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$\DeclareMathOperator\Aut{Aut}\DeclareMathOperator\Ext{Ext}$Let $A$ by $G$ be two non-abelian group. A factor pair of $G$ with coefficients in $A$ is a pair $(\alpha,\varepsilon)$, where $\alpha:G\rightarrow\Aut(A)$ and $\varepsilon:G\times G\rightarrow A$ are maps satisfying the following properties: \begin{eqnarray} \alpha(1)&=&\mathrm{id}_{A},\\ \label{normlized} \varepsilon(h,1)&=&\varepsilon(1,h)=1, \\ \label{map} \alpha(h)\circ\alpha(g)&=&\varepsilon(h,g)\alpha(hg)\varepsilon(h,g)^{-1},\\ \label{Cocycle} \varepsilon(h,g)\varepsilon (hg,k)&=&\alpha (h)(\varepsilon (g,k))\varepsilon (h,gk) \end{eqnarray} for all $g, h, k\in G$. The set of factor pairs of $G$ with coefficients in $A$ is denoted by $\mathcal{Z}^{2}(G,A)$. Let $(\alpha,\varepsilon)$, $(\beta,\varepsilon')\in \mathcal{Z}^{2}(G,A)$, we write $(\alpha,\varepsilon)\sim(\beta,\varepsilon')$ and say that $(\alpha,\varepsilon)$ and $(\beta,\varepsilon')$ are cohomologous if there is a map $t:G\to A$ such that $$\beta(g)=t(g)\alpha(g)t(g)^{-1} \text{ and } \varepsilon'(g,h)=t(g)\alpha(g)(t(h))\varepsilon(g,h)t(gh)^{-1}$$

for all $g$, $h\in G$. The set of cohomology classes of factor pairs of $G$ with coefficients in $A$ is denoted by $\mathcal{H}^{2}(G,A)$.

Now, two extensions $E$ and $E'$ of $A$ by $G$ are said to be equivalent if and only there is a homomorphism $\varphi:E\rightarrow E'$ such that the following diagram commutes \begin{equation*}\require{AMScd} \begin{CD} 1 @>>> A @>>> E @>>>G @>>> 1 \\ @. @| @VV\varphi V @| \\ 1 @>>> A @>>> E' @>>> G @>>> 1. \end{CD} \end{equation*} The set of equivalence classes of group extensions is denoted by $\Ext(G,A)$.

Schreier's Theorem gives us a one-to-one correspondence between $\Ext(G,A)$ and $H^{2}(G,Z(A))$. But I can't find a direct proof for this theorem. In other word, I want to understand how a given extension induces a factor pair $(\alpha,\varepsilon)$ where the $2$-cocycle $\varepsilon$ takes values in the center $Z(A)$ of $A$.

Any help would be appreciated.

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    $\begingroup$ Don't you want $H^3(G,Z(A))$ $\endgroup$
    – Benjamin Steinberg
    Commented Aug 19 at 20:53
  • $\begingroup$ Not exactly. In fact, there is a proof using the free and the transitive action of $H^{2}(G,Z(A))$ on $Ext(G,A)$. Here, I'm looking for a direct proof. More precisely, I want to understand how a given extension induces a factor pair $(\alpha,\varepsilon)$ where the cocycle $\varepsilon$ takes values in the center $Z(A)$ of $A$. $\endgroup$
    – N. SNANOU
    Commented Aug 19 at 21:13
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    $\begingroup$ @N. SNANOU: It is not true that every extension of $A$ by $G$ splits if $Z(A)=1$. For example, let $H$ be the (split) extension of the additive group $N=F_5^+$ by the multiplicative group $C=F_5^\times$ acting naturally (where $F_5$ is the field of $5$ elements). Then $H$ has a subgroup $A$ of index $2$ isomorphic to the dihedral group of order $10$, and $G=H/A\cong Z_2$. Here $Z(A)=1$ but $H$ is not a split extension of $A$ by $G$. You may object that $G$ is abelian, but you can take any nonabelian group $K$ and form $H\times K$, a nonsplit extension of the same $A$ by $Z_2\times K$. $\endgroup$
    – Richard Lyons
    Commented Aug 26 at 4:52
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    $\begingroup$ @N. SNANOU Yes, if $Z(A)=1$ then there is an extension inducing $f$. You can start building it by taking the pullback of $f$ and the canonical homomorphism from $Aut(A)$ to $Out(A)$. Chasing diagrams will show that you have an extension of $Inn(A)\cong A$ by $G$. As for your question, given $f$, there is a way to construct a $3$-cocycle on $G$ with values in $Z(A)$, and such that an extension of $A$ by $G$ exists affording $f$ if and only if the $3$-cocycle is a coboundary. This may give a hint for your desired elementary proof. It is discussed for example in Mac Lane's old text "Homology." $\endgroup$
    – Richard Lyons
    Commented Aug 30 at 23:20
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    $\begingroup$ I think your question is discussed explicitly in a bachelor thesis from 2017: "Classifying group extensions with not necessarily abelian kernel" by Chen Zhang. $\endgroup$
    – Ettore
    Commented Sep 1 at 14:50

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