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$\def\Z{\mathbb{Z}}$ Let $A$ be a finite abelian group and $G$ a finite group acting on $A$. Then the extension $0 \to A \to E \to G \to 1$ is splits if and only if the corresponding $2$-cocycle is trivial in $H^2(G,A)$.

Does there exist such a split extension $E$ such that $E$ can also be seen as a non split extension (for the same action of $G$ on $A$)? Or in other words, can we find two groups $G$ and $A$, and an action of $G$ on $A$ such that there exists $\alpha_1,\alpha_2 \in H^2(G,A)$ which corresponds to the same group $E$, and with $\alpha_1=0$?

With help from user gro-tsen, I can say the following:

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    $\begingroup$ More generally than the abelian case, J. Ayoub proved that a finite group that is a direct product $A\times B$ cannot be written as a non-split extension of $A$ by $B$. $\endgroup$
    – YCor
    Apr 10, 2014 at 21:36
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    $\begingroup$ user.math.uzh.ch/ayoub/PDF-Files/DIRECT.PDF (J. Ayoub, The direct extension theorem. J. Group Theory 9 (2006), 307-316.) $\endgroup$
    – YCor
    Apr 11, 2014 at 10:55
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    $\begingroup$ About an easy infinite example: Let $A$ be the product (of order 8) of cyclic groups of order 2 and 4. Let $G$ be the (restricted or unrestricted) direct product of countably many copies of $A$. Then $G$ is isomorphic to $G\times G$ and can also obviously be decomposed as a non-split extension of $G$ by $G$. $\endgroup$
    – YCor
    Apr 11, 2014 at 10:59
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    $\begingroup$ @Geoff Robinson We want a finite group $E$ with two isomorphic abelian normal subgroups $N_1$ and $N_2$ such that: (i) $E/N_1$ and $E/N_2$ are isomorphic groups and, using this isomorphism, $N_1$ and $N_2$ are equivalent modules under the actions induced by conjugation; and (ii) $N_1$ has a complement in $E$ but $N_2$ does not. $\endgroup$
    – Derek Holt
    Apr 11, 2014 at 13:05
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    $\begingroup$ Perhaps this approach is useful. Assume that $B$ is a subgroup of $G$ which is isomorphic to $A$ as a $G$-module, such an isomorphism can be extended to an isomorphism $\phi : A \rightarrow B$ of $E$-modules. This induces an isomorphism between the abelian groups $Der(E,A)$ and $Der(E,B)$ in the obvious way. I wonder if $\phi$ induce a ring isomorphism or more generally if there is a ring isomorphism between $Der(E,A)$, $Der(E,B)$. If so Then $E$ splits over $B$. (The multiplication in $Der(G,A)$ is the composition of map) $\endgroup$
    – Yassine Guerboussa
    Apr 13, 2014 at 16:36

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