# Is it possible to classify extensions $G$ of an abelian $A$ by an abelian $N$ such that the map $G\rightarrow A$ is abelianization?

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 $$G' = N$$.

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.
• @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. – YCor Oct 11 at 6:57
• 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... – stupid_question_bot Oct 11 at 20:50