In this chapter we consider connections between the short exact sequences of groups of the form
1݄ A
݄ E
݄ G
݄
1with abelian kernel and the cohomology of the group G with coefficients in A. If the sequence splits, then we shall prove that the 1st cohomology group H
1pG� Aq parametrises the splittings. Moreover, we shall also prove that the 2nd cohomology group H
2pG� Aq is in bijection with the isomorphism classes of extensions
1݄ A
݄ E
݄ G
݄
1inducing the given
ZG-module structure on A, and the neutral element of H
2pG� Aq corresponds, under this bijection, to a s.e.s. where E is a semi-direct product of A by G.
References:
[Bro94]
K. S. B����,Cohomology of groups, Graduate Texts in Mathematics, vol. 87, Springer-Verlag, New York, 1994.
[Eve91]
L. E����,The cohomology of groups, Oxford Mathematical Monographs, The Clarendon Press, Oxford University Press, New York, 1991.
[Rot09] J. J.
R�����,An introduction to the theory of groups. Fourth ed., Graduate Texts in Mathema- tics, vol. 148, Springer-Verlag, New York, 1995.
[Wei94] C. A.
W�����,An introduction to homological algebra, Cambridge Studies in Advanced Mathe- matics, vol. 38, Cambridge University Press, Cambridge, 1994.
17 Group Extensions
In Chapter 1, we have seen that if a group G is a semi-direct product of a subgroup N by a subgroup H, then this gives rise to a s.e.s. of the form
1›
Ñ N
݄ G
݄ H
݄
1�This is a special case of a so-called group extension of N by H . Definition 17.1 (Group extension)
A group extension is a short exact sequence of groups (written multiplicatively) of the form
1 //A
� //E
� //G
//1�and, in this situation, we also say that the group E is an extension of A by G.
Convention: We shall always identify A with a normal subgroup of E and assume that
�is simply
61
the canonical inclusion of A in E. Moreover, we shall say that A is the kernel of the extension.
Lemma 17.2
Let
1 //A
� //E
� //G
//1be a group extension, where A is an abelian group. Then A is naturally endowed with the structure of a
ZG-module.Proof : First note that with the above notationpA�¨qis a group written multiplicatively. Next, for each�PG, choose a preimager�PE of�under�, that is�pr�q “�, and define a leftG-action onA via:
˚: GˆA ›Ñ A
p�� �q fiÑ �˚�:“��:“r�¨�¨�r´1,
First, we check that˚is well-defined, i.e. that this definition is independent of the choice of the preimages:
indeed, if�pPE is such that�pp�q “�, then, we have
�`
�r¨p�´1˘
“�¨�´1 “1G
hence�r¨p�´1 Pkerp�q “A, and thus, there exists�PA such thatr�“�p�. Therefore, for every�PA,
�r¨�¨r�´1“���ploomoonp �´1
AúE
�´1 “��´1p��p�´1“p��p�´1� where the last-but-one equality holds becauseA is abelian.
We extend˚byZ-linearity to the whole ofZG, and finally one easily checks thatpA�¨�˚qis aZG-module.
See Exercise 2, Exercise Sheet 10.
Convention: From now on, given a group extension
1 //A
� //E
� //G
//1with A abelian, we always see A as a
ZG-module via the G-action of the proof of Lemma 17.2. We write A
˚ :“ pA�¨� ˚q to indicate that we see A as a
ZG-module in this way.Lemma 17.3
Let
1 //A
� //E
� //G
//1be a group extension with A abelian. Then, A is central in E if and only if A
˚is trivial as a
ZG-module.
Proof : A˚ is a trivialZG-module ñ ��“� @�PA�@�PG ñ r�¨�¨�r´1 “� @�PA�@r�PE
ñ r��“�r� @�PA�@r�PE
ñ AÑZpEq�
Definition 17.4 (Central extension)
A group extension
1 //A
� //E
� //G
//1be a group extension with A abelian satisfying the equivalent conditions of Lemma 17.3 is called a central extension of A by G.
Definition 17.5 (Split extension)
A group extension
1 //A
� //E
� //G
//1splits iff there exists a group homomorphism
�:
G
݄ E such that
�˝
�“
IdG. In this case
�is called a (group-theoretic) section of
�, or asplitting of the extension.
