is one-to-one

={T(x)| x N(a)}

=T(N(a))

19. LetG be a group andTan automorphism ofG. IfN is normal subgroup ofG such thatT(N) N, show how you could use Tto define an automorphismofG/N.Solution:20. Use the discussion following Lemma 2.8.3 to construct(a) a non-abelian group of order 55.(b) a non-abelian group of order 203.Solution:

21. LetGbe the group of order 9 generated by elements a,b, wherea3

=b3

=e.Find all automorphisms ofG.Solution: Again, as in Problem 6, we are going to exploit the fact that anautomorphism leaves the order of elements unchanged. G has four non-trivialsubgroups, which are {e,a,a2}; {e,b,b2}; {e, a2b,ab2}; {e,ab,a2b2}. Now theelement e has not choice but has to map to itself, so only 1 choice. The element


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a has 8 choices for being get mapped as all elements other than e has sameorder as a has. Once image ofa is fixed, the image ofa2 get fixed by itself as

T(a2) =T(a)T(a). So a2 has no choice. Next after fixing image ofe,a,a2, theelement b has only 6 choices left. But fixing the image ofa andbfixes the imageof remaining elements. So b2,ab,a2b2, a2b,ab2 have no choices. Thus the totalautomorphisms ofG are 8 6 = 48. Soo(A(G)) = 48.


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Problems (Page 74)

1. Let G be a group; consider the mappings of G into itself, g, defined forg G by g(x) =xg for all x G. Prove that g is one-to-one and onto, andthatgh = hg.Solution: We start with one-to-one. Suppose g(x) = g(y). But g(x) =g(y) xg = yg x = y. So g(x) = g(y) implies x = y, showing g isone-to-one mapping. Next we tackle onto. Suppose some y G. But then forx = yg1 G, we have g(x) = (yg

1)g = y. This shows x is the inverse-image ofy . Thus g is onto too. Finally, we have gh(x) =xgh = (g(x))h =h(g(x)) = hg(x) for all x G. Thus gh = hg.

2. Letg be defined as in Problem 1,g as in the proof of Theorem 2.9.1. Provethat for any g, h G, the mappings g , h satisfy gh = hg. (Hint: Forx G considergh(x) and hg(x).)Solution: We recap our notation, mappingg : x xg and h : x hx.Now we have for all x G, gh(x) = g(h(x)) = g(hx) = hxg = h(xg) =h(g(x)) = hg(x). Thus gh = hg .

3.If is one-to-one mapping ofG onto itself such that g= g for all g G,prove that = h for some h G.Solution: We are given is an one-to-one and onto mapping with g =g g G. So we haveg(x) = g(x) g, x G (x)g= (xg) g,x G. Since it holds for allx G, in particular must hold for x = e. Thus wehave (e)g= (g). Since (e) G, so we let (e) = h for someh G. Thus we

have (g) = hg g G. But that means = h, hence the result.

4. (a) IfHis a subgroup ofG show that for every g G, gHg1 is a subgroupofG.(b) Prove thatW= intersection of all g Hg1 is a normal subgroup ofG.Solution:(a) Consider gH g1 for some g G. Let x, y gHg1, therefore x = gh1g

1

and y = gh2g1 for some h1, h2 H. But thenxy

1 = g h1g1(gh2g

1)1 =gh1g

1gh12 g1 = gh1h

12 g

1 = gh3g1 for some h3 H. Thus xy

1 gH g1 x, y gH g1. Henceg Hg1 is a subgroup ofG for all g G.

(b) We have W = gG gH g1. Let some x G. Consider xW x1. We

have xW x1 = x(gG

gH g1)x1 =gG

x(gH g1)x1 =gG

xgHg1x1 =gG

(xg)H(xg)1. But since the mapping : G G such that (g) = xg is

an onto mapping, so xW x1 =gG

(xg)H(xg)1 =gG

gHg1 = W. But


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xW x1 =W x G implies W is normal in G.

5. Using Lemma 2.9.1 prove that a group of order p2, where p is a prime num-ber, must have a normal subgroup of order p.Solution: LetG be a group with order p2. Sincep | o(G) andp is a prime num-ber, therefore the Cauchys Theorem guarantees us existence of an element a oforderp. LetH=a. ThereforeHis a subgroup of order p. Now sincep2|p!,therefore we have o(G)|iG(H)!. So using Lemma 2.9.1, we have K = {e},along withK HandK normal inG. But then o(H) = p, a prime number,forcesK = H. Thus His normal in G.

