Chapter 8. Finite Groups

The aim of this chapter is to prove several central results of the theory of finite groups: Theorems of Schur and Zassenhaus and Burnside’s transfer theorem (aslo known as Burnside’s normal �-complement theorem).

Throughout this chapter, unless otherwise stated, G denotes a finite group in multiplicative notation.

References:

[Bro94]

New York, 1994.

[Rot09] J. J.

tics, vol. 148, Springer-Verlag, New York, 1995.

23 Cohomology of Finite Groups

To begin with, we collect in this section a few general results about the cohomology of finite groups.

Lemma 23.1

If G is a finite group, then |G| ¨ H

pG� Mq “

for every �

Proof : Let 1 denote the trivial group. Because 1 is a cyclic group of order one Theorem 16.2 yields H^�p1�Mq – 0 if � • 1 (whereas H⁰p1�Mq – M). Now, the composition of the restriction with the transfer

H^�pG�Mq ^resG1 // Hloooomoooon^�p1�Mq

–0if�•1 tr^G1

//H^�pG�Mq

equals multiplication by the index of|G:1| “ |G|by Proposition 21.3 and factors through0if�•1by the above. Therefore multiplication by|G|is zero inH^�pG�Mqif�•1.

Proposition 23.2

If G is a finite group and M is a finitely generated

H

pG� Mq is a finitely generated abelian group and H

pG� Mq is a finite abelian group of exponent dividing |G| for all �

Proof : Fix�PZ•0.

Claim 1: H^�pG�Mqis a finitely generated abelian group.

79

Indeed: Using the fact that ZG is a noetherian ring as G is finite, we may construct a projectiveZG- resolutionP‚ of Z in which all the modules are finitely generated abelian groups. Now, applying the functor HomZGp´�Mq to P‚ we again obtain complexes of finitely generated abelian groups since for each �•0, HomZGpP��Mq “HomZpP��Mq^G ÑHomZpP��Mq, which is a finitely generated abelian group if bothP� andM are. The cohomology groups of this complex is again finitely generated.

Claim 2: H^�pG�Mqis a finite group if�•1.

Indeed: SinceH^�pG�Mqis a finitely generated abelian group by the first claim and|G| ¨H^�pG�Mq “0 by Lemma 23.1,H^�pG�Mqmust be torsion, hence finite.

Exercise [Exercise 1, Exercise Sheet 12]

If M is a

H

pG� Mq “

for all �

Deduce that H

pG� Mq “

for all �

if M is a projective

Exercise [Exercise 2, Exercise Sheet 12]

Let � be a prime number, let G be a finite group of order divisible by �, and let P be a Sylow

�-subgroup of G. If M is an

G-module, then the restriction map

H

pG� Mq

Ñ H

pP �

pMqq is injective for all �

24 The Theorems of Schur and Zassenhaus

In this section we prove two main results of the theory of finite groups, which are often considered as one Theorem and called the

we differentiate between the abelian and the non-abelian case.

Theorem 24.1 (Schur, 1904)

Let G be a finite group and let A “ pA� ¨� ˚q be a

G-module such that there exists � P

with

�

“

for all � P A. Suppose that

|G|� �

“

(a) Every group extension

A

E

G

inducing the given G-action on A splits.

(b) Any two complements of A in E are E-conjugate.

Proof : We prove that theH^�pG�Aqis trivial for all�•1. For convenience, writeAadditively in this proof.

Thus by assumption we have�¨A“0. By Lemma 23.1, we know that|G| ¨H^�pG�Aq “0for all�•1.

Since�¨A“0, we also have�¨C^�pG�Aq “0, and thus�¨H^�pG�Aq “0. Now, by the Bézout identity there exists�� �PZsuch that

�¨ |G| `�¨�“1� and hence

H^�pG�Aq “�¨ |G| ¨loooooooomoooooooonH^�pG�Aqq

“0

`�¨�loooooomoooooon¨H^�pG�Aq

“0

“0�

Since H²pG�Aq vanishes any extension splits and since H¹pG�Aq vanishes all complements of A are E-conjugate, by Theorem 19.3 and Theorem 18.3(b) respectively.

Theorem 24.2 (Zassenhaus, 1937 )

Let

A

E

G

be an extension of finite groups (where A is not necessarily abelian). If

|A|� |G|

“

Proof : W.l.o.g. we may assume that |G|•1. Then we proceed by induction on the size ofA.

¨ If|A| “1or|A|is prime, thenAis abelian. Moreover,�^|A| “1for all�PA. Thus Schur’s Theorem applies and yields the result.

¨ Suppose now that |A| P Z•2zP. Let � P P be a prime number dividing |A|, let P be a Sylow

�-subgroup ofA, and setN:“NEpPqfor the normaliser ofP inE.

Claim 1: E“AN.

