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Let $G$ be a finite 2-generated metabelian group, and let $S$ be a schur covering group, so that we have an exact sequence $$1\rightarrow M(G)\rightarrow S\rightarrow G\rightarrow 1$$ where $M(G)$ is the Schur multiplier of $G$, and $M(G)\le S'$ (The $'$ denotes derived subgroup). This implies that $S^{ab}= G^{ab}$, and we have an exact sequence $$1\rightarrow M(G)\rightarrow S'\rightarrow G'\rightarrow 1$$ Thus, since $G$ is metabelian, $G'$ is abelian, so $S'$ is also a metabelian group, and is equal to the exterior square $G\wedge G$ of $G$.

It seems reasonable that we should be able to say something about $(G\wedge G)^{ab},(G\wedge G)'$ in terms of $G',G^{ab}$,...etc, though I'm having some difficulty getting an intuitive understanding of what this exterior square "is", and how to work with it.

Is there a good book that discusses nonabelian exterior products of groups?

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    $\begingroup$ I'm afraid that there's no good textbook-format reference for nonabelian homological algebra, but works of G. Ellis and G. Donadze on the subject are pretty self-contained. On your question: if $G$ is finite metabelian and is an extension of $\pi$-group $G_{ab}$ by $\pi'$-group $G'$ of squarefree order (i. e. G' has trivial multiplier), then $G \wedge G$ = $G' \times (G_{ab} \wedge G_{ab})$. As for actual computation (in terms of presentations) of tensor square for finite solvable groups there were A. McDermott thesis, papers by N. Rocco, and some materials on G. Ellis homepage. $\endgroup$
    – Denis T
    Aug 25, 2017 at 10:30
  • $\begingroup$ @DenisT. Thank you for your comment! I'm revisiting this question, and was wondering - what do you mean by a $\pi$-group (or $\pi'$-group)? $\endgroup$ Oct 30, 2018 at 20:12
  • $\begingroup$ @DenisT. (And do you have a specific reference for your statement involving $\pi$-groups?) $\endgroup$ Oct 30, 2018 at 20:20

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