STUDY GUIDE FOR MINOR TOPIC OF GROUP THEORYSTUDY GUIDE FOR MINOR TOPIC OF GROUP THEORY MICHAEL LOPER...

STUDY GUIDE FOR MINOR TOPIC OF GROUP THEORY

MICHAEL LOPER AND GREGORY MICHEL

ABSTRACT. This study guide is meant for any students who have chosen MATH 8245 (GroupTheory) as their minor topic area in their oral exam. In particular, it is aimed at students who havechosen Peter Webb as their minor topic committee member.

CONTENTS

1. Introduction 12. Wreath Products 13. Todd-Coxeter Algorithm 34. Schrier Transversals 85. Groups Acting on Trees 96. Coxeter Groups 167. Low Dimensional Group Homology Theory 188. Crystallography 239. The Burnside Ring and Marks Homomorphism 29

1. INTRODUCTION

This is a study guide to review the important concepts from MATH 8245 Group Theory I andMATH 8246 Group Theory II as taught by Peter Webb at the University of Minnesota during the2015-2016 academic year. As a note, this document contains very few proofs, but the purpose ofthis document is to state the main theorems from the course, understand some of the motivationbehind them, and apply the concepts to simple exercises.

2. WREATH PRODUCTS

Let D and Q be groups, and let Ω be a Q-set, which is to say that Q acts on Ω. Let K =∏ω∈Ω Dω ,that is K is |Ω| many copies of D, which are indexed by the elements of Ω. We define the wreathproduct

D oΩ Q = K oθ Qwhere θ : Q→ Aut(K) is the action of Q on Ω. To say that in another way, the action in thesemidirect product permutes the indices of the elements in K.

Date: 16 January 2018.Peter Webb was invaluable in guiding our studying for this. If you haven’t already, go talk to him.

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2 M. LOPER AND G. MICHEL

Typically, if Ω is an infinite set, we take the restricted wreath product, meaning that all elementsin K are assumed to have only finitely many non-identity elements. Additionally, if no Q-set isspecified, the wreath product D oQ means Ω = Q, with action being given by right multiplication.In the literature, this is often referred to as D or Q, where the r denotes the “regular" wreath product.

If D and Q are finite groups, say |D|= n and |Q|= m, we can compute the order of D oQ. Sincethe set K contains m copies of D, the set K has size nm, so the wreath product has size nm ·m. Fromthis fact, we can see that the wreath product is not associative simply by looking at group orders.If R is a third group with |R|= k, then

|(D oQ) oR|= (nm ·m)r · r = r ·mr ·nmr,

where alternatively|D o (Q oR)|= n(m

r·r) · (mr · r) = r ·mr ·nrmr.

Example 2.1. As an example, lets consider the smallest non-trivial example of a wreath product,the group C2 oC2, which has order 8. In this case, the group K = C2×C2, so the wreath productC2 oC2 = (C2×C2)oC2. First consider the element ((1,0),1) in this group. We can square thiselement

((1,0),1) · ((1,0),1) = ((1,0)+1 · (1,0),1+1)

= ((1,0)+(0,1),1+1)

= ((1,1),0).

And thus the cube of this element is

((1,1),0) · ((1,0),1) = ((1,1)+0 · (1,0),0+1)

= ((1,1)+(1,0),0+1)

= ((0,1),1).

And finally, its fourth power is

((0,1),1) · ((1,0),1) = ((0,1)+1 · (1,0),1+1)

= ((0,1)+(0,1),0)

= ((0,0),0).

Let’s denote this element ρ , which we have shown to have order 4. Second, consider the element((0,0),1), which clearly has order two, and let’s denote this element σ . Note that σρ = ρ3σ =((0,1),0), and thus C2 oC2 can be presented as 〈ρ,σ | ρ4 = σ2 = 1,σρ = ρ3σ〉, which is exactlythe presentation of D8 the dihedral group of order 8, so C2 oC2 ∼= D8.

Remark 2.2. Throughout this document we will denote the dihedral group of order 2n as D2n.

Example 2.3. Another important wreath product is the product Z oZ, in which the group K isthe group of sequences, indexed by the integers, of integers of which only finitely many arenon-zero. As an exercise, the reader can show that that this group is generated by the elementsx = ((. . . ,0,0,0,0,0, . . .),1) and y = ((. . . ,0,0,1,0,0, . . .),0), where the 1 in the sequence in the

GROUP THEORY 3

component of y is in position 0. Therefore, this group is finitely generated. However, this wreathproduct has a copy of K as a subgroup, and K is infinitely generated. Thus, this wreath product con-struction gives an example of a finitely generated group with an infinitely generated subgroup.

Before moving on, consider the following exercise.

Exercise 2.4. Convince yourself that 〈(123),(456),(14)(25)(36)〉 ∼= C3 oC2. Use a similar con-struction to write C4 oC2 as a permutation subgroup of S8 and C2 oC3 as a permutation subgroup ofS6.

Our main application of the wreath product will be to compute the structure of a Sylow p-subgroup of Sm, and in the process find the highest power of p that divides m!. We shall use thefollowing theorem.

Theorem 2.5 (Kaloujnine). If p is a prime, then a Sylow p-subgroup of Spn is a n times iteratedwreath product of the cyclic group Cp. Since the wreath product is not associative, we specifyWn+1 =Wn oCp.

This theory can be expanded to symmetric groups which are not exactly p powers. We canexpress m in base p as m = a0 + a1 p+ a2 p2 + . . .+ at pt , where 0 6 ai 6 p− 1. We then takeX = 1, . . . ,m, the set that is permuted by Sm, and partition it into a0 singletons, a1 p-subsets,a2 p2-subsets, and so on. We can use the theorem above to compute the order of the Sylow p-subgroup of the permutations on each part in the partition, and since disjoint partitions commutethe direct product of all these Sylow p-subgroups will have order pN where N is the highest powerof p dividing m!.

Example 2.6. Let m = 6 and p = 2. We can write 6 = 1 ·22 +1 ·21. Therefore, we can partition1, . . . ,6 into two subsets X1 = 1,2,3,4 and X2 = 5,6. The Sylow 2-subgroup of S22 is thetwice iterated wreath product on C2, which is C2 oC2 ∼= D8, and the Sylow 2-subgroup of S2 is justC2. Therefore, the Sylow 2-subgroup of S6 is isomorphic to D8×C2 with order 8 ·2 = 16. Indeed,we can check that there are four 2’s in the prime factorization of 6! = 2 ·3 ·4 ·5 ·6.

Example 2.7. Let m = 10 and p = 2. We can write 10 = 1 ·23+1 ·21. Therefore, we can partition1, . . . ,10 into two subsets X1 = 1,2,3,4,5,6,7,8 and X2 = 9,10. The Sylow 2-subgroupof S23 is the three times iterated wreath product on C2, which is (C2 oC2) oC2, which has order82 ·2 = 27. Again, the Sylow 2-subgroup of S2 has order 2, and thus the Sylow 2-subgroup of S10is isomorphic to (C2 oC2) oC2×C2 and has order 28. Again one can easily confirm that there are 8instances of 2 in the prime factorization of 10!.

3. TODD-COXETER ALGORITHM

This section will explain the Todd-Coxeter Algorithm which is used to compute cardinality ofa finite group given a presentation in terms of generators and relations. This algorithm is basedentirely on the following theorem.

Theorem 3.1. Let G be a finite group, X a generating set and H a subgroup of G. Suppose⋃ni=1 Hωi is closed under multiplication by elements a ∈ X ∪X−1 where X−1 = x−1|x ∈ X. Then⋃ni=1 Hωi = G and [G : H] = n.


To apply the Todd-Coxeter algorithm, we will consider the special case where H = 1 the trivialsubgroup of G. In this case, the cosets Hωi are exactly the elements of G. So, applying thetheorem, if we can find a set of elements that is closed under multiplication by the generators andthe inverses of the generators, we will have constructed the entire group. While implementingthis algorithm, we will construct group elements named 1,2,3, ... and so on, keeping track of howthe given generators act on the elements we construct, and ensuring that they satisfy the givenrelations. Once we have constructed enough group elements so that we have found a set that isclosed in the sense above, the algorithm stops. The cardinality of the group is the largest integerthat we named. The details of the algorithm are demonstrated in the following example.

Example 3.2. Let G =⟨s, t | s3 = t2 = 1, tst = s2⟩. We will apply the Todd-Coxeter algorithm to

find the cardinality of G. The first step is to make the following three (currently empty) charts ofrelations. As we construct group elements, we will add them to the tables below.

s s s t t

t s t s−1 s−1

We also make the following (also currently empty) key of where s, t and s−1 send each element.We should include t−1 here but t is an involution so we will omit it to save space.

s s−1 t

We are now ready to start the Todd-Coxeter algorithm. We begin by calling our first groupelement 1. We place 1 in the first and last position in the table for each relation. For example, thefirst row tells us that s3 ·1 = 1, but the middle of the table is empty since we still do not know s ·1.Notice that we have written each new piece of information in bold in the tables.

s s s1 1

t t1 1

t s t s−1 s−1

1 1We also place 1 in the first spot in each table of the key.

s1

s−1

1t

1

We are now stuck. So we need to construct another group element and place it in the first emptyspot on our relation table. We will call this element 2 and we will say that s sends 1 to 2. Everytime there is a 1 before a column labeled by s, we will add a 2 to the next blank, since we nowknow that s ·1 = 2. Also, since 2 is a new element we need to put it in the first and last column ofevery relation table.

