nLab
group extension

Idea

A group extension of a group $G$ by a group $A$ is group $\hat{G}$ that sits in an exact sequence $A\to \hat{G}\to G$.

(For now we will put emphasis on the general case i.e. nonabelian group extensions, later we plan a separate entry on abelian group extensions.)

Definition

We say that a sequence of groups

(1)$$1\to K\stackrel{i}{\to }G\stackrel{p}{\to }B\to 1$$1\to K \overset{i}\to G\overset{p}\to B\to 1

is exact if $i$ is monomorphism, $p$ an epimorphism and $\mathrm{Ker}(p)=\mathrm{Im}(i)$. In terms of one-object groupoids, this is equivalent to saying that $BK\to BG\to BB$ is a fibration sequence. If $K$ and $B$ are given groups, then any exact sequence of the form above is called an extension of $B$ by $K$ (some say conversely of $K$ by $B$). Letters are here chosen to suggest that $B$ is a base to which $G$ projects and $K$ is the kernel of $i$ and for $G$ itself we often say that it is a group extension of $B$ by $K$.

If $K$ is abelian then we write $0$ instead of $1$ at the left start of the sequence, and if $B$ is exact we do the same at the right-hand end of the sequence.

We say that two group extensions $1\to K\stackrel{i}{\to }G\stackrel{p}{\to }B\to 1$ and $1\to K\stackrel{i\prime }{\to }G\prime \stackrel{p\prime }{\to }B\to 1$ are equivalent if there is an isomorphism $ϵ:G\to G\prime$ such that $ϵ\circ i=i\prime$ and $p=p\prime \circ ϵ$. In other words, there is a commutative diagram

(2)$$\begin{array}{ccccccc}1\to & K& \stackrel{i}{\to }& G& \stackrel{p}{\to }& B& \to 1\\ & \downarrow =& & \downarrow ϵ& & \downarrow =& \\ 1\to & K& \stackrel{i\prime }{\to }& G\prime & \stackrel{p\prime }{\to }& B& \to 1\end{array}$$\array{ 1\to &K&\stackrel{i}\to &G&\stackrel{p}\to &B&\to 1\\ &\downarrow\mathrlap{=}&&\downarrow\mathrlap\epsilon&&\downarrow\mathrlap=&\\ 1\to &K&\stackrel{i'}\to &G'&\stackrel{p'}\to& B&\to 1 }

Properties

Central group extensions

Let $A\to G\to H$ be a central extension, with $A$ an abelian group included in the center of $G$. Then $A$ is in particular a normal subgroup and hence the morphism

$$(A\to G)$$(A \to G)

may be regarded as a crossed module of groups. This is equivalently a strict 2-group structure on the groupoid whose objects are $G$ and whose morphisms are labeled in $A$

$$(A\to G)=\{g\stackrel{a}{\to }a\cdot g\}.$$(A \to G) = \{ g \stackrel{a}{\to} a \cdot g \} \,.

The delooping of this is the one-object 2-groupoid $B(A\to G)$.

The ω-nerve (or Duskin nerve) $NB(A\to G)\in$ sSet of this is a (3-coskeletal) Kan complex that realizes this is a 2-truncated infinity-groupoid.

The obvious strict 2-functor

$$B(A\to G)\to BH$$\mathbf{B}(A \to G) \to \mathbf{B}H

is an equivalence of 2-groupoids. One way to see this is to notice that this is a k-surjective functor for all $k\in \{0,1,2,3\}$, hence a weak equivalence in the folk model structure on $\omega$-groupoids. Equivalently, under the nerve

$$NB(A\to G)\to NBH$$N\mathbf{B}(A \to G) \to N \mathbf{B}H

is an acyclic Kan fibration, hence a weak equivalence in the standard model structure on simplicial sets.

This weak equivalence is the tool to extend the short exact sequence $A\to G\to H$ to the corresponding long fiber sequence. In the (∞,1)-category of ∞Grpd this goes

$$A\to G\to H\to BA\to BG\to BH\to B^{2}A.$$A \to G \to H \to \mathbf{B}A \to \mathbf{B}G \to \mathbf{B}H \to \mathbf{B}^2 A \,.

The last step is modeled in terms of strict n-functors between strict ω-groupoids by the 2-anafunctor

$$\begin{array}{ccc}B(A\to G)& \to & B(A\to 1)=B^{2}A\\ {}^{\simeq }\downarrow \\ BH\end{array}.$$\array{ \mathbf{B}(A \to G) &\to & \mathbf{B}(A \to 1) = \mathbf{B}^2 A \\ {}^{\mathrlap{\simeq}}\downarrow \\ \mathbf{B}H } \,.

Conversely, this $BH\to B^{2}A$ is the cocycle that classifies the principal 2-bundle $BG\to BA$ (which one may think of as the bundle gerbe over $BH$ classified by $BH\to B^{2}A$).

Torsors

Prosition

For $A\stackrel{i}{\to }\hat{G}\stackrel{p}{\to }G$ a group extension, we have that $p:\hat{G}\to G$ is an $A$-torsor over $G$ where the action of $A$ on $\hat{G}$ is defined by

$$\rho :A\times \hat{G}\stackrel{(i,\mathrm{Id})}{\to }\hat{G}\times \hat{G}\stackrel{\cdot }{\to }\hat{G}.$$\rho : A \times \hat G \stackrel{(i,Id)}{\to} \hat G \times \hat G \stackrel{\cdot}{\to} \hat G \,.

Proof

That $\rho$ is indeed an action over $B$ in that

$$\begin{array}{ccccc}A\times \hat{G}& & \stackrel{\rho }{\to }& & \hat{G}\\ & {}_{p\circ p_2}\searrow & & {\swarrow }_p\\ & & \hat{G}\end{array}$$\array{ A \times \hat G &&\stackrel{\rho}{\to}&& \hat G \\ & {}_{\mathllap{ p \circ p_2}}\searrow && \swarrow_{\mathrlap{p}} \\ && \hat G }

follows from the fact that $p$ is a group homomorphism and that $A$ is in its kernel.

That $A$ is actually equal to the kernel gives the principality condition

$$(\rho ,p_2):A\times \hat{G}\stackrel{\simeq }{\to }\hat{G}{\times }_G\hat{G}.$$(\rho, p_2) : A\times \hat G \stackrel{\simeq}{\to} \hat G \times_G \hat G \,.

For $A$ an abelian group we may understand the $A$-torsor/$A$-principal bundle $\hat{G}$ as the delooping of the $BA$-principal 2-bundle $B\hat{G}\to \mathrm{BG}$ that is classified by (is the homotopy fiber of) the 2-cocycle in group cohomology $c:BG\to B^{2}A$ that classifies the extension.

All this is then summarized by the statement that

$$A\to \hat{G}\to G\to BA\to B\hat{G}\to BG\stackrel{c}{\to }{B}^{2}A$$A \to \hat G \to G \to \mathbf{B}A \to \mathbf{B}\hat G \to \mathbf{B}G \stackrel{c}{\to} \mathbf{B}^2 A

is a fiber sequence in ∞Grpd (or in ∞LieGrpd if we have Lie group extensions, etc).

