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Let $S$ and $R$ be groups and say $\sigma: S \twoheadrightarrow R$ is a group homomorphism that is a central extension; that is, it is surjective (extension) and its kernel is contained in the centre of $S$ (central). Let $\mathbf{ab} :\mathbf{Gp} \rightarrow\mathbf{Ab}$ be the abelianisation functor (the left adjoint to the inclusion of abelian groups into the category of groups). I would like to know of an elementary characterisation of the set of group homomorphisms $\phi: H \rightarrow R$ such that the canonical map $\Theta: S \times_R H \rightarrow S \times_{\mathbf{ab}S} \mathbf{ab}(S \times_R H)$ is an isomorphism.

Notes for the question

  1. I am not assuming that $\phi$ is a surjection. If it were then Lemma 5.2.7 of 'Galois Theories' (Borceux and Janelidze) answers the question. (The answer is that $\Theta$ is iso. iff the commutator subgroups of $S$ and $S \times_R H$ are isomorphic.)

  2. I know that if $\Theta$ is iso. then both $\phi$ and $\pi_2 : S \times_R H \rightarrow H$ are central, but the converse is false. A counterexample is to take $\phi$ to be $e:1 \rightarrow R$ for some $R$ with non-trivial commutator subgroup where $e$ is the identity of $R$.

  3. If $\sigma=\phi$ and $\sigma$ is not necessarily central, then $\Theta$ is an isomorphism iff $\sigma$ is central.

Thanks for any comments!

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I don't have the book "Galois Theories" at hand but it seems to me that the result of the Lemma you mention should hold even if $\phi$ is not surjective.

Indeed, given a commutative diagram $$\require{AMScd}\begin{CD}0 @>>> K @>{k}>> X @>{f}>> Y @>>> 0 \\ & @V{u}VV @V{v}VV @VV{w}V \\ 0 @>>> K' @>>{k'}> X' @>>{f'}> Y' @>>> 0 \end{CD} \tag{1}\label{1}$$ where the rows are short exact sequences, then $u$ is an isomorphism if and only if the canonical arrow $X\to X'\times_{Y'} Y$ is an isomorphism. Indeed, we have a diagram $$\begin{CD}0 @>>> K @>{k}>> X @>{f}>> Y @>>> 0 \\ & @V{u}VV @V{(v,f)}VV @VV{1_Y}V \\ 0 @>>> K' @>{(k',0)}>> X'\times_{Y'}Y @>{\psi_2}>> Y @>>> 0 \\& @V{1_{K'}}VV @V{\psi_1}VV @VV{w}V \\ 0 @>>> K' @>>{k'}> X' @>>{f'}> Y' @>>> 0, \end{CD}$$ where the middle row is exact; hence we can apply the Short Five Lemma to the the two upper rows.

Then it suffices to apply this to the case where the short exact sequences in \eqref{1} are given by the abelianizations of $S\times_HR$ and $S$.

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  • $\begingroup$ Thanks Arnaud. I may be misinterpreting what you are saying, but if you apply with abelianizations of $S \times_R H$ and $S$ respectively (i.e. $X=S \times_R H$, $X'=S$ etc) then the conclusion of the Short Five Lemma (en.wikipedia.org/wiki/Short_five_lemma) is that $S \times_R H \cong S$, which is not the condition. $\endgroup$ – Christopher Townsend Oct 12 '17 at 8:42
  • $\begingroup$ @ChristopherTownsend I didn't mean to apply it directly to the diagram, but to another one. I've edited to explain this more clearly. $\endgroup$ – Arnaud D. Oct 12 '17 at 8:58
  • $\begingroup$ Great - got it. Thank you very much. If you happen to know more about where this class crops up elsewhere, that would be really helpful, but obviously that goes beyond my original question. $\endgroup$ – Christopher Townsend Oct 13 '17 at 19:44

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