GROUP THEORY 5

s s s1 2 12 2

t t1 12 2

t s t s−1 s−1

1 12 2

We also need to keep track of 2 in our table of keys. Notice that since s sends 1 to 2, we alsoknow that s−1 sends 2 to 1.

s1 22

s−1

12 1

t12

Looking back at our tables of relations, we see we can add some new pieces of information intoour tables.

s s s1 2 12 1 2

t t1 12 2

t s t s−1 s−1

1 2 12 2

We again are stuck, so we add a new group element called 3 to the tables. We place 3 in thefirst empty spot, which is to say that s ·2 = 3. But then, reading the first line across, we must haves ·3 = 1. We continue filling in new information into our tables.

s s s1 2 3 12 3 1 23 1 2 3

t t1 12 23 3

t s t s−1 s−1

1 3 2 12 1 3 23 2 1 3


s1 22 33 1

s−1

1 32 13 2

t123

Adding the new group element 4, now into the top row of the second relation table, yields thefollowing.

s s s1 2 3 12 3 1 23 1 2 34 4

t t1 4 12 23 34 1 4

t s t s−1 s−1

1 4 3 2 12 4 1 3 23 2 1 34 1 2 4

s1 22 33 14

s−1

1 32 13 24

t1 4234 1

Continuing in this manner, we add 5 to our tables.

s s s1 2 3 12 3 1 23 1 2 34 5 45 4 5

t t1 4 12 23 5 34 1 45 3 5

GROUP THEORY 7

t s t s−1 s−1

1 4 5 3 2 12 4 1 3 23 5 2 1 34 1 2 5 45 3 1 4 5

s1 22 33 14 55

s−1

1 32 13 245 4

t1 423 54 15 2

And finally, we ad the new group element 6 to the tables.s s s

1 2 3 12 3 1 23 1 2 34 5 6 45 6 4 56 4 5 6

t t1 4 12 6 23 5 34 1 45 3 56 2 6

t s t s−1 s−1

1 4 5 3 2 12 6 4 1 3 23 5 6 2 1 34 1 2 6 5 45 3 1 4 6 56 2 3 5 4 6

s1 22 33 14 55 66 4

s−1

1 32 13 24 65 46 5

t1 42 63 54 15 26 2


At this point, we no longer have any blank places. In fact, if we look at the table of keys, we seethat the group elements 1, . . . ,6 are closed under the set of generators and the set of inverses of thegenerators. Therefore, by the above theorem, G is equal to the union of these singleton elementsso the cardinality of G is 6. Since the group is of order six and is clearly not commutative, it mustbe S3.

As a note, if the group is not finite, then this algorithm will not detect this and the algorithm willnot terminate. As a result, this algorithm is not great for computing the cardinality of very largegroups by hand. The reader is encouraged to try the following two exercises for themselves.

Exercise 3.3. Find the cardinality of the group⟨a,b | a2 = b2 = (ab)2⟩. What group is this if

a2 6= 1. What group is this if a2 = 1?

Exercise 3.4. Find the cardinality of the group⟨x,y | yxy−1 = x4,y2 = 1

⟩. [Remark: This exercise

is nasty and is best done as background during something more entertaining, such as watchingSunday night football.]

4. SCHRIER TRANSVERSALS

For this entire section, let G be a subset of Sn and let H be a subgroup of G. We present atheorem of Schrier that will allow us to find a set of generators for H. We also show another wayto compute the cardinality of a group. This way is especially useful if the group is given as a subsetof Sn. In fact, the first author believes that this is one of the ways the computer algebra softwareGAP computes the cardinality of a group (if G is encoded as a subset of Sn).

We first present the following definition.

Definition 4.1. A right Schrier transversal is a set T of right coset representatives of H with theproperty that, for any product a1a2 · · ·am ∈ T , then the shorter product a1a2 · · ·ai ∈ T for all i6m.

Theorem 4.2 (Schrier). Let X be a set of generators for G, and let T be a right Schrier transversalfor H in G such that the identity element represents the identity coset. For each g ∈ G, let g ∈ Tbe the representative of the coset of H containing g; in other words Hg = Hg. Then, the set ofelements

tg(tg)−1 | t ∈ T, g ∈ X

is a generating set of H.

To better understand this theorem, consider the following example.

Example 4.3. Let G = 〈(15)(26),(13)(46),(23)(45)〉 and let H = StabG(1). We will find both aset of generators for H and the cardinality of G.

For convenience, let’s name the three generators, in order, g1, g2, and g3. Since H is the stabilizerof 1, a coset of H is determined uniquely by where it sends 1, so a transversal should contain 6elements, each of which maps 1 to a different number. We will create a transversal containing g1,which sends 1 to 5, and g2, which sends 1 to 3. Tracking the image of 1, g1g3 sends 1 to 4, g2g3sends 1 to 2, and g2g3g1 sends 1 to 6. These data points combined allow us to create a right Schriertransversal, T = e,g1,g1g3,g2,g2g3,g2g3g1. This data is recorded in the following graph.

GROUP THEORY 9

1 5 4

3 2 6

g1 g3

g2

g3 g1

Now that we have a transversal, we can apply the theorem to construct a generating set of H.Each generator will be of the form tg(tg)−1. To do this, we first compute tg for each pair t ∈ Tand g ∈ g1,g2,g3. That data is given in the chart below. For example, if t = g1g3 and g = g2, thetg = g1g3g2. We can compute that tg(1) = g1g3g2(1) = g1g3(3) = g1(2) = 6. Since the elementin T that sends 1 to 6 is g2g3g1, it follows that g1g3g2 = g2g3g1. The results of the remainingcalculations are shown in the table below.

x \T e g1 g2 g1g3 g2g3 g2g3g1g1 g1 e g2 g1g3 g2g3g1 g2g3g2 g2 g1 e g2g3g1 g2g3 g1g3g3 e g1g3 g2g3 g1 g2 g2g3g1

Now that we know tg for each pair t ∈ T and x∈ g1,g2,g3, we can compute a set of generatorsof H of the form tg(tg)−1. For example, if t = g1g3 and x = g2 as above, then

tg(tg)−1 = g1g3g2(g2g3g1)−1 = (25)(34).

The result of the remaining calculations are shown in the table below.x \T e g1 g2 g1g3 g2g3 g2g3g1g1 e e (24)(35) (23)(45) e eg2 e (24)(35) e (25)(34) (23)(45) (25)(34)g3 (23)(45) e e e e (24)(35)

Clearly, this is not a minimal set of generators, but it shows that H is generated by h1 = (23)(45),h2 = (24)(35), and h3 = (25)(34). Each of these elements has order 2, and h1h2 = h3, so H isisomorphic to the Klein-4 group.

Now the Orbit-Stabilizer Theorem says that |G| = |StabG(1)||Orb(1)|. Since we know that|StabH(1)|= 4 and Orb(1) = 1,2,3,4,5,6, we can conclude that |G|= 24.

Exercise 4.4. Let G = 〈(12)(34)(56),(123)(45)〉. Determine StabG(2).

5. GROUPS ACTING ON TREES

In this section, we will study theory related to groups acting on trees. To begin, we define theCayley graph of a group.

Definition 5.1. Let X be a generating set for a group G. The Cayley graph Γ(G,X) is the directedgraph with vertex set G and an edge g 7→ gx for every g ∈ G and x ∈ X .

Example 5.2. Let G = Z/5Z, and let X = 2. The graph Γ(G,X) is given below.


1

2 0

3 4

The group G acts on its Cayley graph Γ(G,X) by permuting its edges. The action is defined by

g(a 7→ ax) = ga 7→ gax.

Notice that this is a left action. It is important to not confuse the action with the multiplication inthe definition of the Cayley graph, which was right multiplication.

Example 5.3. Consider Example 5.2 above. On the left is the Cayley graph as drawn prior. On theright is the result of acting on the graph by 4 ∈ Z/5Z.

1

2 0

3 4

0

1 4

2 3

Clearly, the action of 4 on this graph is a rotation by 2π

5 .

Example 5.4. Let X = a,b and G = F(X), the free group on two generators. The Cayley graphΓ(G,X) is given below, with an image that is totally stolen from Wikipedia. Note that each vertexhas two edges going out of it, and two edges going in to it corresponding to the two generators.The result is an infinite fractal tree.

GROUP THEORY 11

If we act on this tree by a ∈ G, each edge is translated to the left once in the fractal pattern.

Remark 5.5. As in Example 5.4, if F = F(X) is a free group, then the Cayley graph Γ(F,X) is atree. In general, relations in a group correspond to cycles in the Cayley graph.