Here we may think of $B\hat{G}$ as being the $BA$-principal 2-bundle over $BG$ classified by $c$. See the examples discussed at bundle gerbe.

Schreier theory for nonabelian group extensions

Traditional way

Otto Schreier (1926) and Eilenberg-Mac Lane (late 1940-s) developed a theory of classification of nonabelian extensions of abstract groups leading to the low dimensional nonabelian group cohomology. This is sometimes called Schreier’s theory of nonabelian group extensions.

The traditional Schreier-Mac Lane way to obtain nonabelian group 2-cocycle from a group extension as above starts with choosing a set-theoretic section of $p:G\to B$.

Note. The exposition which follows in this long “traditional” section of this entry is mainly from personal notes of Zoran Škoda from 1997.

Each element $g$ of $G$ defines an inner automorphism $\varphi (g)$ of $K$ by $\varphi (g)(k)={\mathrm{gkg}}^{-1}$. The restriction $\varphi {\mid }_K$ takes (by definition) values in the subgroup $\mathrm{Int}(K)$ of inner automorphisms of $K$. In fact $\varphi :G\to \mathrm{Inn}(G)\subset \mathrm{Aut}(K)$ is a homomorphism of groups.

If $g_1$ and $g_2$ are in the same left coset, that is $g_{1}K=g_{2}K$, then there is $k\in K$, $g_1=g_{2}k$, so that $\forall k\prime \in K$ we have $\varphi (g_{1}k\prime )=\varphi (g_2\mathrm{kk}\prime )=\varphi (g_2)\varphi (\mathrm{kk}\prime )$ and therefore $\varphi (g_{1}K)\subset \varphi (g_2)\mathrm{Int}(K)$. Thus we obtain a well-defined map ${\varphi }_{*}:G/K\to \mathrm{Aut}(K)/\mathrm{Int}(K)$. Choose a set-theoretic section of the projection $p:G\to B$ and let

(3)$$\psi \stackrel{\mathrm{def}}{=}\varphi \circ \sigma :B\to \mathrm{Aut}(K).$$\psi \stackrel{def}{=} \phi \circ \sigma: B \rightarrow Aut(K).

Warning. Unlike $\varphi$, the map $\psi$ is not a homomorphism of groups.

We attempt to reconstruct $G$ from the knowledge of $\psi$ and $K$. As a set, $G$ can be naturally identified with $B\times K$. Indeed, write each element $g\in G$ as $\sigma (b)k,b\in B,k\in K$ by setting $b=p(g),k=\sigma (p(g){)}^{-1}g$. Elements $b\in B$ and $k\in K$ in that decomposition are unique, and we get a bijection

(4)$$B\times K\ni (b,k)\mapsto \sigma (b)k\in G,$$B\times K\ni (b,k)\mapsto\sigma(b)k \in G,

whose inverse is the map $g\mapsto (p(g),\sigma (p(g){)}^{-1}g)$. By means of that bijection, $B\times K$ inherits the group structure from $G$. Let us figure out the multiplication rule on $B\times K.$ If $\sigma (b_1)k_1=g_1$, and $\sigma (b_2)k_2=g_2$, then,

(5)$$g_1{g}_2=\sigma (b_1)k_1\sigma (b_2)k_2=\sigma (b_1)\sigma (b_2)\sigma (b_2{)}^{-1}{k}_1\sigma (b_2)k_2.$$g_1g_2 = \sigma(b_1)k_1\sigma(b_2)k_2 = \sigma(b_1)\sigma(b_2)\sigma(b_2)^{-1}k_1\sigma(b_2)k_2.

Now $p(\sigma (b_1)\sigma (b_2))=p(b_1{b}_2)$ so

$$\chi (b_1,b_2)\stackrel{\mathrm{def}}{=}\sigma (b_1{b}_2{)}^{-1}\sigma (b_1)\sigma (b_2)\in K.$$\chi(b_1,b_2) \stackrel{def}{=} \sigma(b_1b_2)^{-1}\sigma(b_1)\sigma(b_2) \in K.

This formula clearly defines a function $\chi :B\times B\to K$. In this notation,

$$\begin{array}{ccc}{g}_1{g}_2& =& \sigma (b_1{b}_2)\chi (b_1,b_2)\varphi (\sigma (b_2{)}^{-1})(k_1)k_2\\ & =& \sigma (b_1{b}_2)\chi (b_1,b_2)[\psi (b_2{)}^{-1}(k_1)]k_2.\end{array}$$\array{ g_1g_2 & = & \sigma(b_1b_2)\chi(b_1,b_2)\phi(\sigma(b_2)^{-1})(k_1)k_2 \\ & = & \sigma(b_1b_2)\chi(b_1,b_2)[\psi(b_2)^{-1}(k_1)] k_2. }

and using bijection of $G$ with $B\times K$ this can be expressed in terms of elements in $B\times K$ so that

(6)$$(b_1,k_1)(b_2,k_2)=(b_1{b}_2,\chi (b_1,b_2)[\psi (b_2{)}^{-1}(k_1)]k_2).$$(b_1,k_1)(b_2,k_2) = (b_1b_2,\chi(b_1,b_2)[\psi(b_2)^{-1}(k_1)] k_2).

According to this formula, all the information about the multiplication is encoded in functions $\chi :B\times B\to \mathrm{Aut}(K)$ and $\psi :B\to \mathrm{Aut}(K)$, and we may forget about $\sigma$ at this point. However, not every pair $(\chi ,\psi )$ will give some multiplication rule on $B\times K$. Let $a,b,c\in B$, and $e=e_K$ be the unity element in $K$. Then

$$[(a,e)(b,e)](c,e)=(ab,\chi (a,b))(c,e)=(abc,\chi (ab,c)\psi (c{)}^{-1}(\chi (a,b))).$$[(a,e)(b,e)](c,e) = (a b, \chi(a,b))(c,e) = (a b c, \chi(a b,c) \psi(c)^{-1}(\chi(a,b))).

From the other side, this has to be the same, by associativity, to

$$(a,e)[(b,e)(c,e)]=(a,e)(\mathrm{bc},\chi (b,c))=(abc,\chi (a,bc)\chi (b,c))$$(a,e)[(b,e)(c,e)] = (a,e)(bc,\chi(b,c)) = (a b c,\chi(a,b c)\chi(b,c))

where we took into account that expressions like $[{\psi }^{-1}(b)(e)]=e$, because $\psi (b)$ is an automorphism for each $b\in B$.

Comparing the expressions above we obtain

(7)$$\chi (ab,c)\psi (c{)}^{-1}(\chi (a,b))=\chi (a,bc)\chi (b,c),\mathrm{for}\mathrm{all}a,b,c\in B.$$\chi(a b,c)\psi(c)^{-1}(\chi(a,b)) = \chi(a,b c)\chi(b,c), for all a,b,c \in B.

