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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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Of course, $H$ is a normal. –  Aspirin Sep 7 '11 at 1:17
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Hm. Then $G/E$ doesn't make immediate sense to me. How do you get $E$ to sit inside of $G$? –  Dylan Moreland Sep 7 '11 at 1:38
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OK, so let's get this straight: $G$ is a group containing two subgroups $H$ and $E$, both normal, such that $G/H\cong E$. Right? And you want to know if $G/E\cong H$? Then that's not true: Let $G$ be the quaternion group of order $8$, $H$ a subgroup of order $4$, and $E$ a subgroup of order $2$. Then $G/H\cong E$ but $G/E$ is not cyclic. –  user641 Sep 7 '11 at 1:48
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@Steve D We wants to know if there are ANY such groups. And if they are, then what conditions. And sorry for my bad english. –  Ximik Sep 7 '11 at 1:59
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@Alyushin: if Steve is right, please edit the text of the question to include all the details. –  Mariano Suárez-Alvarez Sep 7 '11 at 2:00

1 Answer 1

up vote 2 down vote accepted

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...

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