Proposition 17.6
Let
1 //A
� //E
� //G
//1be a group extension. Then the following are equivalent:
(a) The extension splits.
(b) There exists a subgroup H of E such that
�ˇˇH :H
݄ G is an isomorphism.
(c) There exists a subgroup H of E such that E is the internal semi-direct product of A by H.
(d) There exists a subgroup H of E such that every element
�P E can be written uniquely
�“
��with
�P A and
�P H .
Proof :(a)ñ(b): By (a) there exists a section�:G›ÑEfor�. DefineH:“Im�. Then�ˇˇH is an isomorphism since, on the one hand �ˇˇH˝� “IdG by definition of �, and on the other hand for every �P H, there exists�PGsuch that �“�p�q, so that
p�˝�ˇˇHqp�q “ p�˝�qp�p�qq “�p�q “� and�˝�ˇˇH“IdH.
(b)ñ(c): By (b) there isH§E such that�ˇˇH:H ›ÑG is an isomorphism. Hence t1u “ker`
�ˇˇH˘
“kerp�q XH“AXH� Now, let�PE. Then�p�q PG ñ`
�ˇˇH˘´1
˝�p�q PH and�p�q “�`
�ˇˇ´1H ˝�p�q˘
, so that
�¨´`
�ˇˇH˘´1
˝�p�q¯´1
Pker�“A� Therefore, there exists�PA such that
�“�¨´`
�ˇˇH˘´1
˝�p�q¯ looooooooomooooooooon
PH
PAH
as required.
(c)ñ(d): Was proven in Step 1 of the proof of Proposition 1.3.
(d)ñ(b): We have to prove that�ˇˇH:H ›ÑG is an isomorphism.
Surjectivity: Let�PG. Then by surjectivity of�there exists�PE such that�“�p�q, and by (d),
�can be written in a unique way as�“��with�PA and�PH. Hence�ˇˇH is surjective since
�“�p�q “�p��q “�p�q�p�q “1¨�p�q “�p�q� Injectivity: If�PH is such that�ˇˇHp�q “1, then�Pkerp�q “A, therefore
�“1¨�“�¨1PAH so that by uniqueness, we must have�“1andkerp�ˇˇHq “ t1u.
(b)ñ(a): If �ˇˇH : H ›Ñ G is an isomorphism, then we may define � :“ `
�ˇˇH˘
: G ݄ E. This is obviously a group homomorphism and hence a splitting of the extension.
If the equivalent conditions of the Proposition are satisfied, then there is a name for the subgroup H,
it is called a complement:
Definition 17.7 (Complement of a subgroup)
Let E be a group and A be a normal subgroup of E. A subgroup H of E is called a complement of A in E if E “ AH and A X H “
1, i.e. ifE is the internal semi-direct product of A by H.
Remark 17.8
Unlike short exact sequences of modules, it is not true that
�admits a group-theoretic section if and only if
�admits a group-theoretic retraction. In fact, if
�admits a group-theoretic retraction, then E – A ˆ G. (See Exercise Sheet 10.)
18 H
1and Group Extensions
In order to understand the connexion between the group extensions of the form
1 //A
//E
//G
//1with abelian kernel and H
1pG� A
˚q, first we need to investigate the automorphisms of E.
Definition 18.1 (Inner automorphisms, automorphisms inducing the identity) Let E be a group.
(a) Given
�P E, write
�� :E
›Ñ E� �
fiÑ���´1for the automorphism of E of conjugation by
�. (b) Set
InnpEq:“ t�P
AutpEq | D�P E with
�“
��u .
(c) If A
§G, then set
InnApE q
:“ t�P
AutpEq | D�P A with
�“
��u .
(d) If
1 //A
� //E
� //G
//1is a group extension with abelian kernel, then set
AutA�GpEq
:“ t�P
AutpEq |�|A“
IdAand
�˝
�p�q “�p�q @�P Eu
�We say that the elements
�of
AutA�GpEq induce the identity on both A and G.