If the pair $(\chi ,\psi )$ is constructed as above, then

$$\begin{array}{ccc}\psi (a)\psi (b)k& =& \varphi (\sigma (a))\varphi (\sigma (b))k\\ & =& \sigma (a)\sigma (b)k\sigma (b{)}^{-1}\sigma (a{)}^{-1}\\ & =& \sigma (ab)\chi (a,b)k\chi (a,b{)}^{-1}\sigma (ab{)}^{-1}\\ & =& \psi (ab){\mathrm{Ad}}_K(\chi (a,b))k,\end{array}$$\array{ \psi(a)\psi(b)k & = & \phi(\sigma(a))\phi(\sigma(b))k \\ & = & \sigma(a)\sigma(b)k\sigma(b)^{-1}\sigma(a)^{-1} \\ & = & \sigma(a b)\chi(a,b)k\chi(a,b)^{-1}\sigma(a b)^{-1} \\ & = & \psi(a b) Ad_K(\chi(a,b)) k, }

where ${\mathrm{Ad}}_K$ is the canonical map $K\to \mathrm{Int}(K)$, $k\mapsto k(-)k^{-1}$.

Thus we obtain the relation

(8)$$\psi (a)\psi (b)=\psi (ab){\mathrm{Ad}}_K(\chi (a,b))$$\psi(a)\psi(b) = \psi(a b) Ad_K(\chi(a,b))

Definition

Let $B$ and $K$ be two groups. Let $\chi :B\times B\to K$ and $\psi :B\to \mathrm{Aut}(K)$ satisfy (7) and (8). Then we call that the family $\{\chi (b_1,b_2)\mid b_1,b_2\in B\}$ is a factor system (This term is due Schreier(1924)) or a nonabelian group 2-cocycle with automorphisms, and the family $\{\psi (b)\mid b\in B\}$ – a system of automorphisms

A 2-cocycle $\chi$ is counital if $\chi (b,e)=\chi (e,b)=e$, for all $b\in B$.

If $K$ is commutative, then $\psi$ is always a homomorphism (cf. (8)). Then $K$ is a right $B$-module through $\psi (-{)}^{-1}$. That justifies the sometimes used term “(right) cocycle $B$-module” for $(K,\psi ,\chi )$. If $\psi$ is trivial ($\psi (b)={\mathrm{Id}}_K,\forall b\in B$) then the cocycle condition (7) becomes

(9)$$\chi (ab,c)\chi (a,b)=\chi (a,bc)\chi (b,c).$$\chi(a b,c)\chi(a,b) = \chi(a,b c)\chi(b,c).

Theorem

If formulas (7) and (8) are both satisfied, then the formula (6) for multiplication of pairs defines a group multiplication on $B\times K$. That set, together with multiplication (6) is called the cocycle cross product of $B$ and $K$ with cocycle $\chi$ and action $\psi$. If the cocycle is trivial i.e. $\chi (\cdot ,\cdot )=e_K$, we call it the (external) semidirect product.

Proof

The checking the associativity we have done above for pairs of the form $(a,e)$ etc. This was useful to find the cocycle condition correctly. Now the general associativity should be a similar calculation with general elements. Using (7) and (8) it can be done.

$$\begin{array}{ccc}\psi (a)\psi (e)k& =& \psi (a){\mathrm{Ad}}_K(\chi (a,e))k\\ & =& \psi (a)\chi (a,e)k\chi (a,e{)}^{-1}\end{array}$$\array{ \psi(a)\psi(e)k & =& \psi(a)Ad_K(\chi(a,e))k \\ & = &\psi(a)\chi(a,e)k\chi(a,e)^{-1} }

where we used (8).

Thus ${\mathrm{Ad}}_K(\chi (a,e))=\psi (e)$ and therefore it does not depend on $a$.

Then use (7) with $b=c=e$ to get $\psi (e{)}^{-1}(\chi (a,e))=\chi (e,e),\forall a\in B$.

Thus $\chi (a,e{)}^{-1}(\chi (a,e))\chi (a,e)=\chi (e,e)$, that is $\chi (a,e)$ does not depend on $a$.

Now we claim that the unit element is given by $(e,\chi (e,e{)}^{-1})$. To verify that it is also a right unit we compute

$$\begin{array}{ccc}(a,b)(e,\chi (e,e{)}^{-1})& =& (a,\chi (a,e)\psi (e{)}^{-1}(b)\chi (e,e{)}^{-1})\\ & =& (a,\chi (a,e)\chi (e,e{)}^{-1}b\chi (e,e)\chi (e,e{)}^{-1})\end{array}$$\array{ (a,b)(e,\chi(e,e)^{-1}) & = & (a, \chi(a,e) \psi(e)^{-1}(b)\chi(e,e)^{-1}) \\ & = & (a, \chi(a,e)\chi(e,e)^{-1}b\chi(e,e)\chi(e,e)^{-1}) }

what is equal to $(a,b)$ by just proved statement that $\chi (a,e)$ does not depend on $a$.

Now use (7) with $a=b=e$ to get

(10)$$\psi (c{)}^{-1}(\chi (e,e))=\chi (e,c),\forall c\in B.$$\psi(c)^{-1}(\chi(e,e)) = \chi(e,c), \forall c \in B.

Thus we can verify that $(e,\chi (e,e{)}^{-1})$ is a left unit too by a calculation as follows. Namely

$$(e,\chi (e,e{)}^{-1})(a,b)=(a,\chi (e,a)\psi (a{)}^{-1}(\chi (e,e{)}^{-1})b)$$(e,\chi(e,e)^{-1})(a,b)= (a,\chi(e,a)\psi(a)^{-1}(\chi(e,e)^{-1})b)

by the definition of the product. Then by (10), this equals to

$$(a,\psi (a{)}^{-1}(\chi (e,e))\psi (a{)}^{-1}(\chi (e,e{)}^{-1})b)$$(a,\psi(a)^{-1}(\chi(e,e))\psi(a)^{-1}(\chi(e,e)^{-1})b)

and, because $\psi (a{)}^{-1}$ is an antiautomorphism, this is finally equal to $(a,b)$.

Now check that each element $(a,b)$ can be factorized as $(a,e)(e,\chi (e,e{)}^{-1}b)$. In order to show that $(a,b)$ has an inverse it is then enough to show that both $(a,e)$ and $(e,\chi (e,e{)}^{-1}b)$ have inverses.

Claim: the inverse of $(a,e)$ is

$$(a^{-1},\chi (a,a{)}^{-1}\chi (e,e{)}^{-1}).$$(a^{-1},\chi(a,a)^{-1}\chi(e,e)^{-1}).