6. Show that in a group G of order p2 any normal subgroup of order p must liein the center ofG.Solution: Let H be the normal subgroup of G. Let some h H and someg G. We have ghg1 H as H is normal in G. If h = e, then we haveghg1 = h, or gh = hg g G. Next we assumeh = e, so H = h. Nowsince the o(g) | o(G), therefore o(g) = 1, p , p2. When o(g) = 1, then we haveg = e, and so ghg1 = g, or hg = hg. Wheno(g) = p, we have ghg1 H.SinceH= h, therefore ghg1 =hi for some positive integer i < p. With thishave h = gphgp = gp1(ghg1)g(p1) = gp1(hi)g(p1) = = hi

p

. Sohi

p1 = e, but o(h) = p, therefore p | ip 1, or ip = 1 mod p. But we have,

from Fermat theorem ip = i modp. From these two equations, we concludei= 1 mod p, ori = 1. Thus we have ghg1 =h, orgh= hg, for all h H andfor all g of order p. Finally, wheno(g) = p2, we have G = g, therefore G isabelian and we have gh = hg in this case too. Thus we concludedgh = hg for

all h Hand for all g G. And so h Z h H, where Z is the centersubgroup ofG. HenceH Z.

7. Using the result of Problem 6, prove that any group of order p2 is abelian.Solution: Let G be a group with order p2. Problem 5 implies that there exista normal subgroup H of order p. But since p is a prime number, thereforeH = a for some a H. But since o(G) = p2, therefore there exist b Gsuch that b / H. Let K = b. Now order of b divides order ofG, thereforeo(b) = 1 orp or p2. Ifo(b) = 1, then we have b= e H, which is contravene ourassumption. So o(b) = 1. Next ifo(b) = p, then since o(G)|iG(K)!, we haveKtoo normal in G. Therefore, Problem 6 implies H, K Z. And soH K Z.Buto(HK) = o(H)o(K)/o(H K) = p2 asH K= {e}. Thereforeo(Z) p2,

forcingZ=G, making G an abelian group. Finally ifo(b) =p2

, then we haveG = b and hence abelian in this case too. So we concludedG is an abeliangroup.

8. If p is a prime number, prove that any group G of order 2p must have a


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subgroup of order p, and that this subgroup is normal in G.Solution: Cauchy Theorem guarantees the existence of an element a G such

thato(a) = p. But then H= a is a subgroup ofG with o(H) = p. Also since2p|2!, therefore o(G)|iG(H)!. But then there exists a normal subgroupK ofG inside Hwith K={e}. So we have o(K) | o(H) and with o(H) being primeforcesK=H. Thus His normal in G.

9. Ifo(G) ispqwherep and qare distinct prime numbers and ifG has a normalsubgroup of order p and a normal subgroup of order q, prove that G is cyclic.Solution: We are given a group G with o(G) = pq, where p, q are primenumbers. Also we are given H, Knormal subgroups ofG with o(H) = p ando(K) = q. So we have H = a for some a H; and K = b for someb K. Also since p and q are distinct primes, so we have HK = {e}.We claim ab = ba. To establish our claim, consider aba1b1. We haveaba1b1 =a(ba1b1) = ah for someh H. Thereforeaba1b1 H. Again,aba1b1 = (aba1)b1 =kb1 for some k K. Thereforeaba1b1 K. Soaba1b1 H K = {e}. That is aba1b1 = e ab = ba. Next we claimo(ab) = pq, so that G = ab, i.e. a cyclic group. We have (ab)pq = apqbpq,as ab = ba. Therefore (ab)pq = apqbpq = (ap)q(bq)p = eqep = e. Therefore

pq| o(ab). Suppose for some positive integer t with t < pq, we have (ab)t = e.But (ab)t = e atbt = e at = bt atq = (bq)t atq = e o(a) | tq

p | tq p | t as gcd(p, q) = 1. Similarly, we have q | t. But then we have pq| t,implying t > pqas p = 0, which is against our assumption. So we havepqasthe smallest positive integer for which (ab)pq = e, implying o(ab) = pq. ThusG= aband is cyclic.