If a group G acts on a tree, we say that this action is properly discontinuous if the only groupelement that fixes any vertex is the identity element (which fixes all vertices). The followingproposition will be crucial in our study of groups acting on trees.

Proposition 5.6. A group G is free if and only if G acts properly discontinuously on a tree.

The following fact about free group follows immediately from this proposition.

Corollary 5.7. Subgroups of free groups are free.

Proof. Let F = F(X) be a free group and H be a nontrivial subgroup of F . By Proposition 5.6, Facts properly discontinuously on a tree, so the only element in F that fixes any vertex is the identity.However, the subgroup H inherits an action on the same tree. Since no non-identity element of Ffixes a vertex, no non-identity element of H fixes a vertex. Therefore, the action of H on this treeis properly discontinuous, so using Proposition 5.6 again, H is free.

If a group G acts on a graph Γ, we can get a new graph by quotienting out by the action of G inthe following way.

Definition 5.8. If G acts on a graph Γ, the graph G\Γ is the graph with vertices that are orbits ofvertices and edges that are orbits of edges.

Example 5.9. Let Γ = Γ(F(a,b,a,b) as in the previous example and let G = 〈a〉 ⊂ F(a,b).Notice that in the original tree, each edge corresponds to a word in the letters a, b, a−1, and b−1. Forexample, the edge b 7→ ba corresponds to the word ba. Any edge corresponding to a word startingin a, for example the edge ab 7→ aba is the image of some other edge, in this case a(b 7→ ba).Remember that the action of G on this tree is a left action and the multiplication in the Cayleygraph is right multiplication. Therefore, in the quotient graph G\Γ, any edge corresponding to aword starting with an a is identified with another edge. Therefore, from the picture above, the rightand left fractal branches out of the identity vertex in the middle are omitted as all edges in thesebranches start with an a and are therefore identified with some other edge in the tree. The othertwo fractal branches are not effected at all. In the place of the branches we omitted, we place aloop. The vertices . . . ,a−1,1,a, . . . are all identified, since they are in the same orbit under theaction of G, so the edge 1 7→ a becomes a loop in the quotient graph. Therefore, the quotient graphhas one loop coming in to and out of the identity vertex along with two infinite fractal trees comingout of the identity vertex, each resembling the original tree from Example 5.4.

Thinking of these graphs as topological spaces (CW-complexes containing only 0-cells and 1-cells), we can compute the fundamental group of these graphs. In particular, we have the followingresult about the fundamental group of a quotient graph.

Lemma 5.10. If G acts properly discontinuously on a tree Γ, then π1(G\Γ,x0)∼= G.


For instance, the graph of Exercise 5.9 deformation retracts to a circle so the fundamental groupis isomorphic to Z∼= 〈a〉.

Example 5.11. As in the previous example, let F(x,y) be a free group on generators x and y andlet Γ = Γ(F(x,y),x,y) be the Cayley graph with respect to these generators (which is a tree). Foreach n≥ 0, let Hn = 〈x, yx, y2

x, ..., yn−1x〉 as a subgroup of F(x,y), where yx = yxy−1.

Just as in Example 5.9, each conjugate of x that is in Hn removes a pair of left and right branchesof the fractal tree in the quotient graph and replaces them with a loop. For example, in H2\Γ, thevertices y and yx are identified, so the edge y→ yx becomes a loop. Accepting this, the graph Hn\Γstill has two infinite fractal trees and n loops. Therefore, Hn\Γ is homotopy equivalent to a wedgeof n circles, so π1(Hn\Γ) is free of rank n. As an especially useful example, we can considerH∞ := 〈yi

x | i ∈ N〉. The quotient graph H∞\Γ has infinitely many loops with only one fractalbranch. The branch coming downwards from the vertex corresponding to the identity remains.This tree is homotopy equivalent to an infinite wedge of circles, and therefore its fundementalgroup has infinite rank. So, even though F(x,y) is a free group with rank two, it contains aninfinite rank subgroup.

5.1. Finding a free group of finite index in SL(2,Z). Now we will use the theory of groupsacting on trees to show that there is a free group of finite index in SL(2,Z). We start with thefollowing definition.

Definition 5.12. Suppose G = 〈xi | r j〉 and H = 〈yk | sl〉 where the xi and yk are generators andthe r j and sl are the relations for their respective groups. The free product G ∗H is defined as〈xi,yk | r j,sl〉. Furthermore, if K ≤ G and α : K → H is an injective homomorphism, then we candefine the normal subgroup N of G∗H as

N := 〈kα(k)−1〉 ≤ G∗H for k ∈ K.

N is normal, so we can define the amalgamated free product to be

G∗K H = (G∗H)/N.

Example 5.13. For example, C2 6C4, and there is a natural inclusion C2 →C6, so we can definethe amalgamted free product of these cyclic groups. If α is a generator of C4 and β is a generatorof C6, then α2 and β 3 are identified in the amalgamated free product, so that

C4 ∗C2 C6 = 〈α,β | α4 = β6 = α

2β

3 = 1〉.

Note that in this amalgamated free product, because α2 = β 3 and powers of a generator commutewith other powers of that generator, the element α2 commutes with both generators, so it is in thecenter of this group. Therefore, we can name−1 := α2 and think of words in C4 ∗C2 C6 as± wordsin C2 ∗C3.

Now that we have defined amalgamated free products, we state this total random fact.

Remark 5.14. There is a homomorphism C4 ∗C2 C6C12 defined by α 7→ 3 and β 7→ 2. Noticethat αβ−1 7→ 1, so this homomorphism is surjective.

GROUP THEORY 13

Now we turn our attention to SL(2,Z). By Gaussian elimination, it can be shown that thefollowing two matrices generate SL(2,Z):

α =

[0 −11 1

], β =

[1 −11 0

].

By a quick check, we note that α has order 4, β has order 6, and that α2 = β 3 = −1. Therefore,there is a surjection C4 ∗C2 C6 SL(2,Z). Quotienting out both groups by their centers, there isalso a surjection C2 ∗C3 SL(2,Z)/〈−I2〉= PSL(2,Z). Looking at a tree on which SL(2,Z) acts,we will prove that these surjections are isomorphisms.

Recall that SL(2,Z) acts on C by Möbius transformations, that is[a bc d

].z =

az+bcz+b

.

Recall that Möbuis transformations send circles and lines to circles and lines. We will use thisaction of SL(2,Z) to construct a graph that is embedded in the complex plane. The action ofSL(2,Z) on this tree will be inherited from the action of SL(2,Z) on C. Let ω be the sixth root of

unity1+ i√

32

. We can compute that the β fixes ω , and we take for granted that Stab(ω) = 〈β 〉.Similarly, we can check that α fixes i, and we take for granted that Stab(i) = 〈α〉. Drawing anedge labeled e from i to ω that is the arc of a circle, we can act on this edge by powers of α and β .This edge is fixed by α3 = β 2 =−I2, so by acting on this edge, we attain a tree with edges labeledby the elements in C2 ∗C3. The reader should not stress over the exact structure of the tree; whatis important is that this construction gives a tree that SL(2,Z) acts on. Here is a picture of a partof this infinite tree taken from Peter Webb’s notes. Notice the initial edge e is near the middle, andeach other edge is the image of the action of some element in C2 ∗C3 on e.

After staring at this tree for a while, one sees that only the empty word in C2 ∗C3 fixes e.Moreover, given an arbitrary edge labeled by a word of length n in α , β , and β 2 in this tree, if youwere to construct a path from the initial edge e to that edge in the tree, that path would pass throughexactly n edges. As a result, the surjective homomorphism C2 ∗C3→ PSL(2,Z) is injective, andtherefore C2 ∗C3 ∼= PSL(2,Z). Further, the image of C2 is a normal subgroup in C4 ∗C2 C6 that has


quotient C2 ∗C3. Turning to the other surjective map C4 ∗C2 C6 → SL(2,Z), we can easily checkthat this map is an isomorphism between the normal subgroup C2 and the subgroup generated by−I2. We just showed that this map is an isomorphism on quotient C2 ∗C3, so it follows that thismap is an isomorphism and SL(2,Z)∼=C4 ∗C2 C6. To conclude, we need the following lemma.

Lemma 5.15. Let G be a finite group acting on tree (without exchanging the ends of any edges).Then G stabilizes a vertex.

To use this Lemma, we must determine which elements of C2 ∗C3 stabilize a vertex on our tree.The only vertex stabilizers are the conjugates of 〈α〉 and 〈β 〉. Now, recall the non sequitur fromRemark 5.14 where we defined a surjective homomorphism SL(2,Z)C12. Denote this surjectionρ and let K = ker(ρ). By the first isomorphism theorem, the index of K in SL(2,Z) is 12. Nowconsider an arbitrary vertex stabilizer (which has the form of either x−1α ix or x−1β ix). Since C12is abelian, ρ(x−1α ix) = ρ(α i) and ρ(x−1β ix) = ρ(β i). As a result, no vertex stabilizer, other thanthe identity, is in the kernel of ρ . Therefore, no nonidentity element of K stabilizes a vertex of thetree which means K acts properly discontinuously on the tree. Proposition 5.6 then implies that Kis free. Hence, K is a free subgroup of finite index in SL(2,Z). The ideas from Section 4 can nowbe used to determine that K is free of rank 2.