To this aim, we calculate

$$(a,e)(a^{-1},\chi (a,a^{-1}{)}^{-1}\chi (e,e{)}^{-1})=(e,\chi (a,a^{-1})\psi (a^{-1}{)}^{-1}(e)\chi (a,a^{-1}{)}^{-1}\chi (e,e{)}^{-1})=(e,\chi (e,e^{-1}),$$(a,e)(a^{-1},\chi(a,a^{-1})^{-1}\chi(e,e)^{-1}) = (e,\chi(a,a^{-1})\psi(a^{-1})^{-1}(e)\chi(a,a^{-1})^{-1}\chi(e,e)^{-1}) =(e,\chi(e,e^{-1}),

because $\psi (a{)}^{-1}(e)=e$. Furthermore,

$$(a,e)(a^{-1},\chi (a,a^{-1}{)}^{-1}\chi (e,e{)}^{-1})=(e,\chi (a,a^{-1})\psi (a^{-1}{)}^{-1}(e)\chi (a,a^{-1}{)}^{-1}\chi (e,e{)}^{-1})=(e,\chi (e,e^{-1}),$$(a,e)(a^{-1},\chi(a,a^{-1})^{-1}\chi(e,e)^{-1}) = (e,\chi(a,a^{-1})\psi(a^{-1})^{-1}(e)\chi(a,a^{-1})^{-1}\chi(e,e)^{-1}) = (e,\chi(e,e^{-1}),

because $\psi (a{)}^{-1}(e)=e$. Next,

$$(a^{-1},\chi (a,a^{-1}{)}^{-1}\chi (e,e{)}^{-1})(a,e)=(e,\chi (a^{-1},a)\psi (a{)}^{-1}(\chi (a,a^{-1})\chi (e,e{)}^{-1}))$$(a^{-1},\chi(a,a^{-1})^{-1}\chi(e,e)^{-1})(a,e) = (e,\chi(a^{-1},a)\psi(a)^{-1}(\chi(a,a^{-1}) \chi(e,e)^{-1}))

what equals $(e,\chi (e,e{)}^{-1})$.

Indeed, (7) with $a=a,b=a^{-1},c=a$ reads $\chi (e,a)\psi (a{)}^{-1}(\chi (a,a^{-1}))$ $=\chi (a,e)\chi (a^{-1},a)$.

Then apply (10) and take inverse of both sides to obtain

$$\psi (a{)}^{-1}(\chi (a,a^{-1}{)}^{-1}\chi (e,e{)}^{-1}))=\chi (a^{-1},a{)}^{-1}\chi (a,e{)}^{-1}.$$\psi(a)^{-1}(\chi(a,a^{-1})^{-1}\chi(e,e)^{-1})) = \chi(a^{-1},a)^{-1}\chi(a,e)^{-1}.

Then recall that $\chi (a,e)$ does not depend on $a$ and multiply by $\chi (a^{-1},a)$ from the left.

Claim: the inverse of $(e,\chi (e,e{)}^{-1}k)$ is $(e,\chi (e,e)k^{-1})$. Here the verification is symmetric ($k$ vs. $k^{-1}$) for the left and for the right inverse and immediate.

Given groups $K$ and $B$ and any maps $\chi$ and $\psi$ satisfying (7) and (8), needed to define a cocycle cross product $B{\times }_{\chi }K$ of $K$ and $B$, one defines the map $i:K\to B{\times }_{\chi }K$ by $k\mapsto (e,\chi (e,e{)}^{-1}k)$. Then $i$ is a monomorphism of groups, $i(K)$ is a normal subroup of the cocycle cross product of $B$ and $K$, and there is a canonical isomorphism $B\cong G/K$. We define the set-theoretic maps $\sigma \prime ,\chi \prime$ and $\psi \prime$ as follows. $\sigma \prime :B\to B\times K$ is defined by $\sigma \prime (b)=(b,e)$ , for all $b\in B$. Then $\chi \prime :B\times B\to i(K)$ is defined by $\chi \prime (b_1,b_2)=\sigma \prime (b_1{b}_2{)}^{-1}\sigma \prime (b_1)\sigma \prime (b_2)$ and $\psi \prime :B\to \mathrm{Aut}(i(K))$ is defined by $\psi \prime (b)i(k)=\sigma \prime (b)i(k)\sigma \prime (b{)}^{-1}$. Using the natural identifications $i:K\cong i(K)$, and $i_{\mathrm{Aut}}:\mathrm{Aut}(i(K))\cong \mathrm{Aut}(K)$, we have $\psi \prime =i_{\mathrm{Aut}}\circ \psi$ and $\chi \prime =i\circ \chi$. Now

$$\begin{array}{ccc}\chi \prime =i\circ \chi & \iff & (b_1,e)(b_2,e)(e,\chi (e,e{)}^{-1}k)=(b_1{b}_2,e)(e,\chi (e,e{)}^{-1}\chi (b_1,b_2)k)\\ & \iff & (b_1{b}_2,\chi (b_1,b_2))(e,\chi (e,e{)}^{-1}k)=(b_1{b}_2,\chi (b_1{b}_2,e)\chi (e,e{)}^{-1}\chi (b_1,b_2)k)\\ & \iff & \chi (b_1{b}_2,e)\psi (e{)}^{-1}(\chi (b_1,b_2))\chi (e,e{)}^{-1}k=\chi (b_1{b}_2,e)\chi (e,e{)}^{-1}\chi (b_1,b_2)k\end{array}$$\array{ \chi'=i\circ\chi &\Leftrightarrow&(b_1,e)(b_2,e)(e,\chi(e,e)^{-1}k) =(b_1 b_2,e)(e,\chi(e,e)^{-1}\chi(b_1,b_2)k)\\ &\Leftrightarrow& (b_1b_2,\chi(b_1,b_2))(e,\chi(e,e)^{-1}k) = (b_1 b_2,\chi(b_1 b_2,e)\chi(e,e)^{-1}\chi(b_1,b_2)k)\\ &\Leftrightarrow& \chi(b_1 b_2,e)\psi(e)^{-1}(\chi(b_1,b_2))\chi(e,e)^{-1}k = \chi(b_1 b_2,e)\chi(e,e)^{-1}\chi(b_1,b_2)k }

for all $b_1,b_2\in B$ for all $k\in K$ in all these lines. The last line is true by (7).

Similarly, $\psi \prime =i_{\mathrm{Aut}}\circ \psi$ iff $(b,e)(e,\chi (e,e{)}^{-1}k)=(e,\chi (e,e{)}^{-1}\psi (b)k)(b,e)$ for all $b$ and $k$.

Here the LHS computes as $(b,k)$ using $\chi (b,e)=\chi (e,e)$, and the RHS is

$$(e,\psi (b)k)(b,e)=(b,\chi (e,b)\psi (b{)}^{-1}(\chi (e,e{)}^{-1}\psi (b)(k)))=(b,k)$$(e,\psi(b)k)(b,e) = (b, \chi(e,b)\psi(b)^{-1}(\chi(e,e)^{-1}\psi(b)(k))) = (b, k)

by (10).

Proposition. The following are equivalent

• (i) extension (1) is split

• (ii) for extension (1) there is a subgroup $B_1\subset G$ such that $B_1\cap i(K)=1$ and $B_{1}i(K)=G$ ($G$ is an internal semidirect product of $K$ and $B_1$).

• (iii) extension (1) is isomorphic to an external semidirect product of $K$ and $B$.