Recall (e.g. from the Einführung in die Algebra-lecture) that:
InnpEqEAutpEq, as
�˝�
�˝�
´1“
��p�qfor every
�P E and every
�P
AutpEq, and the quotient AutpEq{
InnpEqis called the outer automorphism group of E . Moreover,
InnpEq –E{Z pEq. It is also obvious that
AutA�GpEq
§AutpEq.Theorem 18.2 (H
1 and automorphisms)Let
1 //A
� //E
� //G
//1be a group extension with abelian kernel. Then:
(a) H
1pG� A
˚q –
AutA�GpEq{
InnApE q; and
(b) if, moreover, the extension is a central extension then
H
1pG� A
˚q –
AutA�GpEq
�Proof :
(a) Claim 1: InnApEqEAutA�GpEq.
Indeed, clearly for each�PA, ��|A“IdA becauseAis abelian and, moreover,
�˝��p�q “�`
���´1˘
“�p�q�`
��´1�´1˘
�p�q “�p�q
for every� PE, so that �˝�� “�. Therefore InnAE §AutA�GpEq, and it is a normal subgroup, because
�˝��˝�´1“��p�q“��
for every�PA, every�PAutA�GpEqas�|� “IdA. Claim 2: AutA�GpEq –Z1pG�A˚q.
We aim at defining a group isomorphism
α: AutA�GpEq ›Ñ Z1pG�A˚q .
¨ To begin with, we observe that given�PAutA�GpEqand�PE, we can write�p�q “�p�q� for some element�p�q PE. This defines a map (of sets)
�: E ›Ñ E
� fiÑ �p�q�´1, such that Imp�qÑA“kerp�qbecause for every�PE,
�p�p�qq “�p�p�q�´1q “�p�p�qqlooomooon
“�p�q
�p�´1q “1G
since�induces the identity onG. Moreover,�is constant on the cosets ofEmoduloAbecause
�p��q “�p��q ¨ p��q´1“�p�q ¨loomoon�p�q
“�
¨�´1¨�´1 “�p�q�´1“�p�q�
Therefore � induces a map� :G ›ÑA� �fiÑ�p�q:“�pr�qwhere we may chooser�arbitrarily in�´1p�q. This is a1-cocycle since for all�� �PG, we may choose��ÄP�´1p��q,r�P�´1p�q, and r�P�´1p�qsuch that��Ä“r�r�, and hence
�p��q “�`��Ģ
“�`
r�r�q “�pr�q ¨�pr�q ¨r�´1¨r�´1
“�pr�q ¨r�´1¨r�¨�p�q ¨r�´1“�p�q��p�q� which is the1-cocycle identity in multiplicative notation.
¨ As a consequence, we set
αp�q:“`
� :G›ÑA˘
�
To prove that this defines a group homomorphism, let �1� �2PAutA�GpEqand respectively let
�1� �2:G›ÑA be the associated1-cocycles, i.e. αp�1q “�1 and αp�2q “�2. Then
�1pr�q “�1p�qr�� �2pr�q “�2p�qr� @�PGwith�rP�´1p�q� and hence using the fact that Ais abelian yields
αp�1˝�2qp�q “ p�1˝�2qpr�qr�´1“�1`
�2p�qr�˘ r�´1
“�2p�q�1pr�qr�´1
“�2p�q�1p�qr�r�´1
“�2p�q�1p�q
“αp�1qp�q ¨αp�2qp�q “`
αp�1q ¨αp�2q˘ p�q� as required.
¨ In order to prove thatα is an isomorphism, we define β: Z1pG�A˚q ›Ñ AutA�GpEq
� fiÑ βp�q:E›ÑE��rfiÑ�p�q� �r where�“�pr�q.
First, we check that βp�q is indeed a group homomorphism: for r��r� P E with the above notation, we have
βp�q`
�r¨r�˘
“�p��qr�r� 1-cocycle id.
“ �p�q ¨��p�q ¨r�r�
“�p�qr��p�qr�´1�r�r
“�p�qr��p�qr�
“βp�qpr�q ¨βp�qp�qr � Next, if r�PA“kerp�q, then�“1G and therefore
βp�qpr�q “�p1q ¨�r“1¨r�“r� �
where we use the fact that a 1-cocycle is always normalised (indeed �p1Gq “ 1A, since for
� P G, �p1G¨�q “ �p1Gq ¨p1Gq�p�q “ �p1Gq�p�q by the 1-cocycle identity). Thus we have proved thatβp�q|A“IdA.