Indeed, if�PE, thenP and�P�^´1 are Sylow�-subgroups of A, henceA-conjugate, so that there exists �PA such that�P�^´1 “�P�^´1. Thus `

�^´1�qP`

�^´1�˘´1

“P, i.e. �^´1�PNEpPq “N, and therefore�“�`

�^´1�˘ PAN.

Claim 2: Ahas a complement inE.

We split the proof of this claim in two cases:

Case 1: N‰E.

In this case, restricting�toN yields the group extension

1 //AXN ^� //N ^�|N //G //1

whereAXNàAbecauseG–N{pAXNq –AN{A“E{A. Thus, by the induction hypothesis, this extension splits. Hence, by Proposition 17.6, there exists a complementH ofAXN in N. Since

|H| “ |G|, we have HXA“ t1u, and thereforeH is a complement ofAinE.

Case 2: N“E.

LetZ :“ZpPqbe the centre ofP, which is a non-trivial subgroup ofP becauseP is a�-group.

SinceZ is characteristic inP (i.e. invariant under all automorphisms ofP), and sinceP is normal inN, we deduce thatZ is normal in E. Thus, by the universal property of the quotient,�induces a group homomorphism�:E{Z›ÑG� �Z ›Ñ�p�q, whose kernel isA{Z. In other words, there is a group extension of the form

1 //A{Z //E{Z //G //1�

Now |Z| ‰ 1implies that |A{Z| †|A|, hence, by the induction hypothesis again, this extension splits. So letF be a complement ofA{Z inE{Z. By the Correspondence Theorem, there exists a subgroupFr §E containingZ such thatF “Fr{Z. In other words, there is a group extension

1 //Z //Fr //F //1� SinceF –GandZ §P§A, we have`

|Z|�|F|˘

“1and therefore this extension splits by Schur’s Theorem. Thus, there is a complementH of Z inFr. But|H| “ |G| impliesHXA“1, so thatH is also a complement ofA inE. The second claim is proved.

We conclude that the extension splits using Proposition 17.6 .

Remark 24.3

Notice that both Schur’s and Zassenhaus’ Theorems can be stated in easy terms not involving

cohomology, but their proofs rely on cohomological methods.

25 Burnside’s Transfer Theorem

Throughout this section, we let H be a subgroup of G of index |G

H | “: � and A be a

G- module. Our first aim is to understand the action of the transfer homomorphism on H

pG� Aq. So first recall that H

pG� Aq “ Z

pG� Aq “

pG� Aq by Example 10, and hence we see the transfer as a homomorphism

tr^G_H : HomGrp

pH� Aq

Ñ

pG� Aq � Lemma 25.1

Let � P

and let

be the �-th term of the bar resolution of

as a

a projective resolution of

as a

F -module by restriction. Fix a right transversal S “ t�

� � � � � �

u of H in G. Then the comparison maps between this and the bar resolution of

as a

H-module are given by the canonical inclusion

�

ÑZG

and by the map

�

:

G

Ñ

H

p�

� � � � � �

q

p�

� � � � � �

q , where, for every

�

�, �

“ �

�

for some �

P H and some �

P S.

Proof : Using the definition of the differential maps of the bar resolution, we see that there are commutative diagrams

ZH^�`1

��

✏✏

�� //

ö

ZG^�`1

��

✏✏

ZH

ε

✏✏

�0 //

ö

ZG�

ε

✏✏ZH^� _��´1 //ZG^� Z _Id //Z

and ZG^�`1

��

✏✏

��

//

ö

ZH^�`1

��

✏✏

ZG

ε

✏✏

�0

//

ö

ZH�

ε

✏✏ZG^� _��´1 //ZH^� Z _Id //Z

(where�•1). Thus the Comparison Theorem yields the result.

Proposition 25.2

Fix a right transversal S “ t�

� � � � � �

u of H in G. Then the transfer for H

is described as follows:

tr^G_H : HomGrp

pH � Aq

Ñ

pG� Aq

�

Ñ

˜

tr^G_H

p�q

�

Ñ

�“1

� p�

q

¸

�

where �

� “ �

� with �

P H for all

�

�.

Proof : On the one hand

HomGrppH�Aq “H¹pH�Aq –H¹`

HomZHpZH²�Aq˘

� via the bar resolution. On the other hand,

HomGrppH�Aq “H¹pH�Aq –H¹`

HomZHpZG²�Aq˘

�

via the bar resolution forGrestricted toH. Now, transfer is defined using the second resolution, therefore, we need to compare these two resolutions. But

H¹pH�Aq “Z¹pH�Aq�

becauseB¹pH�Aq “0sinceH acts trivially onA, and

Z¹pH�AqÑC¹pH�Aq –HomZH`

ZH²�A˘

� If for a given� PHomGrppH�Aq,˜� denotes the image of � inHomZH`

ZH²�A˘

, then for�PH set

˜� :ZH² ›ÑA p1� �q “ r�sfi›Ñ�p�q and thus for each� PH,

˜�`

p�� ��q˘

“˜�`

�¨ p1� �q˘

“�¨�p�q “�p�q�

becauseH acts trivially on A, and we extend this map by Z-linearity to the whole of ZH². Using the comparison map�1:ZG²›ÑZH² of the previous lemma yields˜�˝�1:ZG²›ÑA, which isZH-linear.