Proof. (i) $\Rightarrow$ (ii) If the extension is split then there is a homomorphism $\sigma :B\to G$ such that $p\circ \sigma ={\mathrm{id}}_B$. Let $B_1=\sigma (B)$. By exactness of (1)), all elements in $i(K)$ map $p$ sends to 1, and by $p\circ \sigma ={\mathrm{id}}_B$ map $p{\mid }_{B_1}$ is injection, therefore the only element in $i(K)$ which belongs to $B_1$ is 1.

$B_{1}i(K)=G$ is also obvious: e.g. for given $g\in G,$ $p(g)=p\sigma p(g)$, so that $p((\sigma p(g){)}^{-1}g)=1$ what means $(\sigma p(g){)}^{-1}g\in \mathrm{Ker}(p)$ so that $g=(\sigma p(g))i(k)$ for some $k\in K$ by exactness.

(ii) $\Rightarrow$ (iii) Our previous elaborate discussion of cocycle cross products makes it obvious: choosing a section $\sigma$ which is a homomorphism gives $\chi (a,b)=1$, and we can construct equivalent external semidirect product as a cocycle cross product with trivial $\chi$.

(iii) $\Rightarrow$ (i) Equivalence of extensions preserves the property of the corresponding short exact sequence to be split. Every external semidirect product is as a set $K\times B$ and the product is given by formula (6) without a cocycle. The map $\sigma :B\to G$, $B\ni b\mapsto (1_K,b)\in K\times B$, splits the sequence.

Definition. Extension (1) is Abelian iff $K$ is Abelian. An Abelian extension (1) is central iff it is isomorphic to a cocycle cross product extension with all the automorphisms $\psi (b),b\in B$ trivial. We say that the extension (1) is Abelian iff $G$ is Abelian.

Remarks. (i) Note that (8) implies that $\psi$ is a homomorphism if $K$ in the case of Abelian extensions (for any choice of set-theoretic section $\sigma$.

(ii) If $G$ is Abelian then (1) is central, but not every central extension is corresponding to an Abelian $G$. Abelian extensions in terms of the above definition trivially (strictly!) include both central extensions and extensions with $G$ central. By abuse of language one sometimes says for $G$ to be an extension of $K$ what leads to strange expression that not every Abelian extension (as extension – in terms of the definition above) is Abelian (as a group).

Comparing different extensions; 2-coboundaries

Let us now investigate when two extensions $G_1$ and $G_2$ of $B$ by $K$, given by $\psi ,\chi$ and $\psi \prime ,\chi \prime$ respectively, are equivalent, cf. diagram (2).

We know that $ϵ{\mid }_K:i(k)\stackrel{ϵ}{\mapsto }i\prime (k)$, for all $k\in K$. The formula for $i$ in \luse{crossform} says that whenever we represent an extension as a cocycle extension we have $i(k)=(e,\chi (e,e{)}^{-1}k).$ Thus $ϵ(e,\chi (e,e{)}^{-1}k)=(e,\chi \prime (e,e{)}^{-1}k)$, for all $k\in K.$ Also recall (or recalculate) that every element $(a,k)$ in $G$ can be factorized as $(a,e)(e,\chi (e,e{)}^{-1}k)$. By the definition $ϵ$ is a homomorphism of groups, so $ϵ(a,k)=ϵ(a,e)ϵ(e,\chi (e,e{)}^{-1}k)$. Also the cosets are preserved, so $ϵ(a,e)=(a,\lambda (a))$ where $\lambda :B\to K$ is some set-theoretic map. Thus

$$\begin{array}{ccc}ϵ(a,k)& =& (a,\lambda (a))(e,\chi \prime (e,e{)}^{-1}k)\\ & =& (a,\chi \prime (a,e)\psi \prime (e{)}^{-1}(\lambda (a))\chi \prime (e,e{)}^{-1}k)\\ & =& (a,\lambda (a)k).\end{array}$$\array{ \epsilon(a,k) & = & (a,\lambda(a))(e,\chi'(e,e)^{-1}k) \\ & = & (a, \chi'(a,e)\psi'(e)^{-1}(\lambda(a))\chi'(e,e)^{-1}k) \\ & = & (a, \lambda(a)k). }

Now multiply more general elements in $G$:

$$\begin{array}{c}ϵ((b_1,k_1)(b_2,k_2))=(b_1,\lambda (b_1)k_1)(b_2,\lambda (b_2)k_2)\\ =(b_1{b}_2,\chi \prime (b_1,b_2)\psi \prime (b_2{)}^{-1}(\lambda (b_1)k_1)\lambda (b_2)k_2)\end{array}$$\array{ \epsilon((b_1,k_1)(b_2,k_2)) = (b_1,\lambda(b_1)k_1)(b_2,\lambda(b_2)k_2) \\ = (b_1b_2,\chi'(b_1,b_2)\psi'(b_2)^{-1}(\lambda(b_1)k_1)\lambda(b_2)k_2) }

what should be the same as

(11)$$ϵ((b_1{b}_2,\chi (b_1,b_2)\psi (b_2{)}^{-1}(k_1)k_2))=(b_1{b}_2,\lambda (b_1{b}_2)\chi (b_1{b}_2)\psi (b_2{)}^{-1}(k_1)k_2)$$\epsilon((b_1b_2,\chi(b_1,b_2)\psi(b_2)^{-1}(k_1)k_2)) = (b_1b_2,\lambda(b_1b_2)\chi(b_1b_2)\psi(b_2)^{-1}(k_1)k_2)

In a special case, when $k_1=e_K$ we have therefore

(12)$$\chi (b_1,b_2)=\lambda (b_1{b}_2{)}^{-1}\chi \prime (b_1,b_2)\psi \prime (b_2{)}^{-1}(\lambda (b_1))\lambda (b_2)$$\chi(b_1,b_2) = \lambda(b_1b_2)^{-1}\chi'(b_1,b_2) \psi'(b_2)^{-1}(\lambda(b_1))\lambda(b_2)

In order to obtain a relation between $\psi \prime (b)(k)$ and $\psi (b)(k)$ note that

(13)$$ϵ((e,\chi (e,e{)}^{-1}k)(b,e))=(e,\chi \prime (e,e{)}^{-1}k)(b,\lambda (b)).$$\epsilon ((e,\chi(e,e)^{-1}k)(b,e)) = (e, \chi'(e,e)^{-1}k)(b,\lambda(b)).