To conclude this subsection, we will find a finite index free subgroup of SL(2,Z) of rank greaterthan 2. It suffices to find such a subgroup for K. We apply the following theorem.

Theorem 5.16 (Subgroup Formula). If H is a subgroup of a free group F of finite index and rk(F)is finite, then

rk(H)−1 = [F : H](rk(F)−1).

If we let the generators of K be called x and y, then we can define a surjection ϕ : F Cn byx 7→ 1, y 7→ 0. By the first isomorphism theorem, kerϕ has index n in K. Using the SubgroupFormula, this kernel has rank n+1.

In this way it is possible to find free subgroups of finite index in SL(2,Z) of any rank.

5.2. HNN-Extensions. To begin our final subsection related to groups acting on trees, recall thedefinition of an amalgamated free production from 5.12. What follows is a generalization of thisdefinition.

Definition 5.17. An HNN-extension for a group G with generating set S and a subgroup H ≤ Gand a homomorphism ϕ : H→ G is

G∗H,ϕ := 〈S, t | tht−1 = φ(h)〉.

This definition lets us understand the notion of a graph of groups, and the fundamental group ofa graph of groups.

Definition 5.18. A graph of groups is a graph with a group attached to each vertex and a groupattached to each edge so that if we have an edge labeled by H that connects the vertices G and K,then there exist homomorphisms H → G and H → K.

Example 5.19. Let Γ be the graph below:

GROUP THEORY 15

C4 C6C2

We will formally define the notion of a fundamental group of a graph of groups later, but for now,to compute the fundamental group of this graph of groups, start at one vertex, say C4 and using thefact that there is an inclusion of C2 into both C4 and C6 we can take the amalgamated free productC4 ∗C2 C6. Set π1(Γ) =C4 ∗C2 C6.

Example 5.20. Now consider a more complicated graph of groups Γ:

S4 GL(3,2)

D14

D8

C7C2

To compute π1 start at an arbitrary vertex, say S4. As in the previous example, we can take theamalgamated free product S4 ∗D8 GL(3,2). Continuing around the graph, we can continue to usethe graph of group structure to take amalgamated free products and get S4 ∗D8 GL(3,2) ∗C7 D14.However, now we have reached a circuit in our graph and we have this C2 edge that has not beenaccounted for. However, C2 is a subgroup of D12 and the edge in the graph of groups gives aninjection ϕ : C2 → S4, so we can take an HNN-extension, (S4 ∗D8 GL(3,2) ∗C7 D14)∗C2,ϕ . Definethis to be π1(Γ).

In these examples, we have worked through the intuition of computing the fundemental groupof a graph of groups–just take iterated free products at each edge and take an HNN-extensionwhenever the graph contains a circuit. The formal definition of this constuction is as follows.

Definition 5.21. If Γ is a graph of groups with Gx the group for vertex x (this is not a coset, it isan index) and if T is the spanning tree, then π1(Γ) is freely generated by the Gx with an additionalelement y for each edge such that

• if y and y correspond to reverse edges then y = y−1

• yϕy,0(x)y−1 = ϕy,1(x) for x in the group corresponding to the edge for y and ϕy,0 and ϕy,1are the two morphisms• if y ∈ T,y = id.

Now, we can state the following theorem.

Theorem 5.22 (Fundamental Theorem of Bass–Serre Theory). For a group G, the following areequivalent:

(i) G is the fundamental group of a graph of groups.(ii) G acts on a tree without inverting edges.

Example 5.23. In the previous subsection, we showed that SL(2,Z) acts on an tree without invert-ing edges and in Example 5.19, we showed that SL(2,Z) is the fundamental group of a graph ofgroups. To give some insight about the origin of this result, observe that if we quotient our tree by


the action of SL(2,Z), then we are left with our initial edge e. If the edges and vertices are labeledwith their vertex and edge stabilizers, then we obtain the graph of groups whose fundamental groupis SL(2,Z).

6. COXETER GROUPS

A Coxeter system (W,S) is group W with a minimal generating get S, along with natural numbersm(s,s′) for each pair s, s′ ∈ S so that (ss′)m(s,s′) = 1. We allow for the case that m(s,s′) = ∞ and weenforce that m(s,s′) = 1 if and only if s = s′, which is to say that each s ∈ S has order 2.

For each Coxeter system, we can draw a Coxeter graph: a graph with vertices in bijection withthe generating set S and an edge drawn between s and s′ labeled by m(s,s′). Conventionally, ifm(s,s′) = 2, the edge is omitted, and if m(s,s′) = 3, an unlabeled edge is drawn.

Example 6.1. The group PGL(2,Z) is isomorphic to the group with Coxeter system given byS = s1,s2,s3 where m(s1,s2) = 3, m(s2,s3) = ∞ and m(s2,s3) = 2. The explicit isomorphismbetween PGL(2,Z) and this group is given by

s1 7→[

0 11 0

]s2 7→

[−1 1

0 1

]s3 7→

[−1 0

0 1

].

The Coxeter graph of this system is drawn below

s1 s2 s3∞

Example 6.2. The dihedral group of order 2m is isomorphic to the Coxeter system represented inthe graph below. Here s1 and s1s2 are the generators of order 2 and m, respectively, in the standarddefinition of the dihedral group.

s1 s2m

Corresponding to a Coxeter system, we can define an |S|- dimensional vector space over R,spanned by basis vectors αs which are in bijection with the elements in S. On this vector space wecan define a symmetric bilinear form

B(αs1,αs2) =−cos(

π

m(s1,s2)

).

Note that B(αs,αs) = 1 and that B(αs1,αs2) 6 0 if s1 6= s2. We specify that if m(s1,s2) = ∞, thenB(αs1,αs2) =−1. From this bilinear form, we can define the linear operator σs ∈ GL(V ) as

σs(λ ) = λ −2B(αs,λ )αs.

Notice that σs(αs) =−αs and that if λ ∈ α⊥s , which is to say that B(αs,λ ) = 0, then σαs(λ ) = λ .For the proofs of most ideas in this unit, it is important that there is a well-defined length function

` on a Coxeter group W so that the length of an element is its minimal length as a word in thegenerators si. Since we are omitting most proofs, we can safely omit the discussion of the length

GROUP THEORY 17

function, but it behaves intuitively. For example, with a length argument, it can be proven that therepresentation s 7→ σs is faithful, which is to say W maps to GL(V ) injectively.

Let V ∗ = HomR(V,R) denote the dual space, and if f ∈V ∗ and v ∈V , then let 〈 f ,v〉= f (v) bethe natural pairing. Define σ∗ : W → GL(V ∗) by

〈σ∗(w)( f ),λ 〉= 〈 f ,σ(w)−1λ 〉.

If we define the actions of W on V and V ∗ as w. f := σ∗(w) f and w.λ := σ(w)λ , this definitionensures that 〈 f ,λ 〉 = 〈w. f ,w.λ 〉. As a result, if σ(w) has matrix T , then σ∗(w) will have matrix(T−1)t .

Let fs be the basis element dual to αs. Then define As = f ∈ V ∗|〈 f ,αs〉 > 0 and C =⋂

s As.This space C will be imperative in the coming example. We will use C to understand the cardinalityof a Coxeter group W . This is because for an arbitrary x ∈C, the orbit Wx is a discrete set that isin bijection with W . This is a result of the following lemma.

Lemma 6.3. If x ∈C and wx ∈C, then w = 1.

Now, we state the main theorem of this section which tells us precisely when a Coxeter group isfinite.

Theorem 6.4. If the bilinear form B associated to a Coxeter group W is positive definite, then Wis finite.

Proof. If B is positive definite, then we can identify V ∗ with Euclidean space. Since W preservesB, W acts on V ∗ preserving vector length. Therefore, considering an arbitrary x ∈C, the orbit Wxis contained within the ball of radius ‖x‖. However, we previously asserted that W is in bijectionwith Wx, and since Wx is a discrete subset of a compact set, the ball of radius ‖x‖, Wx is finite,and therefore W is finite.

Remark 6.5. The converse of Theorem 6.4 is also true. But this result is highly non-trivial, so weomit the proof.

Example 6.6. We will show that the order of the Coxeter group given by the Coxeter graph belowhas infinite order.

s1 s2

s3

Recall that this means that m(si,s j) = 3 for i 6= j and therefore B(αs1,αs2) = −cos( π

m(s1,s2)).

Then we can compute

σs1(αs1) =−αs1

σs1(αs2) = αs2 +αs1

σs1(αs3) = αs1 +αs3


So s1 acts on V and V ∗ respectively by the matrices−1 1 10 1 00 0 1

and

−1 0 01 1 01 0 1

.Similarly, the reader can work out how s2 and s3 act on V ∗. We now note that the actions of s1, s2and s3 preserve the hyperplane E := f ∈V ∗ | 〈 f ,α1 +α2 +α3〉= 1.