Furthermore, since�p�q PA“kerp�q,�p�p�qq “1G and we get
`�˝βp�q˘
pr�q “�`
�p�q ¨�r˘
“�`
�p�q˘ looomooon
“1G
¨�pr�q “�pr�q and so�˝βp�q “�, or in other wordsβp�qinduces the identity onG.
Finally, using Exercise 2(c), Exercise Sheet 10, we obtain that any group homomorphism E ›ÑE inducing the identity on A and on G must be an isomorphism. Therefore, we have proved thatβp�q PAutA�GpEqfor every�PZ1pG�A˚q.
¨ It remains to prove thatα andβ are inverse to each other. Firstly,
`pα˝βqp�q˘
p�q “βp�qpr�q ¨r�´1“�p�qr�r�´1 “�p�q @�PG�@�PZ1pG�A˚q� so thatα˝βis the identity on Z1pG�A˚q. Secondly,
`pβ˝αqp�q˘
pr�q “ pαp�qqp�q ¨�r´1“�p�qr�r�´1 “�p�q @r�PE�@�PAutA�GpEq� so thatβ˝α is the identity on AutA�GpEq.
Claim 3: InnApEq –B1pG�A˚q.
¨ Let�PA and��PInnApEq. Then for every�PG, αp��qp�q “��pr�q ¨�r´1“�¨�loooooomoooooonr¨�´1¨�r´1
PAúE
“r��´1�r´1¨�“�p�´1q�“�˚1p�´1qp�q and therefore αp��q PB1pG�A˚q, i.e. α`
InnAE˘
ÑB1pG�A˚q.
Conversely, if �PAand�˚1p�q PB1pG�A˚q, then�˚1p�qp�q “�� ¨�´1 and β`
�˚1p�q˘
pr�q “�˚1p�qp�q ¨�r“��¨�´1¨r�“r�¨�¨r�loooooomoooooon´1¨�´1¨r�
PAúE
“r�loomoon¨r�´1
“1
�´1r�¨�“��´1pr�q�
Hence β`
�˚1p�q˘
“ ��´1 P InnApEq, and β`
B1pG�A˚q˘
Ñ InnApEq. It follows that InnApEq corresponds to AutA�GpEqunder the bijection given byα andβ, and we obtain
H1pG�A˚q “Z1pG�A˚q{B1pG�A˚q –AutA�GpEq{InnApEq�
(b) If A is a central subgroup ofE, then for every�PA the conjugation automorphism by� is given by�� :E›ÑE� �fiÑ���´1 “��´1�“�, i.e. the identity on E. Thus
InnApEq “ t��:E›ÑE|�PAu “ tIdEu and it follows from (a) that
H1pG�A˚q –AutA�GpEq{InnApEq “AutA�GpEq�
We are now ready to parametrise the slpittings of split group extensions with abelian kernel:
Theorem 18.3 (H
1 and splittings)Let
�‚ :“ p1 //A
� //E
� //G
//1q be a split group extension with abelian kernel. Then the following holds:
(a) There is a bijection between H
1pG� A
˚q and the set
�of A-conjugacy classes of splittings of the given extension.
(b) There is a bijection between H
1pG� A
˚q and the set of E-conjugacy classes of complements of A in E.
Proof :
(a) Choose a splitting�0 :G›ÑE and define a map
α: AutA�GpEq ›Ñ tsplittings of�‚u
� fiÑ �˝�0.