That is equivalent to any in the following chain of formulas:

$$\begin{array}{ccc}ϵ(b,\chi (e,b)\psi (b{)}^{-1}(\chi (e,e{)}^{-1}k))& =& (b,\psi \prime (b{)}^{-1}(\chi \prime (e,e{)}^{-1}k)\lambda (b))\\ \iff \lambda (b)\chi (e,b)\psi (b{)}^{-1}(\chi (e,e{)}^{-1}k))& =& \chi \prime (e,b)\psi \prime (b{)}^{-1}(\chi \prime (e,e{)}^{-1}k)\lambda (b)\end{array}$$\array{ \epsilon (b,\chi(e,b)\psi(b)^{-1}(\chi(e,e)^{-1}k)) &=& (b, \psi'(b)^{-1}(\chi'(e,e)^{-1}k)\lambda(b)) \\ \Leftrightarrow \lambda(b)\chi(e,b)\psi(b)^{-1}(\chi(e,e)^{-1}k)) &=& \chi'(e,b)\psi'(b)^{-1}(\chi'(e,e)^{-1}k)\lambda(b) }

Then by (10), it follows that

$$\begin{array}{ccc}\lambda (b)\chi (e,b)\chi (e,b{)}^{-1}\psi (b{)}^{-1}(k)& =& \chi \prime (e,b)\chi \prime (e,b{)}^{-1}\psi \prime (b{)}^{-1}(k)\lambda (b)\\ \iff \lambda (b)\psi (b{)}^{-1}(k))& =& (\psi \prime (b{)}^{-1}(k))\lambda (b)\\ \iff ({\mathrm{Ad}}_K(\lambda (b))\circ \psi (b{)}^{-1})(k)& =& \psi \prime (b{)}^{-1}(k)\end{array}$$\array{ \lambda(b)\chi(e,b)\chi(e,b)^{-1}\psi(b)^{-1}(k) &=& \chi'(e,b)\chi'(e,b)^{-1}\psi'(b)^{-1}(k)\lambda(b) \\ \Leftrightarrow \lambda(b)\psi(b)^{-1}(k)) &=& (\psi'(b)^{-1}(k))\lambda(b) \\ \Leftrightarrow (Ad_K(\lambda(b)) \circ\psi(b)^{-1})(k) &=& \psi'(b)^{-1}(k) }

Now invert the maps in $\mathrm{Aut}(K)$ to obtain

(14)$$\psi \prime (b)=\psi (b){\mathrm{Ad}}_K(\lambda (b{)}^{-1})$$\psi'(b) = \psi(b)Ad_K(\lambda(b)^{-1})

Thus we obtain

Theorem. Two extensions of a group $B$ by group $K$ with corresponding systems $(\psi ,\chi )$ and $(\psi \prime ,\chi \prime )$ are equivalent iff there is a homomorphism $\lambda :B\to K$ such that the relations (12) and (14) are valid.

If function $\lambda$ takes values in the center of $B$ then (14) implies that $\psi \prime =\psi :B\to \mathrm{Aut}(K)$ and conversely.

If instead of functions $\psi$ and $\psi \prime$ we consider the respective maps into the group of external automorphisms (cosets of automorphisms with respect to the group of internal homomorphisms) $[\psi ],[\psi \prime ]:~B\to \mathrm{Aut}(K)/\mathrm{Int}(K)$, then the equivalent extensions define the same maps. By (8) these maps are actually homomorphisms (unlike e.g.$\psi$). For a given $\psi$ if there is $\chi$ so that $(\psi ,\chi )$ does define an extension of $B$ by $K$ we say that the extension is associated to (the homomorphism) $[\psi ]$. That does not mean that any given homomorphism in ${\mathrm{hom}}_{\mathrm{Group}}(B,\mathrm{Aut}(K)/\mathrm{Int}(K))$ is associated to any extension, nor it means that if a homomorphism is associated to some extension, that every its representative in ${\mathrm{hom}}_{\mathrm{Set}}(B,\mathrm{Aut}(K))$ is a part of a pair $(\psi ,\chi )$ defining an extension. To see that situation in more detail we start with a given automorphism, which we call $\theta$ , and choose an element $\psi (a)\in \theta (a)$, the representative of a coset in $\mathrm{Aut}(K)/\mathrm{Int}(K)$; that choice should be specified for all $a\in B$. Note that for any $\rho \in \mathrm{Aut}(K),a\in K$ we have, by direct inspection, $\rho {\mathrm{Ad}}_K(a){\rho }^{-1}={\mathrm{Ad}}_K(\rho (a))$. Thus there is a well-defined function

$${\mathrm{Ad}}_K\circ h:B\times B\to \mathrm{Int}(K),({\mathrm{Ad}}_K\circ h)(a,b):=\psi (ab{)}^{-1}\psi (a)\psi (b)$$Ad_K \circ h : B \times B \rightarrow Int(K), \,\,\, (Ad_K\circ h)(a,b) := \psi(a b)^{-1}\psi(a)\psi(b)

• indeed

(15)$$\psi (a){\mathrm{Ad}}_K(r_1)\psi (b){\mathrm{Ad}}_K(r_2)=\psi (a)\psi (b){\mathrm{Ad}}_K(\psi (b{)}^{-1}(r_1)r_2)$$\psi(a)Ad_K(r_1)\psi(b)Ad_K(r_2) = \psi(a)\psi(b)Ad_K(\psi(b)^{-1}(r_1)r_2)

so choosing $\psi (ab)\in [\psi (ab)]$ is the same as choosing it in $[\psi (a)][\psi (b)]$ and guarantees that $\psi (ab{)}^{-1}\psi (a)\psi (b)$ is in $\mathrm{Int}(K)$. Let us choose some $h$ so that ${\mathrm{Ad}}_K\circ h$ is interpretable as a genuine composition.

$$\begin{array}{ccc}(\psi (a)\psi (b))\psi (c)& =& \psi (ab){\mathrm{Ad}}_K(h(a,b))\psi (c)\\ & =& \psi (ab)\psi (c)\psi (c{)}^{-1}{\mathrm{Ad}}_K(h(a,b))\psi (c)\\ & =& \psi (abc){\mathrm{Ad}}_K(h(ab,c)){\mathrm{Ad}}_K(\psi (c{)}^{-1}h(a,b))\\ & =& \psi (abc){\mathrm{Ad}}_K(h(ab,c)\psi (c{)}^{-1}h(a,b))\end{array}$$\array{ (\psi(a)\psi(b))\psi(c) & = & \psi(a b)Ad_K(h(a,b))\psi(c) \\ & = & \psi(a b)\psi(c)\psi(c)^{-1}Ad_K(h(a,b))\psi(c) \\ & = & \psi(a b c)Ad_K(h(a b,c))Ad_K(\psi(c)^{-1}h(a,b)) \\ & = & \psi(a b c)Ad_K(h(a b,c)\psi(c)^{-1}h(a,b)) }

what is by associativity the same as

$$\begin{array}{ccc}\psi (a)(\psi (b)\psi (c))& =& \psi (a)\psi (bc){\mathrm{Ad}}_K(h(b,c))\\ & =& \psi (abc){\mathrm{Ad}}_K(h(a,bc)){\mathrm{Ad}}_K(h(b,c))\\ & =& \psi (abc){\mathrm{Ad}}_K(h(a,bc)h(b,c)).\end{array}$$\array{ \psi(a)(\psi(b)\psi(c)) & = & \psi(a)\psi(b c)Ad_K(h(b,c)) \\ & = & \psi(a b c)Ad_K(h(a,b c))Ad_K(h(b,c)) \\ & = & \psi(a b c)Ad_K(h(a,b c)h(b,c)). }

Thus ${\mathrm{Ad}}_K(h(ab,c)\psi (c{)}^{-1}h(a,b))={\mathrm{Ad}}_K(h(a,bc)h(b,c)).$ Two elements of $K$ generate the same automorphism iff they differ by a central element. Thus

(16)$$h(ab,c)\psi (c{)}^{-1}h(a,b)=h(a,bc)h(b,c)z(a,b,c)$$h(a b,c)\psi(c)^{-1}h(a,b) = h(a,b c)h(b,c)z(a,b,c)

for a unique central element $z(a,b,c)\in Z(K).$ The correspondence $z:(a,b,c)\mapsto z(a,b,c)$ maps $B\times B\times B$ into $Z(K)$.