Next we determine C. Thinking of V ∗ as three dimensional Euclidean space with dual basisfsi , each Asi is the half of V ∗ in which the corresponding coordinate is positive. Therefore, C, theintersection of the three Asi’s is the first octant in V ∗. To determine the cardinality of W , we willtrace the orbit of an arbitrary x ∈ C. Since we already stated that the action of W preserves thehyperplane E, we will choose an x ∈C∩E. The orbit Wx will be contained within the plane E andwill be in bijection with W . Under this Euclidean interpretation of V in which C is the first octant,E is the plane x+ y+ z = 1. Therefore, C∩E is an equilateral triangle, where the action of s1, s2,and s3 correspond to the reflections over the three sides of this triangle. We leave it to the reader tocheck that, given a point inside this triangle, the composition of reflections s1s2s1s3 is a translationwithin E. Therefore, we can continue to apply powers of (s1s2s1s3) to x. Since this translation hasinfinite order within the plane E, the orbit Wx is infinite, so W is an infinite group.

Finally, we state Sylvester’s Criterion.

Theorem 6.7. An n×n Hermitian matrix B is positive definite if and only if the determinant of theupper left i× i corner of B is nonzero for all 16 i6 n.

Example 6.8. Continuing the example above, we can also show this Coxeter group is infiniteby computing the matrix of the bilinear form B. This matrix has (i, j) entry B(αsi,αs j). In thisexample, the matrix is

B =

1 −1/2 −1/2−1/2 1 −1/2−1/2 −1/2 1

.The bilinear form is positive definite if and only if the corresponding matrix is positive definite.However, this matrix has det(B) = 0, so by Sylvester’s criterion B is not positive definite. ThereforeTheorem 6.4 implies the Coxeter group W is infinite.

7. LOW DIMENSIONAL GROUP HOMOLOGY THEORY

In this section, we’ll introduce group homology and group cohomology. At the time of writ-ing, both authors had studied basic homological algebra. If the reader has not at least seen thedefinitions of Ext and Tor, some supplemental reading may be required.

7.1. Preliminaries and Degree 0 Homology Theory. For a group G, we denote its integral groupring ZG as the free abelian group with elements of G as a Z-basis with ring multiplication on thebasis elements given by the group multiplication. If G acts on a set M, we can equivalently viewM as having the structure of a ZG-module. We can give Z the structure of a ZG-module with atrivial action, which is to say that gn = n for all g ∈ G and n ∈ Z. With this construction, we can

GROUP THEORY 19

define group homology and cohomology in the canonical way as the derived functors of M⊗−and Hom(−,M) respectively. Formally,

Hn(G,M) := ExtnZG(Z,M),

Hn(G,M) := TorZGn (M,Z).

As a note, we will typically work with left modules. Given two left modules A and B, we canmake sense of their tensor product by defining a new right module Ar where, for m ∈M and g ∈G,the action mg := g−1m. Therefore by A⊗ZG B, we mean Ar⊗ZG B.

Let ε : ZG→ Z denote the augmentation map generated by sending each group element to1. Formally, ε(∑λgg) = ∑λg. This is a homomorphism of rings and of ZG-modules, and wedenote its kernel IG, which we call the augmentation ideal. Since ε is a surjection, we note thatZ= ZG/IG. When computing Ext and Tor, we can always start projective resolutions with

ZG ε−→ Z→ 0.

As some more preliminaries, for a finite group, we denote N = ∑g∈G g ∈ ZG as the norm ele-ment. If M is a ZG-module, then the we denote the set of fixed points of the G action on M byMG = m∈M | gm = m for all g∈G. Moreover, we denote by the fixed quotient (also sometimesreferred to as the co-fixed points of the G action on M) by MG =M/〈gm−m |m∈M,g∈G〉. Sincewe forced gm−m to be zero, G acts trivially on the co-fixed points, and the co-fixed points are thelargest quotient on which G acts trivially.

The following are some preliminary results in this area,(1) The set g− 1 | 1 6= g ∈ G is a Z basis for IG, the augmentation ideal. Furthermore, if

G is generated by n elements, say g1, . . . ,gn, then IG is generated as ZG module by nelements, namely g1−1,g2−1, . . . ,gn−1.

(2) By definition, the degree 0 cohomology is H0(G,M) = HomZG(Z,M). By a fact fromhomological algebra, this is isomorphic to the colon module 0 :M IG, the submodule of ele-ments in M that are annihilated by IG. If m ∈M is annihilated by IG, then m is annihilatedby g−1∈ IG for each g∈G, so (g−1)m= 0 and it follows that gm=m. Since elements ofthe form g−1 form a basis for IG, these elements that satisfy gm = m completely describethis cohomology group, so H0(G,M)∼= MG.

(3) By definition, the degree 0 homology group is H0(G,M) = Z⊗ZG M. Thinking of Z asZG/IG, we can express this as ZG/IG⊗ZG M = M/(IG ·M). This is exactly the co-fixedpoints, so H0(G,M)∼= MG.

(4) If G is a finite group, the fixed points of the entire group ring ZG are exactly the multiplesof the norm element, which is to say that (ZG)G = NZ∼= Z. If G is an infinite group, then(ZG)G = 0. The co-fixed points (ZG)G = ZG/IG∼= Z.

Before we proceed, let’s consider some examples.

Example 7.1. Let G denote the cyclic group of order n. In this example, we will compute thegroup cohomology H∗(G,M). To do this, we first compute a projective resolution of ZG-modules

· · · → ZGg−1−−→ ZG N−→ ZG

g−1−−→ ZG ε−→ Z→ 0.


This resolution is 2-periodic, which tells us that, for n > 1, we have H2n+1(G,M)∼= H1(G,M) andH2n(G,M)∼= H2(G,M). To compute these cohomology groups, we apply the contrvariant functorHomZG(−,M) to this projective resolution. Each entry in the new resolution is HomZG(ZG,M)∼=M, and the application of Hom on the multiplication map by g−1 or N yields the action of g−1or N respectively on M. Therefore, the new resolution becomes

· · · ←Mg−1←−−M N←−M

g−1←−−M.

Now we can compute H2(G,M) and H1(G,M) by taking kernel mod image. Specifically,

H2(G,M) =ker(M

g−1−−→M)

N ·M=

MG

N ·M, and

H1(G,M) =ker(M N−→M)

(g−1) ·M=

ker(M N−→M)

IG ·M.

Example 7.2. Let G denote the free group of rank 2. In Example 5.4, we determined the Cayleygraph for a free group of rank 2. In this graph, the vertices are in bijection with the elements of thegroup G, and for each vertex, there are two edges coming out of the vertex. Thinking of this graphas a simplical complex,

C0 = Zvertices= ZG, and

C1 = Zedges= (ZG)2, and

Cn = 0, for all n> 2.

This translates to a complex

0→ (ZG)2 ∂−→ ZG ε−→ Z→ 0,where ∂ denotes the boundary map. Since the action of G on the tree commutes with all thehomomorphisms in this complex, these have the strucutre of ZG-module homomorphisms and thisis a free (and therefore projective) resolution of Z in the category of ZG-modules. This constructionextends to free groups of rank d 6= 2, the only exception being that C1 = (ZG)d in the general case.

This shows free groups have projective resolutions of length 1 and it follows immediately thatboth Hn(G,M) and Hn(G,M) are 0 for all n > 1.

7.2. Degree 1 Homology. To study the degree 1 homology of a group, consider the short exactsequence 1→ IG→ ZG→ Z→ 0. This gives a long exact squence in homology

· · · → H1(G,ZG)→ H1(G,Z)→ IG⊗Z→ ZG⊗Z→ Z⊗Z→ 0,

where these tensors products can be thought of as degree 0 homology groups. Since ZG is a freeZG-module, and hence a flat ZG-module, H1(G,ZG) = 0. The map ZG⊗Z→ Z⊗Z can beidentified as the identity map Z→ Z with trivial kernel. Since this map has trivial kernel, themap IG⊗Z→ ZG⊗Z in the long exact sequence is the 0 map, so we can replace it with 0. Oursequence becomes

0→ H1(G,Z)→ IG⊗Z→ 0,

GROUP THEORY 21

so H1(G,Z)∼= Z⊗ZG IG. Since Z= ZG/IG, we can rewrite this as

H1(G,Z)∼= Z⊗ZG IG∼= ZG/IG⊗ZG IG∼= IG/(IG)2.

Moreover, there is a bijection between this quotient and the abelianization of G, G/G′. Explicitly,this map is (g− 1)+ IG2 7→ gG′ with inverse gG′ 7→ (g− 1)+ IG2. Hence we have proven thefollowing proposition.

Proposition 7.3. The degree 1 homology with integer coefficients, H1(G,Z) is isomorphic to theabelianization G/G′.

7.3. Degree 1 Cohomology. A mapping d : G→M is called a derivation if d(gh) = gd(h)+d(g)for all g,h ∈G. If the reader has seen derivations before, it may help to view M as a bimodule withtrivial right action. This way the above definition will agree with the definition that the reader hasprobably seen before: d(gh)= gd(h)+d(g)h. The set of all derivations Der(G,M) is a group underpointwise addition. For any m ∈M, we can define a derivation dm defined by dm(g) = (g− 1)m.This is called the principal derivation associated to m. The set of principal derivations is denotedP(G,M) and is a subgroup of Der(G,M).