10. Leto(G) be pq, p > qare primes, prove(a)G has a subgroup of order p and a subgroup of orderq.(b) Ifq|p 1, then G is cyclic.(c) Given two primes p, q,q |p 1, there exists a non-abelian group of order pq.(d) Any two non-abelian groups of order pqare isomorphic.Solution:(a)


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Problems (Page 80)

1. Find the orbit and cycles of the following permutations:(a)

1 2 3 4 5 6 7 8 92 3 4 5 1 6 7 9 8

(b)

1 2 3 4 5 66 5 4 3 1 2

Solution:

(a) Clearly

1 2 3 4 5 6 7 8 92 3 4 5 1 6 7 9 8

= (1, 2, 3, 4, 5)(6)(7)(8, 9). So orbit

of 1, 2, 3, 4 and 5 is the set {1, 2, 3, 4, 5}; orbit of 6 is 6; orbit of 7 is 7; orbit of8 and 9 is the set {8, 9}. Also (1, 2, 3, 4, 5) and (8, 9) are its cycles.

(b) Again

1 2 3 4 5 66 5 4 3 1 2

= (1, 6, 2, 5)(3, 4). So the orbit of 1, 2, 5 and 6

is the set {1, 2, 5, 6}; and the orbit of 3 and 4 is the set {3, 4}. Also (1, 6, 2, 5)and (3, 4) are its cycles.

2. Write the permutation in the Problem 1 as the product of disjoint cycles.

Solution: We have

1 2 3 4 5 6 7 8 92 3 4 5 1 6 7 9 8

= (1, 2, 3, 4, 5)(6)(7)(8, 9)

and

1 2 3 4 5 66 5 4 3 1 2

= (1, 6, 2, 5)(3, 4).

3. Express as the product of disjoint cycles:(a) (1, 5)(1, 6, 7, 8, 9)(4, 5)(1, 2, 3).(b) (1, 2)(1, 2, 3)(1, 2).

Solution:(a) Let (1, 5)(1, 6, 7, 8, 9)(4, 5)(1, 2, 3) = . So we have = 1234, where1 = (1, 5), 2= (1, 6, 7, 8, 9), 3 = (4, 5) and 4 = (1, 2, 3). Now

(1) =1234(1)

=1(2(3(4(1))))

=1(2(3(2)))

=1(2(2))

=1(2)

= 2

Repeating analogously, we have (2) = 3;(3) = 6;(6) = 7;(7) = 8;(8) = 9;(9) = 5;(5) = 4; and (4) = 1. Thus we have =

1 2 3 4 5 6 7 8 92 3 6 1 4 7 8 9 5

=

(1, 2, 3, 6, 7, 8, 9, 5, 4).

(b) Proceeding as in part (a), we have (1, 2)(1, 2, 3)(1, 2) = (1, 3, 2).


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4. Prove that (1, 2, . . . , n)1 = (n, n 1, n 2, . . . , 2, 1).Solution: One can easily check (1, 2, . . . , n)(n, n 1, . . . , 1) = I, whereIis theidentity permutation. Hence (1, 2, . . . , n)1 = (n, n 1, . . . , 1).

5. Find the cycle structure of all the powers of (1, 2, . . . , 8).Solution: Let (1, 2, 3, 4, 5, 6, 7, 8) = . So we have

2 = = (1, 2, 3, 4, 5, 6, 7, 8)(1, 2, 3, 4, 5, 6, 7, 8)

= (1, 3, 5, 7)(2, 4, 6, 8)

3 =2= (1, 3, 5, 7)(2, 4, 6, 8)(1, 2, 3, 4, 5, 6, 7, 8, 9)

= (1, 4, 7, 2, 5, 8, 3, 6)

4 =3= (1, 4, 7, 2, 5, 8, 3, 6)(1, 2, 3, 4, 5, 6, 7, 8, 9)

= (1, 5)(2, 6)(3, 7)(4, 8)

5 =4= (1, 5)(2, 6)(3, 7)(4, 8)(1, 2, 3, 4, 5, 6, 7, 8, 9)

= (1, 6, 3, 8, 5, 2, 7, 4)

6 =5= (1, 6, 3, 8, 5, 2, 7, 4)(1, 2, 3, 4, 5, 6, 7, 8, 9)

= (1, 7, 5, 3)(2, 8, 6, 4)

7 =6= (1, 7, 5, 3)(2, 8, 6, 4)(1, 2, 3, 4, 5, 6, 7, 8, 9)

= (1, 8, 7, 6, 5, 4, 3, 2)

8 =7= (1, 8, 7, 6, 5, 4, 3, 2)(1, 2, 3, 4, 5, 6, 7, 8, 9)

= (1)(2)(3)(4)(5)(6)(7)(8) =I

So for i Z, we have i =i mod 8

group_theory Herstein Sol.pdf

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