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Let $H$ and $E$ be normal subgroups of a group $G$ such that $G/H \cong E.$ Under what sort of conditions would we also have $G/E \cong H?$

Thanks.

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    Is $G$ finite? For infinite, I think you're likely to be sunk; e.g., take a non-Hopfian group $G$, take $H$ a nontrivial normal subgroup such that $G/H\cong G$, and take $E=G$. That would be pretty much as different as $G/H$ and $G/E$ can be.2011-09-07

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This holds if $G=H\times E$, almost by definition.

Another example is $G=H \rtimes H$ for suitable $H$. $\mathbb{Z}$ works, for example, $\langle a, b; aba^{-1}=b^{-1}\rangle$ as does $C_6$: $\langle a, b; a^6, b^6, aba^{-1}=b^{-1}\rangle$ as $Aut(C_6)\cong C_2$ (because $\varphi(6)=2$). Indeed, any group $H$ such that $H\rightarrow Aut(H)$ non-trivially works.

It would be interesting to see if there exist (finite) groups $H$ and $E$ and homomorphisms $\phi$ and $\varphi$ such that $H\rtimes_{\phi} E\cong H\ltimes_{\varphi} E$. However, I haven't found any examples yet...