Lemma 7.4. If d : G→M is a mapping, we can define δ : IG→M by δ (g−1) = d(g). The mapd is a derivation if and only if δ is a ZG-module homomorphism. Therefore, HomZG(IG,M) ∼=Der(G,M).

Now, we can apply the long exact sequence in cohomology to the short exact sequence 1→Z→ZG→ IG→ 1 to get

· · · → HomZG(Z,M)→ HomZG(ZG,M)→ HomZG(IG,M)→ H1(G,M)→ H1(ZG,M).

Working from left to right, HomZG(Z,M) is exactly the 0th cohomology H0(G,M)∼= MG the fixedpoints of the G action. The second term in this sequence is HomZG(ZG,M)∼= M. The third term isHomZG(IG,M), which we just asserted to be isomorphic to Der(G,M) in Lemma 7.4. The fourthterm is what we set out to compute, and the final term is the first cohomology group of a free-ZGmodule, so it is trivial. Therefore, this sequence becomes

MG→M→ Der(G,M)→ H1(G,M)→ 0.

To complete this computation, we use the following lemma.

Lemma 7.5. A derivation d ∈Der(G,M) is a principal derivation if and only if the correspondingmap δ : IG→M from Lemma 7.4 is in the image of HomZG(ZG,M)→ HomZG(IG,M) as in ourexact sequence.

It follows from this lemma and the exact sequence above that H1(G,M)∼= Der(G,M)/P(G,M),so the first cohomology group can be computed as the group of derivations quotiented out by thegroup of principal derivations. However, there is yet another way to make sense of this cohomologygroup.

Definition 7.6. Given a short exact sequence of groups 1→ M → Ep−→ G→ 1, a map of sets

s : G→ E is a section if p s = idG. A section that is also a group homomorphism is called asplitting.


The following lemma gives the connection between derivations and splittings.

Lemma 7.7. Let s : G→ E = M o G be a section so that s(g) = (d(g),g) for some mappingd : G→M. Then s is a splitting if and only if d is a derivation. In other words, Der(G,M) is inbijection with the set of splittings G→ E.

Definition 7.8. Two splittings s1,s2 : G→ E = MoG are conjugate if there is an element m ∈Msuch that (m,1)s1(g)(m,1)−1 = s2(g) for all g ∈ G.

Putting these two ideas together, it can be shown that two splittings are conjugate if and only iftheir corresponding derivations differ by a principal derivation.This gives our final result

Theorem 7.9. If M is a ZG-module, then the M conjugacy classes of splittings biject with the firstcohomology group H1(G,M).

To conclude, we can think of the first cohomology group in two different ways, either as thegroup of derivations quotiented out by the principal derivations, or as the conjugacy classes ofsplittings G→MoG.

7.4. Degree 2 Cohomology. To begin this section, we define an extension of a group G by a groupN as a short exact sequence of groups

1→ Nϕ−→ E→ G→ 1,

such that ϕ(N) is normal in E. If N is abelian, then there is an action of G on N by conjugation inE, where gng−1 where g is a lift of g in E. Moreover, two extensions E1 and E2 are equivalent ifthere is an isomorphism φ : E1→ E2 such that the following diagram commutes:

1 N E1 G 1

1 N E2 G 1.

id φ id

As a note, given a ZG-module M, there is always an extension 1→ M → M oG→ G→ 1,so there is always at least one equivalence class of extensions. The following theorem explicitlyrelates extensions to the degree 2 cohomology of a group, which will be used extensively in thenext section of this guide.

Theorem 7.10. Let M be a ZG-module. There the is a bijection between H2(G,M) and the equiv-alence classes of extensions of G by M.

7.5. Degree 2 Homology. The second homology group H2(G,Z) is called the Schur multiplier ofG, and is described by the following theorem.

Theorem 7.11 (Hopf Formula). Let 1→ R→ F → G→ 1 be a presentation of G. The the Schurmultiplier is given by H2(G,Z)∼= (R∩F ′)/[R,F ], and is independent of the choice of presentation.

While the Schur multiplier is generally difficult to compute, there is a nice bound on the numberof generators of H2(G,Z).

GROUP THEORY 23

Proposition 7.12. If G is a finite group with a presentation using d generators and r relations,then the minimum number of generators of H2(G,Z) is bounded above by r−d.

Example 7.13. As an example, we will compute H2(S3,Z) and H2(SL(2,5),Z). We write a pre-sentation for S3 as S3 =

⟨x,y | x2 = 1,xyx−1 = y2⟩. This presentation has two generators and two

relations, so the proposition above tells us that the minimal number of generators of H2(G,Z) iszero, so it is the trivial group.

Similarly, there is a presentation SL(2,5) =⟨x,y | x2 = y3 = (xy)5⟩. Therefore, we can write

F = 〈x,y〉 and R =⟨x2y−3,x2(xy)−5⟩, and as with S3, the Schur multiplier has no generators and

is therefore trivial.

We finish this section on Low-Dimensional Group Homology by stating this nice result of Schurand Zassenhaus. It is a corollary of a result that we did not include, but it is useful for understandingextensions of a finite group.

Corollary 7.14 (Schur-Zassenhaus). Let 1→M→ E→G→ 1 be a short exact sequence of finitegroups such that the orders of G and M are relatively prime. Then, the extension is split, soE ∼= MoG, and all subgroups of E of order |G| are conjugate.

8. CRYSTALLOGRAPHY

Let En represent Euclidean n-space. For any element v∈En, we define a translation tv : En→En

by tv(x) = x+ v. We denote the group of translations T = tv | v ∈ En and clearly T ∼= En. LetO(n) be the group of orthogonal transformations. This consists of rotations and rotations composedwith reflections that fix 0∈En. We denote the group of all distance preserving maps of En as R(n),called the group of rigid transformations. We can further describe the structure of R(n) using thefollowing proposition.

Proposition 8.1. The group of rigid transformation is isomorphic to the semi-direct product

R(n)∼= T oO(n),

where the fact that T is normal in R(n) follows from the fact that, for any x∈ R(n), the compositionxtvx−1 = tx(v). Notice that R(n) is a subgroup of Rn oGL(n,R).

We begin our study of crystallography with the following definition.

Definition 8.2. A crystal structure C is a subset of Euclidean n-space that satisfies the followingproperties

(1) C is preserved by n independent translations.(2) There is some δ > 0 such that for every translation tv that preserves C, we have |v|> δ .

The second condition is a technical condition to ensure discreteness in our crystal structures. Wewill usually use small checkmark patterns to denote crystal structures. The crystal structures aresometimes called "wallpaper patterns." Consider the following examples. In all examples in thissection, the pattern is assumed to continue infinitely in all directions.


Example 8.3. The following 2-dimensional crystal structure is preserved by two independenttranslations, t1 and t2 as shown in the diagram below. To reiterate, the pattern is assumed tocontinue infinitely in each direction. There are no other rigid transformations that preserve thispattern other than those generated by t1 and t2.

t2

t1

Example 8.4. The following is not a crystal structure. It is is preserved by two independent trans-lations, but since any horizontal translation preserves the structure, it fails the second condition ofbeing a crystal structure.

t1

Before we proceed, we need a few more definitions to better understand these crystal structures.

Definition 8.5. Let C be a crystal structure.

• The space group S(C) = γ ∈ R(n) | γ(C) = C is the group of all rigid transformationsthat preserve C.• The translation subgroup T (C) = S(C)∩T is the group of translations that preserve C. By

the remark in the previous proposition, this subgroup is normal in T (C).• The point group P = S(C)/T (C).

We will now compute the space group, translation subgroup, and the point group for a numberof crystal structures.

Example 8.6. Using the crystal structure from Example 8.3, T (C) is the free abelian group gener-ated by t1 and t2. Since these are the only rigid motions that preserve C, the space group S(C) andthe translation subgroup T (C) are equal. Therefore P is the trivial group.

Example 8.7. Consider the following crystal structure.

GROUP THEORY 25

This crystal structure has two independent translations and two different types of reflections,up to translation. We denote the reflections over the two dashed lines s1 and s2 in the followingdiagram.

t2

t1

s1

s2

However, by tracking the image of the pieces of the pattern under these maps, it is not hardto verify that s2 = s1 t−1

2 . So, the space group S(C) =⟨t1, t2,s1 | s2

1 = 1⟩

and the point groupP = S(C)/T (C) has order 2.

Example 8.8. Consider the following crystal structure.

