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Suppose we have a given action $\varphi : A\rightarrow\text{Aut}(N)$ with $A,N$ abelian groups. Is it possible describe the isomorphism classes of extensions $G$ of $A$ by $N$ realizing $\varphi$ such that the map $G\rightarrow A$ is abelianization (ie, such that $N = G'$)?

(Without the requirement $N = G'$ this is classified by the group cohomology $H^2(A,N)$)

Some remarks:

Nontrivial central extensions certainly give examples where classes in $H^2(A,N)$ do not necessarily satisfy the condition that $G' = N$.

I wonder if there is a condition on $\varphi$ which guarantees that every class in $H^2(A,N)$ represented by $G$ satisfies $G' = N$.

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When $A$ acts trivially on $N$, this $H^2(A,N)$ has a canonical map $\Phi$ into the group $\mathrm{Hom}(\Lambda^2A,N)$ ($\Lambda^2A$ being the second exterior power, quotient of $A\otimes_\mathbf{Z}A$ by the subgroup generated by elements of the form $x\otimes x$ when $x$ ranges over $A$), induced by the commutator map. Given a central extension with cocycle $c$, the resulting extension has $N$ as derived subgroup iff the corresponding element $\Phi(c)\in\mathrm{Hom}(\Lambda^2A,N)$ is surjective.

(Remark: there's a canonical inclusion $\Psi$ of $\mathrm{Hom}(\Lambda^2A,N)$ into $H^2(A,N)$ and $\Phi\circ\Psi=2\mathrm{Id}$.)

In general, let $N/M$ be the co-invariants of the $A$-action on $N$ (i.e., $M$ is generated as a group by the $\varphi(g)h-h$, when $h$ ranges over $N$ and $g$ over $A$). Then after modding out by $M$, every extension as given yields a cocycle in $H^2(A,N/M)$ and hence, taking $\Phi$ an element in $f\in\mathrm{Hom}(\Lambda^2A,N/M)$; then $N$ is the derived subgroup iff $f$ is surjective.

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  • $\begingroup$ Would you happen to have a reference for this? (or perhaps a slick way to check this without doing a bunch of computations with cocycles?) $\endgroup$
    – stupid_question_bot
    Commented Oct 10, 2018 at 22:35
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    $\begingroup$ @rtz my initial statement was not correct: for instance if the whole extension is abelian, the extension can be non-split. So I rephrased. Also there's no need of reference: the only thing is to observe, given a central extension $Z\to G\to G/Z$, that the commutator map $G\times G\to G$ factors though $G/Z\times G/Z\to G$, and if moreover $G/Z$ is abelian, it yields a map $G/Z\times G/Z\to Z$, which is clearly bilinear and alternating. $\endgroup$
    – YCor
    Commented Oct 11, 2018 at 6:57
  • $\begingroup$ Ah, the thing I was missing was that $(a-1)n$ is a commutator! ($a\in A, n\in N$). I guess there is some relation between this and the Schur multiplier... $\endgroup$
    – stupid_question_bot
    Commented Oct 11, 2018 at 20:50

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