It is obvious thatαis well-defined, i.e. that�˝�0is a splitting of the extension as���0“��0“IdG. Define a second map
β: tsplittings of�‚u ›Ñ AutA�GpEq
� fiÑ `
ψ�:E ›ÑE� ��0p�qfiÑ��p�q˘ ,
where by Proposition 17.6 an arbitrary element�PEcan be written in a unique way as�“��0p�q with�PA and�PG. We check thatβis well-defined. Firstly, ψ� is a group homomorphism: for every�1 “�1�0p�1q� �2“�2�0p�2q PE, we have
ψ�p�1¨�2q “ψ�`
�1�0p�1q ¨�2�0p�2q˘
“ψ�`
�1¨�1�2¨�0p�1�2q˘
“�1¨�1�2¨�`
�1�2˘
“�1�p�1q ¨�2�p�2q
“ψ�`
�1�0p�1q˘
¨ψ�`
�2�0p�2q˘
“ψ�p�1q ¨ψ�p�2q� Secondly,ψ�|A “IdA by definition. Thirdly,�ψ�“�since for�“��0p�q PE, we have
p�˝ψ�qp�q “ p�˝ψ�qp��0p�qq “�p��p�qq “loomoon�p�q
“1
¨�p�p�qqloomoon
“IdGp�q
“�“�p��0p�qq�
Finally, the fact that ψ� is an isomorphism follows again from Exercise 2(c), Exercise Sheet 10 becauseψ� induces the identity on bothAand G. Whenceβ is well-defined.
Next, we check thatα andβ are inverse to each other. On the one hand, pα˝βqp�q “αpψ�q “ψ�˝�0 @�P tsplittings of�‚u
but for every�PG, pψ�˝�0qp�q “ψ�p1A¨�0p�qq “1G�p�q “�p�q, henceα˝β is the identity on the set of splittings of�‚. On the other hand, for every�PAutA�GpEq, we have
pβ˝αqp�q “βp�˝�0q “ψ�˝�0
and for each�“��0p�q PE (with �PAand �PG), we have ψ�˝�0
`��0p�q˘
“�¨ p�˝�0qp�q �|A““IdA �p�q ¨ p�˝�0qp�q “�`
�¨�0p�q˘
� henceβ˝α is the identity onAutA�GpEq.
Therefore,
AutA�GpEq α // tsplittings of�‚u oo β
are bijections (of sets). Finally, we determine the behaviour ofInnApEqunder these bijections. Let
�PAutA�GpEqand��PInnApEqwith�PA. Let�1“��˝�. Then αp�q “�˝�0 and αp��˝�q “��˝�˝�0�
Hence a coset moduloInnApEqis mapped viaα to an equivalence class for the action by conjugation ofAon splittings
Aˆ tsplittings of�‚u ›Ñ tsplittings of�‚u p�� �q fiÑ ��˝�.
Thus passing to the quotient (group quotient on the left hand sideAutA�GpEq, and orbits ofInnApEq on the right hand side) yields a bijection
AutA�GpEq{InnApEq „ //
OO
pThm.18�2q –
✏✏
tA-conjugacy classes of splittings of�‚u
H1pG�A˚q as required.
(b) By Proposition 17.6, a splitting�of the extension corresponds to a complement�pGqofAinE, and conversely, a complementH of A inE corresponds to a splitting `
�ˇˇH˘´1 : G ›ÑH. Moreover, theA-conjugacy class of H is the same as theE-conjugacy class of H, because every�PE may be written in a unique way as�“��with�PA and�PH and so �H�´1 “�H�´1. The claim follows.
19 H
2and Group Extensions
Convention: In this section all group extensions are assumed to have abelian kernel.
Definition 19.1 (Equivalent group extensions)
Two group extensions
1 //A
� //E
� //G
//1and
1 //A
�1 //E
1 �1 //G
//1with abelian kernels are called equivalent if there exists a group homomorphism
� :E
݄ E
1such that the following diagram commutes
1 //
A
� //IdA
✏✏ ö
E
� //�
✏✏ ö
G
//IdG
✏✏
1 1 //
A
�1 //E
1 �1 //G
//1�Remark 19.2
(a) In the context of Definition 19.1, the homomorphism
�is necessarily bijective. However an isomorphism of groups does not induce an equivalence of extensions in general. In other words, the same middle group E can occur in non-equivalent group extensions with the same kernel A, the same quotient G and the same induced
ZG-module structure onA.
(b) Equivalence of group extensions is an equivalence relation.
Notation: If G is a group and A
˚ :“ pA�¨� ˚q is a
ZG-module (which may see simply as an abelian group), then we let
�pG� A
˚q denote the set of equivalence classes of group extensions
1 //
A
� //E
� //G
//1inducing the given
ZG-module structure on A.