Again, this structure has two independent translations t1 and t2. There is another rigid motionthat preserves the pattern, but it is much less obvious than in the previous example. The pattern ispreserved by the map s1 that is a rotation by π around the dot in the drawing below followed by areflection about the dotted line.

t2

t1

By inspection, we can confirm that s21 = t1, and thus S(C) =

⟨t1, t2,s1 | s2

1 = t1⟩ ∼= 〈t2,s1〉, and

T (C) = 〈t1, t2〉. So P = S(C)/T (C) again has order 2. In each of these examples, it is obvious that the translation subgroup T (C) is isomorphic to a

free abelian group and that the point group P is finite. Additionally, using the ideas of Subsection7.4, there is an extension

1→ T (C)→ S(C)→ P→ 1,and P acts on T (C) by conjugation in S(C), which is well defined since T (C) is abelian. Thisaction turns out to be faithful, which leads to an alternative definition of a space group.

Theorem 8.9. A group G is a space group if and only if there is a normal subgroup T 6 G suchthat T ∼= Zn so that P := G/T is finite and P acts faithfully on T .


For the remainder of this section, we will discuss what it means for two space groups to beequivalent. First, we state the two equivalent definitions of this notion.

Definition 8.10. Let G1 and G2 be space groups acting on En with extensions 1→ T1 → G1 →P1→ 1 and 1→ T2→ G2→ P2→ 1. The groups G1 and G2 are said to be equivalent if they meeteither of the two equivalent conditions below.

• G1 and G2 are conjugate as subgroups of R(n)∼= Rn oGL(n,R), see Proposition 8.1.• There is a commutative diagram

1 T1 G1 P1 1

1 T2 G2 P2 1such that the map G1→ G2 is an isomorphism.

Moreover, if C1 and C2 are crystal structures with equivalent space groups, then we say that C1 andC2 are equivalent crystal structures.

With this definition, we will classify all two dimensional space groups up to equivalence. First,we want to determine all possible point groups P, which is to say that we want to determine allfinite groups with a faithful action on Z2. This is equivalent to determining all finite subgroups ofGL(2,Z). We begin with the following, very helpful lemma.

Lemma 8.11. If g is an automorphism of T ∼= Z2 of finite order, then g has order 1, 2, 3, 4, or 6.

As a corollary of this fact, it follows that no two-dimensional crystal structure has 5-fold sym-metry. This is closely related to the fact that you cannot tile the plane with regular pentagons.

Definition 8.12. A Bravais Lattice in dimension n is a subgroup Zn ∼= T 6 En together with itsfull orthogonal automorphism group Q = q ∈ O(n) : qT = T acting on it. In this context, Q isreferred to as the Bravais Point group, and two Bravais lattices (T1,Q1) and (T2,Q2) are said to beequivalent if there is some α ∈ GL(n,R) such that T2 = αT1 and Q2 = α−1Q1α .

We can study the possible point groups P and translation subgroups T by looking at Bravaislattices, according to this proposition.

Proposition 8.13. Any faithful ZP-module T with T ∼= Zn as abelian groups and P finite is iso-morphic as ZP-modules to a Bravais lattice (T,Q) where P is a subgroup of Q.

The upshot of this result is that we can study the possible point groups P and translation groupsT that appear in the extensions for space groups by studying 2-dimensional Bravais lattices, andconveniently, by a case study, there are only five possible Bravais lattices, which are outlined inthe following chart.

GROUP THEORY 27

Bravias lattice T Bravais point group Q

D12

D8

C2

C2×C2

C2×C2


We label the Bravais lattices in this table B1, B2, B3, B4 and B5. Note that the lattices for B1 andB5, and also for B2 and B4 are the same, but the Bravais point groups are different. We name thethree lattices that appear in the left column T1, T2, and T3 from top to bottom. According to theprior proposition, we can use this table to study space groups. Given any space group G given bythe extension 1→ T → G→ P→ 1, the translation group T will embed in E2 as one of the threelattice T1, T2, or T3, and P must be a subgroup of the corresponding Bravais lattice point group.Since P must be a subgroup of these Bravais point groups, the following is immediate.

Corollary 8.14 (da Vinci). Any finite subgroup of real 2×2 matrices is either cyclic or dihedral.

Now that we have studied all possible pairs T and P that can appear in the extensions of spacegroups, we can complete the classification of space groups up to equivalence. According to Sub-section 7.4, the possible extensions up to equivalence are given by the cohomology group H2(P,T ).We first consider the case in which the point group is cyclic. In Example 7.1, we computed that,for a cyclic group P, we have for n> 1,

H2(P,T ) = H2n(P,T ) = T P/NT,

where T P denotes the fixed points of the action of P on T and N denotes the norm element of thegroup ring, N = ∑g∈G g. If P is cyclic of order 3, 4, or 6, then P is a rotation of E2. Rotations haveno non-zero fixed points, so in these cases, H2(P,T ) = 0. Therefore all space groups with pointgroup C3 are equivalent, and the same holds for C4 or C6.

If P = C2, we have a lot less information as C2 can appear as a subgroup of any of the fiveBravais point groups from the table. Now, we can consider how C2 acts on the three lattices, T1, T2,and T3. Let Z denote the trivial C2Z-module and Z denote the C2Z-module which is isomorphicto Z as an abelian group yet on which the generator of C2 acts as −1. Then, one can compute thatT1 = Z⊕ Z, and T2 = Z⊕ Z, and T3 = ZC2, a free module of rank 1. There are no fixed points ofT1 under the C2 action, so H2(C2,T1) = 0. Similarly, T3 is the regular representation, and ZC2 hasno zero-divisors, so there are also no fixed points of T3 under the C2 action and H2(C2,T3) = 0.However, by direct computation, H2(C2,T2)∼= Z/2Z.

Therefore, for any pair (P,T ) where P is cyclic and T embeds in E2 as one of the three latticesfrom the chart of Bravais lattices, there is only one equivalence class of space groups for (P,T ),unless P = C2 and T embeds in E2 as T2. In this exceptional case, we can determine if the spacegroups are equivalent using the following proposition.

Proposition 8.15. Two space groups which are extensions of P by T are equivalent if and onlyif their cohomology classes in H2(P,T ) belong to the same orbit in the action of the normalizerNGL(T )P.

We also note that, in the case in which P is a dihedral group, this proposition also applies, butthe overall calculation of H2(P,T ) is significantly more difficult.

This entire section has been an application of what is called the Zassenhaus algorithm, whichworks to determine the equivalence classes of crystal structures of arbitrary dimension.

Example 8.16. Let’s apply these ideas to determine the equivalence of the crystal structures fromExample 8.7 and 8.8. Looking back at these examples, in both cases we computed that the point

GROUP THEORY 29

group P was cyclic of order 2. Looking at the lattices in the table of Bravais lattices, since thecheckmark patterns are aligned as in T2, the translation subgroup T will embed in E2 as T2. Re-viewing our case analysis, this was unfortunately the only case in which our point group was cyclicbut there are two equivalence classes of space groups. Therefore, it is possible that these spacegroups are not equivalent. However, rather than compute the relevant normalizer and its actionon the second cohomology group, we will note that in Example 8.7, we computed S(C) = 〈t2,s1〉and in Example 8.8, we computed S(C) =

⟨t1, t2,s1 | s2

1 = 1⟩. Namely, the first space group is

torsion free while the second has an element of finite order. Therefore, the space groups are notisomorphic, so they certainly cannot be equivalent, so the underlying crystal structures are also notequivalent.

9. THE BURNSIDE RING AND MARKS HOMOMORPHISM

Throughout this section, G will denote a finite group and X a finite G-set. Recall the statementof the Orbit-Stabilizer theorem.

Theorem 9.1 (Orbit-Stabilizer Theorem). Let G be a finite group that acts on a finite set X. Fixx ∈ X. Then there is a G-set isomorphism

Orb(x)∼= G/Stab(x).

We can rewrite X as a disjoint union of orbits⊔

Xi. If we choose an element xi ∈ Xi, we canfurther rewrite

X =⊔

G/Gi,

where Gi = Stab(xi). This decomposition is not independent of the choice of the xi, but it is upto the conjugacy class of the subgroups Gi since if two elements belong to the same orbit Xi, theirstabilizers are conjugate. This is motivation to define the Burnside ring.

Definition 9.2. The Burnside ring, denoted b(G) is the ring of isomorphism classes of finite G-setswith addition as disjoint union and multiplication as Cartisian product.

As a result of the remarks preceding the defintion, an arbitrary element of b(G) can be expressedas

n

∑i=1

ai[G/Gi]

where ai ∈ Z, the Gi are a complete set of representatives for the conjugacy classes of subgroupsof G, and [G/Gi] denotes the G-set isomorphism class of G/Gi.

Definition 9.3. If G acts on X and H 6 G, then the mark of H on X is defined as

mX(H) = |XH |= #x ∈ X | hx = x for all h ∈ H.

Studying marks will give us a much clearer understanding of the Burnside ring. First, noticethat if H and K are conjugate, the is a natural bijection between the fixed point sets XH and XK ,and therefore mX(H) = mX(K). Additionally, it can be easily checked that fixed points worknicely with disjoint union and cartesian products. So the map b(G)→ Z defined for a fixed H byX 7→ mX(H) is a ring homomorphism. Since the G-sets G/Gi form an additive basis of b(G), we


will consider the case where the G-set X is G/K for some subgroup K. With this as motivation, fora pair of subgroups K, H of G, we define

m(K,H) := mG/K(H) = |(G/K)H |= #gK | HgK = gK.