Proposition. $z$ is an (Abelian) 3-cocycle with values in ${}_{/}Z(K{)}_{{\psi }^{-1}}$ ($Z(K)$ understood as trivial-${\psi }^{-1}$ $B$-bimodule):

(17)$$z(b,c,d)z(a,bc,d)\psi (d{)}^{-1}z(a,b,c)=z(a,b,cd)z(ab,c,d)$$z(b,c,d)z(a,b c,d)\psi(d)^{-1}z(a,b,c) = z(a,b,c d)z(a b,c,d)

To see this we calcuate

$$\begin{array}{cc}h(abc,d)[\psi (d{)}^{-1}h(ab,c)\psi (c{)}^{-1}h(a,b)]& =h(abc,d)[\psi (d{)}^{-1}h(a,bc)h(b,c)z(a,b,c)]\\ & =h(a,bcd)h(bc,d)z(a,bc,d)[\psi (d{)}^{-1}h(b,c)z(a,b,c)]\\ & =h(a,bcd)h(b,cd)h(c,d)z(b,c,d)z(a,bc,d)\psi (d{)}^{-1}z(a,b,c)\end{array}$$\array{ h(a b c,d)[\psi(d)^{-1}h(a b,c)\psi(c)^{-1}h(a,b)] & = h(a b c,d)[\psi(d)^{-1}h(a,b c)h(b,c)z(a,b,c)] \\ & = h(a,b c d)h(b c,d)z(a,b c,d)[\psi(d)^{-1}h(b,c)z(a,b,c)] \\ & = h(a,b c d)h(b,c d)h(c,d)z(b,c,d)z(a,b c,d)\psi(d)^{-1}z(a,b,c) }

Compare

$$\begin{array}{cc}h(abc,d)[\psi (d{)}^{-1}h(ab,c)\psi (c{)}^{-1}h(a,b)]& =h(abc,d)[\psi (d{)}^{-1}h(a,bc)]\psi (d{)}^{-1}\psi (c{)}^{-1}h(a,b)\\ & =h(abc,d)[\psi (d{)}^{-1}h(ab,c)]h(c,d{)}^{-1}[\psi (cd{)}^{-1}h(a,b)]h(c,d)\\ & =h(abc,d)h(ab,cd)z(ab,c,d)[\psi (cd{)}^{-1}h(a,b)]h(c,d)\\ & =h(a,bcd)h(b,cd)h(c,d)z(a,b,cd)z(ab,c,d)\end{array}$$\array{ h(a b c,d)[\psi(d)^{-1}h(a b,c)\psi(c)^{-1}h(a,b)] & = h(a b c,d)[\psi(d)^{-1}h(a,b c)]\psi(d)^{-1}\psi(c)^{-1}h(a,b) \\ & = h(a b c,d)[\psi(d)^{-1}h(a b,c)]h(c,d)^{-1}[\psi(c d)^{-1}h(a,b)]h(c,d) \\ & = h(a b c,d)h(a b,c d)z(a b,c,d)[\psi(c d)^{-1}h(a,b)]h(c,d) \\ & = h(a,b c d)h(b,c d)h(c,d)z(a,b,c d)z(a b,c,d) }

Proposition. (i) If we choose a different $h$ such that

$${\mathrm{Ad}}_K(h(a,b))=\psi (ab{)}^{-1}\psi (a)\psi (b),$$Ad_K(h(a,b)) = \psi(a b)^{-1}\psi(a)\psi(b),

then $z$ will change only up to a 3-coboundary $df,$ i.e. there is a function $f:B\times B\to Z(K)$, such that $z\prime =(df)z$ where

(18)$$(df)(a,b,c)=f^{-1}(b,c)f^{-1}(a,bc)f(ab,c)\psi (c{)}^{-1}f(a,b),\mathrm{for}\mathrm{all}a,b,c\in B.$$(d f)(a,b,c) = f^{-1}(b,c)f^{-1}(a,b c)f(a b,c)\psi(c)^{-1}f(a,b),\,\,\,\,\, for all a,b,c \in B.

(ii) Conversely, if $z$ is a 3-cocycle obtained from $\psi$ as above and $df$ is a 3-coboundary, then there is a $h\prime$ determining the same inner automoprhism of $K$ such that the corresponding 3-cocycle $z\prime$ is equal to $(\mathrm{df})z$.

(iii) Let $\psi ,\psi \prime :B\to \mathrm{Aut}(K)$ be two set-theoretic sections so that $[\psi ]=[\psi \prime ]=\theta :B\to \mathrm{Aut}(K)/\mathrm{Int}(K)$, then (for arbitrary choice of $h$, $h\prime$) the cocycles $z$ and $z\prime$ obtained as above differ only up to a 3-coboundary. $\parallel$

Proof. (i) Choose two different $h\prime ,h:B\times B\to K$ such that ${\mathrm{Ad}}_K(h\prime )={\mathrm{Ad}}_K(h)$. Then $h\prime (a,b)=h(a,b)f(a,b)$ where $f:B\times B\to Z(K)$ is some function with values in center of $K$. A direct comparison of (16) written for $h,z$ and $h\prime ,z\prime$ respectively proves the assertion.

(ii) Trivial: Any $f:B\times B\to Z(K)$ such that $h\prime =\mathrm{hf}$ will not change the inner automorphism. Thus any central 3-coboundary $\mathrm{df}$ can be obtained by changing a choice of $h$.

(iii) $[\psi \prime ]=[\psi ]$ implies that exists $k:B\to K,\psi \prime (a)=\psi (a){\mathrm{Ad}}_K(k(a)).$ Then

$$\begin{array}{ccc}\psi \prime (ab){\mathrm{Ad}}_K(h\prime (a,b))& =& \psi \prime (a)\psi \prime (b)=\psi (a){\mathrm{Ad}}_K(k(a))\psi (b){\mathrm{Ad}}_K(k(b))\\ & =& \psi (a)\psi (b){\mathrm{Ad}}_K([\psi (b{)}^{-1}k(a)]k(b))\\ & =& \psi (ab){\mathrm{Ad}}_K(h(a,b)[\psi (b{)}^{-1}k(a)]k(b))\\ & =& \psi \prime (ab){\mathrm{Ad}}_K(k(ab{)}^{-1}h(a,b)[\psi (b{)}^{-1}k(a)]k(b)).\end{array}$$\array{ \psi'(a b)Ad_K(h'(a,b)) & = & \psi'(a)\psi'(b) = \psi(a)Ad_K(k(a))\psi(b)Ad_K(k(b))\\ & = & \psi(a)\psi(b)Ad_K([\psi(b)^{-1}k(a)]k(b)) \\ & = & \psi(a b)Ad_K(h(a,b)[\psi(b)^{-1}k(a)]k(b)) \\ & = & \psi'(a b)Ad_K(k(a b)^{-1}h(a,b)[\psi(b)^{-1}k(a)]k(b)). }

