3
$\begingroup$

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 $G' = N$.

$\endgroup$
4
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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 Oct 10 '18 at 22:35
  • 1
    $\begingroup$ @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 Oct 11 '18 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 Oct 11 '18 at 20:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.