If we let G1,G2, . . . ,GN be a set of representatives for the conjugacy classes of subgroups of G,then we can define a ring homomorphism ~m : b(G)→ ZN by mi(X) = m(Gi,X). If we take ZN tobe a ring under pointwise addition and multiplication, then this is a ring homomorphism.

As an important note, from the definition of m(G,K), a coset gK is a fixed point wheneverHgK = gK. When this is satisfied, gHg−1 ≤ K. Therefore, if there is a fixed point of the action ofH on G/K, then H is conjugate to a subgroup of K. Therefore if H is not conjugate to a subgroupof K, there are no such fixed points, so m(K,H) = 0. Without loss of generality, assume that weordered our conjugacy classes representative such that if Gi is conjugate to a subgroup of G j theni≤ j. This can easily be done by ordering according to the cardinality of the subgroup.

From this we construct a matrix with (i, j)-entry as m(Gi,G j) called the table of marks. Therows of this matrix are the row vectors ~m(Gi). By our assumed ordering on the conjugacy classes,this matrix will be lower triangular. Additionally, for a given i, since the identity coset of G/Giwill always be fixed by the action of Gi, the entries on the diagonal of the table of marks are atleast 1.

Example 9.4. As an example, let’s compute the table of marks for G = Z/6Z. Consider thesubgroups G1 = 1, G2 = Z/2Z, G3 = Z/3Z, and G4 = G. We label the rows of our table of marksby the G-sets G/Gi and the columns by the subgroups G j. For now, we only fill in the fact that Gis lower triangular.

1 Z/2Z Z/3Z GG/1 0 0 0G/(Z/2Z) 0 0G/(Z/3Z) 0G/G

Since every element of every G-set is fixed by the action of the trivial subgroup 1, the leftmostcolumn is the size of the G-set. Additionally, the trivial G-set G/G contains only one elementwhich must be fixed by the action of every group element, so the bottom row is all 1’s. This givesus

1 Z/2Z Z/3Z GG/1 6 0 0 0G/Z/2Z 3 0 0G/Z/3Z 2 0G/G 1 1 1 1

.

For the final three entries, we first note that Z/2Z is not conjugate to a subgroup of Z/3Z,so the remaining non-diagonal entry is 0. For the (2,2) entry, the right cosets G/(Z/2Z) are0,3, 1,4 and 2,5. The left action of group Z/2Z= 0,3 fixes all three of these cosets, som(Z/2Z,Z/2Z) = 3. By a similar argument, m(Z/2Z,Z/2Z) = 2. These computations are notparticularly exciting since G is abelian. Therefore, the table of marks is

GROUP THEORY 31

1 Z/2Z Z/3Z GG/1 6 0 0 0G/(Z/2Z) 3 3 0 0G/(Z/3Z) 2 0 2 0G/G 1 1 1 1

.

Example 9.5. As a second example, let’s compute the table of marks for G = S3. Since S3 andZ/6Z have the same subgroups, we again set G1 = 1, G2 = Z/2Z, G3 = Z/3Z, and G4 = G. Theleftmost column and the bottom row are exactly as in the last example.

1 Z/2Z Z/3Z GG/1 6 0 0 0G/(Z/2Z) 3 0 0G/(Z/3Z) 2 0 0G/G 1 1 1 1

However, this time, to compute m(Z/2Z,Z/2Z), we denote Z/2Z as (),(12). Then, theright cosets S3/(Z/2Z) are (),(12), (13),(123), and (23),(132). The left action of (12)fixes the first coset but exchanges the second two, and therefore there is only one fixed coset, som(Z/2Z,Z/2Z) = 1. We leave it to the reader to confirm that m(Z/3Z,Z/3Z) = 2, and thereforethe table of marks is

1 Z/2Z Z/3Z GG/1 6 0 0 0G/(Z/2Z) 3 1 0 0G/(Z/3Z) 2 0 2 0G/G 1 1 1 1

.

We now state the following theorem of Burnside which explains the close relation between theBurnside ring of a group and its table of marks.

Theorem 9.6 (Burnside). Let M denote the table of marks for the Burnside ring b(G). Given aarbitrary finite G-set X, we can express X as a sum

n

∑i=1

ai[G/Gi]

where~aM = ~m(X).

Since M is lower triangular with non-zero entries on the diagonal, M is an invertible matrix, sothe coefficient vector ~a can be recovered from the mark vector ~m(X). As a final preliminary, since~m is a ring homomorphism, it follows that the multiplication and addition in b(G) can be moreeasily understood by studying the table of marks. Consider the following example.

Example 9.7. In b(S3), we will compute the product [S3/(Z/3Z)] · [S3/(Z/2Z)]. The mark vectorsfor these two elements are (3,1,0,0) and (2,0,2,0), whose componentwise product is (6,0,0,0).


By the remarks above, the coefficients of this product as a sum of basis vectors can be recov-ered from the table of marks. Notice that this row vector is exactly the top row of the table, so[S3/(Z/3Z)] · [S3/(Z/2Z)] = [S3/1].

For the remainder of this section, we will discuss how to find idempotents in the Burnside ring.Recall that that an element e in a ring is idempotent if e2 = e. An idempotent is primitive ifwhenever e = e1 + e2 is a sum of idempotents, then e = e1 or e = e2.

Also recall that a group G is perfect if G = G′ = [G,G], that is G is equal to its commutatorsubgroup. A group is solvable if there is a series of subgroups 1 = G0 ≤G1 ≤ . . .Gn = G such thatGi−1 ≤Gi and Gi/Gi−1 is abelian. Equivalently, G is solvable if and only if the derived series of Gterminates at the trivial group. As a simple exercise, the reader can check that G is solvable if andonly if the trivial group is the only perfect subgroup of G. We also note that for any n, the dihedralgroup D2n is solvable which means S3 ∼= D6 is solvable.

To study idempotents in the Burnside ring, let B(G) denote the rational Burnside ring Q⊗Zb(G).By the functoriality of Q⊗Z−, the homomorphism ~m induces a ring homomorphism ~m : B(G)→Q⊗ZZN = QN , where N denotes the number of conjugacy classes of subgroups of G. It can beshown that this induced map is a ring isomorphism. Since Z embeds naturally in Q, we can thinkof b(G)⊆ B(G).

Since this induced map is surjective, define the element eH ∈ B(G) as the element such that~m(eH) = (0,0, . . . ,0,1,0, . . . ,0) where the 1 is in the position indexed by the conjugacy class of H.Since ~m is a ring isomorphism, if e is an idempotent in B(G), the vector ~m(e) consists only of 0’sand 1’s. Therefore, any such idempotent e can be written as a sum

e = ∑H⊆H

eH ,

where H is some indexing set such that no two elements of H are conjugate in G (which is to saythat no eH is repeated). The following theorem describes when such an idempotent of B(G) willlie within b(G).

Theorem 9.8 (Dress). An idempotent ∑H∈H

eH ∈ B(G) lies in b(G) if and only if whenever we have

two subgroups H ≤H ′≤G where H is normal in H ′ and H ′/H is cyclic, then H ∈H ⇐⇒ H ′ ∈H.

And, as a result of this theorem, the following corollary can be deduced, which completelydescribes the primitive idempotents of b(G).

Corollary 9.9 (Dress). The primitive idempotents of B(G) that are also contained in b(G) are inbijection with perfect subgroups of G. More specifically, if M ≤G is perfect, the element ∑H∈H eHis a primitive idempotent in b(G) where H is the set of all conjugacy classes of subgroups H of Gwhose derived series terminates at M.

Example 9.10. As an example, we shall find a primitive idempotent in b(S3). By the previouscorollary, the primitive idempotents of b(G) are in bijection with the perfect subgroups of S3.However, S3 is solvable, so the only perfect subgroup of S3 is 1. Since the derived series ofevery subgroup of S3 terminates at 1, the primitive idempotent is the sum of all non-conjugateeH elements in B(G). Therefore, the primitive idempotent in S3 is the element e with ~m(e) =

GROUP THEORY 33

(1,1,1,1). This is exactly the multiplicative identity 1 = [G/G] ∈ b(G). Since 1 is a primitiveidempotent in b(G), it follows that 1 is the only non-zero idempotent in b(G). Indeed, if e 6= 1were another nonzero idempotent, then (1− e) is idempotent also and 1 = e+(1− e) contradictsthe primitivity of 1.

The example above can be generalized through the following corollary.

Corollary 9.11. Let G be a finite group. The following are equivalent.(i) G is solvable.

(ii) 1 is a primitive idempotent of b(G).(iii) The only idempotents in b(G) are 0 and 1.

MIKE LOPER: SCHOOL OF MATHEMATICS, UNIVERSITY OF MINNESOTA, MINNEAPOLIS, MINNESOTA, 55455,UNITED STATES OF AMERICA; [email protected]

GREG MICHEL: SCHOOL OF MATHEMATICS, UNIVERSITY OF MINNESOTA, MINNEAPOLIS, MINNESOTA,55455, UNITED STATES OF AMERICA; [email protected]

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