Let G and H be finitely generated groups. Let G' = [G,G] and H' = [H,H] be the corresponding derived groups (commutator subgroups) of G and H. we know that there exist groups G and H such that G' is isomorphic to H' and G/G' is isomorphic to H/H' but G and H are not isomorphic. with which additional condition on G,H can we obtain G and H are isomorphic? (for example residually finite, FC-group, m-generators with n-expanent,...)
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5$\begingroup$ Certainly residual finiteness is not enough. For instance, consider a semi-direct product $\Gamma_\phi=\mathbb{Z}^2\rtimes_\phi\mathbb{Z}$, where the eigenvalues of $\phi$ are real and distinct. Then $\Gamma_\phi$ is residually finite with $[\Gamma_\phi,\Gamma_\phi]\cong\mathbb{Z}$ and the abelianisation isomorphic to $\mathbb{Z}$ but $\Gamma_\phi$ and $\Gamma_\psi$ are isomorphic if and only if $\phi$ and $\psi$ are $\mathbb{Z}$-conjugate. $\endgroup$– HJRWCommented Apr 16, 2012 at 9:00
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$\begingroup$ Sorry, there's a typo - the commutator subgroup is isomorphic to $\mathbb{Z}^2$. $\endgroup$– HJRWCommented Apr 16, 2012 at 10:10
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1$\begingroup$ I think that there exist finite groups $G, H$ with $G'\cong H'$ and $G/G' \cong H/H'$, but $G, H$ are not isomorphic. So all your additional conditions, as finiteness conditions, are not enough. $\endgroup$– Wei ZhouCommented Apr 16, 2012 at 13:14
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5$\begingroup$ I think $Q_8$ and $D_8$ are such examples. $\endgroup$– Wei ZhouCommented Apr 16, 2012 at 13:32
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1 Answer
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Wei Zhou's answer already gives a very good example and I agree with the above comments. However for a "way out" - I want to give you two classes of obtacles, if you additionally remove the first, you're on your way to a well-known equivalence, which is weaker than isomorphy but helps e.g. in $p$-groups.
1) In my opinion the serious thing: You loose knowledge of the commutator map $G/G'\times G/G'\rightarrow G'/G^{(2)}$ and only preserve THAT elements appear as commutators.
Take as example the extraspecial groups $p^{2n+1}_\pm$ (such as the direct product with centers identified $D_4\ast\ldots\ast D_4$) compared to e.g. these one: $p_\pm^{2\cdot 1+1}\times \mathbb{Z}_p^{2(n-1)}$ (such as $D_4\times \mathbb{Z}_2^{2(n-1)}$). Both are central/stem-extensions of $\mathbb{Z}_p^{2n}$ by $\mathbb{Z}_p$, but the second is a lot "more commutative" ;-)
2) A lot more tricky to detect but maybe not-so-serious: Even if the commutator maps match, it remains unclear how elements powered-up/"fused" to the divided-out commutators.
Best examples are certainly the different extraspecials $p_+^{2\cdot n+1}$ vs. $p_-^{2\cdot n+1}$, especially $D_4$ vs. $Q_8$ (as Zhou said), that can only be distinguished by how many elements powering to the central commutator.
The latter behaviour is sometimes described as the groups being isoclinic (there are different definitions!) and this is also responsible e.g. for the three different nonabelian isomorphy types in order $2^n$. It is fairily mild and used e.g. in the classification efforts on $p$-groups. Most prominent examples you get from different Schur covers of a group being isoclinic (again $D_4/Q_8$ but also the different 2-covers of $S_n$). In this situation by construction all isoclinics are isomorphic as grouprings $k[D_4]\cong k[Q_8]$ and hence have the same representation theory, although I have no sources confirming this for nonabelian extensions!!
Hope that helps ;-)