Theorem 19.3
Let G be a group and let A
˚:“ pA�¨� ˚q be a fixed
ZG-module (written multiplicatively). Then, thereis a bijection
H
2pG� A
˚q
oo „ // �pG� A
˚q
�Moreover, the neutral element of H
2pG� A
˚q corresponds to the class of the split extension.
Proof : We want to define a bijection�pG�A˚q›ÑH2pG�A˚q.
¨ To begin with, fix an extension
1 //A � //E � //G //1
inducing the given action˚onA, and we choose a set-theoretic section�:G›ÑE for�, i.e. such that�˝�“IdG. Possibly�is not be a group homomorphism, but we may write
�p�q ¨�p�q “�p�� �q ¨�p��q for some element�p�� �q PE. This defines a map
�: GˆG ›Ñ E
p�� �q fiÑ �p�� �q:“�p�q ¨�p�q ¨�p��q´1. Furthermore, notice that�p�� �q PA“kerp�qbecause
�`
�p�� �q˘
“�`
�p�q�p�q�p��q´1˘
“�`
�p�q˘
¨�`
�p�q˘
¨�`
�p��q˘´1
“���´1�´1“1G
for every�� �PG. Hence� PHomSetpGˆG�Aq, and as a matter of fact,� is a2-cocycle because:
`�p�q ¨�p�q˘
¨�p�q “�p�� �q ¨�p��q ¨�p�q “�p�� �q ¨�p��� �q ¨�p���q and
�p�q ¨`
�p�q ¨�p�q˘
“�p�q ¨�p�� �q ¨�p��q “�p�q ¨�p�� �q ¨�p�q´1¨�p�q ¨�p��q
“��p�� �q ¨�p�� ��q ¨�p���q�
Therefore, by associativity inE, we obtain
�p�� �q ¨�p��� �q “��p�� �q ¨�p�� ��q�
which is precisely the2-cocycle identity in multiplicative notation.
Now, we note that if we modify�by a 1-cochain�:G›ÑAand define
�1: G ›Ñ E
� fiÑ �1p�q:“�p�q ¨�p�q, then the corresponding2-cocycle is given by
�1p�� �q “�1p�q ¨�1p�q ¨�1p��q´1
“�p�q ¨�p�q ¨�p�q ¨�p�q ¨�p��q´1¨�p��q´1
“�p�q ¨�p�q ¨�p�q ¨�p�q´1�p�q ¨�p�q ¨�p��q´1¨�p��q´1
“�p�q ¨�p�q ¨�p�q ¨�p�´1q ¨�p�� �q ¨�p��q´1
“�p�q ¨��p�q ¨�p��q´1¨�p�� �q asAis abelian
“��p�q ¨�p��q´1¨�p�q ¨�p�� �q asAis abelian
“ p�˚2p�qqp�� �q ¨�p�� �q @�� �PG�
To sum up, we have modified the2-cocycle�by the2-coboundary�˚2p�q. Therefore, the cohomology class r�s :“�B2pG�A˚qof � inH2pG�A˚q is well-defined, depending on the given extension, but does not depend on the choice of the set-theoretic section�. Hence, we may define a map
ξ: �pG�A˚q ›Ñ H2pG�A˚q r1 //A �//E �//G //1s fiÑ r�s.
¨ We check thatξ is well-defined. Suppose that we have two equivalent extensions r1 //A �//E �//G //1s “ r1 //A �1//E1 �1//G //1s P�pG�A˚q� that is a commutative diagram of the form
1 //A � //
IdA
✏✏
ö
E � //
�
✏✏
ö
G //
IdG
✏✏
1
1 //A �1 //E1 �1 //G //1
where� is an isomorphism ofE›ÑE1. As above, we choose a set-theoretic section�:G›ÑE of�, and it follows that�˝� is a set-theoretic section for�1, since�1˝�˝�“�˝�“IdG. The corresponding2-cocycle is given by
�1p�� �q “ p�˝�qp�q ¨ p�˝�qp�q ¨ p�˝�qp��q´1 “�`
�p�q ¨�p�q ¨�p��q´1˘
“�`
�p�� �q˘
“�p�� �q @�� �PG as�|A“IdA. Henceξ is well-defined.