Now, computing the transfer using its definition yields for every�PZG²: tr^G_H`˜�˝�1˘

p�q “ÿ^�

�“1�^´1_� `˜�˝�1˘

p���q ““ÿ^�

�“1

`˜� ˝�1˘ p���q because �^´1₁ � � � � � �^´1_� (

is a left transversal ofHinGandAis trivial. We want to view this as a1-cocycle forG, that is evaluate this on an elementr�s “ p1� �q PG²ÄZG² :

tr^GHp�qp�q “tr^GH`˜�˝�1˘

p1� �q “ÿ^�

�“1

`˜�˝�1˘

p��� ���q “ÿ^�

�“1

˜�` 1� ��˘ because��“1¨�� and ���“���σp�q. So we obtain

tr^GHp�qp�q “ÿ^�

�“1

˜�p1� ��q “ ÿ�

�“1�p��q� as required.

Lemma 25.3 (Choice of transversal for a fixed

P

) Fix � P G.

(a) There exists a right transversal of H in G of the form

S “ �

� �

�� � � � � �

�

� �

� �

�� � � � � �

�

� � � � � �

� �

�� � � � � �

�

with �

` � � � ` �

“ |G

H| “ � and �

�

�

P H for all

�

�.

(b) If � P

pH� Aq and � P G, then

p�qp�q “

�“1

�

�

�

�

.

Proof :

(a) The element� acts on the right on right cosetsH�, viaH�fi›ÑH��. Decompose the set of right cosets into�-orbits. Let�be the number of�-orbits and letH�1� � � � �H��be representatives of the

�-orbits. We get all the right cosets of H by applying powers of �to each H�� and we suppose that H���^��´1� “H�� (that is, �� is the cardinality of the orbit). With this choice, we obtain a right transversal with the required properties.

(b) Proposition 25.2 together withpaqyield for1§�§�:

���� ���²� � � � � ���^��´1 belong to the right transversalS, so that`

���^�˘

¨�“1¨`

���^�`1˘

PH¨S for0§� §��´2and

`���^��´1�˘

“`

���^����^´1˘

�� PH¨S. Therefore tr^G_Hp�qp�q “ÿ^�

�“1�`

���^���_�^´1˘

� because all the other elements ofH appearing are 1and �p1q “0.

Theorem 25.4 (Burnside’s transfer theorem (or Burnside’s normal p-complement theorem), 1911) Let G be a finite group, let � be a prime number such that � | |G| and let P be a Sylow �-subgroup of G. If P is abelian and C

pPq “ N

pPq, then there exists a normal complement N to P in G, i.e.

G “ N ¸ P .

Proof :

Claim: If there exist�PG and�PP such that�� ���^´1PP, then���^´1 “�.

Indeed: using the assumptions, we have�P�^´1P�, which is abelian, and therefore bothP and�^´1P�

are Sylow �-subgroups of CGp�q. Thus P and �^´1P� are conjugate in CGp�q, so that there exists

�PCGp�qsuch that�P�^´1“�^´1P�, that is��Pp��q^´1“P and hence��PNGpPq “CGpPq. Finally,

���^´1“ p��q�p��q^´1“�because�PP, as required.

Now consider the identity mapIdP PHomGrppP�Pqand tr^GPpIdPq PHomGrppG�Pq. The previous lemma yields for a fixed�PP,

tr^GPpIdPqp�q “π^�

�“1IdP`

���^����^´1

looomooon

PP

˘�

Now using the claim yields���^����^´1 “�^�� for each1§�§�, hence tr^GPpIdPqp�q “π^�

�“1�^�� “�^|G:P|�

In particular, this proves that tr^GPpIdPq: G ⇣P is a surjective group homomorphism, because for each

� PP, there exists�PP such that�^|G:P| “� by the Bézout identity. (Indeed, Bézout implies that there exist�� �PZsuch that�|P| `�|G:P| “1, hence� “�¹“ p�^�q^|G:P| and we choose�“�^�.)

Finally, setN:“ker´

tr^GPpIdq¯

, so that we have a group extension

1 //N //G^trGPpIdPq//P //1

with a section given by _|G:P|¹ ¨�, where � : P ›Ñ G is the canonical inclusion. It follows that N is a normal complement ofP inG.