Thus $h\prime (a,b)=k(ab{)}^{-1}h(a,b)[\psi (b{)}^{-1}k(a)]k(b),$ for appropriate choice of $h\prime$ - what can change $z\prime$ up to coboundary - using the freedom from (i). If we want formula involving $\psi \prime$ instead than we use $\psi \prime (a)=\psi (a){\mathrm{Ad}}_K(k(a))$ to obtain $k(ab)h\prime (a,b)=h(a,b)k(b)[\psi \prime (b{)}^{-1}k(a)]$. Using that and previous identities,

$$\begin{array}{ccc}k(abc)h\prime (ab,c)\psi \prime (c{)}^{-1}h\prime (a,b)& =& h(ab,c)k(c)[\psi \prime (c{)}^{-1}k(ab)]\psi \prime (c{)}^{-1}h\prime (a,b)\\ & =& h(ab,c)k(c)\psi \prime (c{)}^{-1}k(ab)h\prime (a,b)\\ & =& h(ab,c)k(c)\psi \prime (c{)}^{-1}h(a,b)k(b)[\psi \prime (b{)}^{-1}k(a)]\\ & =& h(ab,c)[\psi (c{)}^{-1}h(a,b)]k(c)\psi \prime (c{)}^{-1}k(b)[\psi \prime (b{)}^{-1}k(a)]\\ & =& h(a,bc)h(b,c)z(a,b,c)k(c)[\psi \prime (c{)}^{-1}k(b)][\psi \prime (c{)}^{-1}\psi \prime (b{)}^{-1}k(a)]\\ & =& h(a,bc)h(b,c)k(c)[\psi \prime (c{)}^{-1}k(b)][\psi \prime (c{)}^{-1}\psi \prime (b{)}^{-1}k(a)]z(a,b,c)\\ & =& h(a,bc)k(bc)h\prime (b,c)[\psi \prime (c{)}^{-1}\psi \prime (b{)}^{-1}k(a)]z(a,b,c)\\ & =& h(a,bc)k(bc)h\prime (b,c)h\prime (b,c{)}^{-1}[\psi \prime (bc{)}^{-1}k(a)]h\prime (b,c)z(a,b,c)\\ & =& h(a,bc)k(bc)[\psi \prime (bc{)}^{-1}k(a)]h\prime (b,c)z(a,b,c)\\ & =& k(abc)h\prime (a,bc)h\prime (b,c)z(a,b,c)\end{array}$$\array{ k(a b c)h'(a b,c)\psi'(c)^{-1}h'(a,b) &=& h(a b,c)k(c)[\psi'(c)^{-1}k(a b)]\psi'(c)^{-1}h'(a,b) \\ &=& h(a b,c)k(c)\psi'(c)^{-1}k(a b)h'(a,b) \\ &=& h(a b,c)k(c)\psi'(c)^{-1}h(a,b)k(b)[\psi'(b)^{-1}k(a)] \\ &=& h(a b,c)[\psi(c)^{-1}h(a,b)]k(c)\psi'(c)^{-1}k(b)[\psi'(b)^{-1}k(a)] \\ &=& h(a,b c)h(b,c)z(a,b,c)k(c) [\psi'(c)^{-1}k(b)][\psi'(c)^{-1}\psi'(b)^{-1}k(a)] \\ &=& h(a,b c)h(b,c)k(c)[\psi'(c)^{-1}k(b)] [\psi'(c)^{-1}\psi'(b)^{-1}k(a)]z(a,b,c) \\ &=& h(a,b c)k(b c)h'(b,c)[\psi'(c)^{-1}\psi'(b)^{-1}k(a)]z(a,b,c) \\ &=& h(a,b c)k(b c)h'(b,c)h'(b,c)^{-1}[\psi'(b c)^{-1}k(a)]h'(b,c)z(a,b,c)\\ &=& h(a,b c)k(b c)[\psi'(b c)^{-1}k(a)]h'(b,c)z(a,b,c)\\ &=& k(a b c)h'(a,b c)h'(b,c)z(a,b,c) }

for all $a,b,c\in B$. Thus $h\prime (ab,c)\psi \prime (c{)}^{-1}h\prime (a,b)=h\prime (a,bc)h\prime (b,c)z(a,b,c)$ i.e. our choice of $h\prime$ insured no change in $z$. Of course that means that in arbitrary choice of $h\prime$ we do not miss more than a coboundary by (i).

Corollary. A given homomorphism $\theta :B\times B\to \mathrm{Aut}(K)/\mathrm{Int}(K)$ is associated to some extension of $B$ by $K$ iff $z$ is a 3-coboundary.

Proof. Indeed, if $\theta$ is associated to an extension, then we know that there is an isomorphism of the extension with a cross product given by some cocycle $\chi$ and some automorphism $\psi$ such that $[\psi ]=\theta$. But using the identification, $\chi =h$ for that particular choice of $\psi$, so that $z=1$. By the proposition, every other $z$ obtained from $\theta$ is in the same cohomology class, thus every such $z$ is a coboundary. Conversely, if $z$ is a coboundary, then by the proposition, we can change it to $z=1$, and then we have all the conditions for a cross product extension satisfied.

$n$POV

One may regard the above from the nPOV as a special case of the way cocycles in the general notion of cohomology classify their homotopy fibers. More on this is at

• group cohomology

• nonabelian group cohomology.

Related concepts

• group cohomology

  • nonabelian group cohomology, groupoid cohomology

• group extension

• Lie group cohomology

  • ∞-Lie groupoid cohomology

References

• Samuel Eilenberg, Saunders MacLane, Cohomology theory in abstract groups. II. Group extensions with a non-Abelian kernel. Ann. of Math. (2) 48, (1947). 326–341 jstor

• Saunders MacLane, Cohomology theory in abstract groups. III. Operator homomorphisms of kernels. Ann. of Math. (2) 50, (1949). 736–761.

• Lawrence Breen, Théorie de Schreier supérieure, Ann. Sci. École Norm. Sup. (4) 25 (1992), no. 5, 465–514 numdam.

• A. G. Kurosh, Theory of groups

• Kenneth S. Brown, Cohomology of groups, Graduate Texts in Mathematics, 87, Springer-Verlag, New York-Berlin, 1982.

• J-P. Serre, Cohomologie galoisienne, Lecture Notes in Mathematics, 5 (Fifth ed. 1994), Springer-Verlag, MR1324577

• M. Bullejos, A. Cegarra, A 3-dimensional non-abelian cohomology of groups with applications to homotopy classification of continuous maps Canad. J. Math., vol. 43, (2), 1991, p. 1-32.

• Antonio M. Cegarra, Antonio R. Garzón, A long exact sequence in non-abelian cohomology, Proc. Int. Conf. Como 1990., Lec. Notes in Math. 1488, Springer 1991.

See also references of Dedecker listed here.