¨ Remark: We may choose � : G ›Ñ E is such that �p1q “ 1, and the associated 2-cocycle is normalised. Now if we modify � by a normalised 1-cochain � : G ›Ñ A (i.e. such that
�p1q “ 1), then �˚2p�q is a normaised2-coboundary. Therefore, we may as well use normalised cocycles/cochains/coboundaries.
¨ Surjectivity ofξ:
LetαPH2pG�A˚qand choose a normalised2-cocycle� :GˆG›ÑAsuch thatα “ r�s. Construct E�:“AˆG(as a set), which we endow with the product
p�� �q ¨ p�� �q “`
�¨��¨�p�� �q� �¨�˘
@�� �PA�@�� �PG�
ThenpE��¨qis a group whose neutral element isp1�1q. (Exercise, Exercise Sheet 11) Clearly there are group homomorphisms:
�:A›ÑE�� �fi›Ñ p��1q�
�:E�›ÑG� p�� �qfi›Ñ�
such thatkerp�q “Imp�q, thus we get a group extension 1 //A � //E� �
//G //1�
We need to prove that the cohomology class of the 2-cocycle induced by this extension via the above construction is preciselyr�s. So consider the set-theoretic section�:G›ÑE�� �fi›Ñ p1� �q and compute that for all�� �PG, we have
�p�q ¨�p�q ¨�p��q´1 “ p1� �q ¨ p1� �q ¨ p1� ��q´1
“`
1¨�1¨�p�� �q� ��˘
¨`p��q´1
�p���p��q´1q´1�p��q´1˘
“ p�p�� �qp��qp��q´1�p���p��q´1q�p��qp��q´1q
“`
�p�� �q�1˘ as required.
¨ Injectivity ofξ: Let
r1 //A � //E � //G //1s� r1 //A ˜� //E˜ ˜� //G //1s
be two classes of group extensions in�pG�A˚q. Choose, respectively,�:G›ÑE and˜�:G›ÑE˜ two set-theoretic section with corresponding2-cocycles� and˜� respectively. Now, assume that
r�s ““˜�‰
PH2pG�A˚q�
Then ˜� “ �˚2p�q ˝� for some 1-cochain � : G ›Ñ A. Changing the choice of ˜� by defining
˜˜� : G ›Ñ E˜� � fiÑ �p�q´1¨˜�p�q modifies˜� into �˚2p�q´1˝˜� by the first part of the proof. But
�˚2p�q´1˝˜� “�, therfore, we may assume without loss of generality that the two 2-cocycles are the same. Compute the group law inE: each element ofE can be written uniquely as�¨�p�qfor
�PAand �PGbecause�:G›ÑE is a section for�:E›ÑG. Hence the product is
��p�q ¨��p�q “��p�q��p�q´1�p�q�p�q
“����p�q�p�q
“�loooomoooon���p�� �q
PA
�p��q
which is exactely the group law inE�. HenceE� –E (viap�� �qfiÑ�¨�p�q) as groups, but also as extensions, because the latter isomorphism induces the identity on bothA andG. Similarly, we get thatE˜ –E�, as group extensions. The injectivity ofξ follows.
¨ Finally notice that the image underξ of the split extension
1 //A //A¸G //G //1
where the action of G on A is given by ˚, and where the first map is the canonical inclusion and the second map the projection onto G, is trivial. This is because we can choose a section
� : G ›Ñ A¸G� � fiÑ p1� �q, which is a group homomorphism. Therefore the corresponding 2-cocycle is� :GˆG›ÑA�p�� �qfiÑ1. This proves the 2nd claim.
Remark 19.4
(a) In the above proof, if we choose
�:G
݄ E such that
�p1q “1, then we obtain a normalised 2-cocycle. If we modify � :G
݄ A by a
1-cocycle � :G
݄ A such that
�p1q “ 1(a normalized
1-cochain), then ��is a normalized
2-coboundary. So we see that we can usenormalized cochains, cocycles and coboundaries throughout.
(b) If the group A is not abelian, then H
3`G� Z pAq
˘comes into play for the classification of the extensions. This is more involved.
Example 11
For example, if we want to find all
2-groups of order 2�(�
•3) with a central subgroup of order2and a corresponding dihedral quotient, then we have to classify the central extensions of G
:“D
2�´1by A
:“C
2. By Theorem 19.3 the isomorphism classes of central extensions of the form
1›
