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Suppose I've a group $G$ and a normal subgroup $H$. I know the structure of $G/H$ as well as the structure of $H$. Is it possible to recover the original structure of G from this?

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    $\begingroup$ Dear Mohan, This question and its answers are quite relevant. Regards, $\endgroup$ – Matt E May 30 '13 at 2:12
  • $\begingroup$ An infinite groups counter-example: suppose $G$ is a free group of arbitrary rank greater than one, then every (non-trivial) normal subgroup of infinite index is isomorphic to the free group on countably many generators. Thus, if $G/H$ is any infinite group and $H$ is the free group of countably-infinite rank, then $G$ can be any free group of rank greater than one. $\endgroup$ – user1729 May 30 '13 at 10:18
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There is one important piece of data that you have omitted: namely, if $H$ is normal in $G$, then $G$ acts on $H$ by conjugation, and so there is a homomorphism $G/H \to Out(H)$ (the group of outer automorphisms of $H$); without remembering this, there would be no chance of reconstructing $G$.

Now in general giving $G/H$, $H$, and a map from $G/H$ to $Out(H)$, there is still no chance of reconstructing $G$. E.g. even if $G$ and $H$ are abelian, so that the homomorphism $G/H \to Out(H)$ is trivial, Zev's answer gives a counterexample.

However, if $G/H$ and $H$ are finite of coprime order, then in fact any extension of $G/H$ by $H$ is necessarily a semidirect product, and so in this case your question (suitably modified by remembering the conjugation action) has a positive answer. This is the Schur--Zassenhaus theorem.

This is why counterexamples (such as in Zev's answer) tend to focus on $p$-groups.

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Unfortunately, no; the simplest example is $G_1=\mathbb{Z}/4\mathbb{Z}$ and $H_1=2\mathbb{Z}/4\mathbb{Z}$, as compared with $G_2=(\mathbb{Z}/2\mathbb{Z})\times(\mathbb{Z}/2\mathbb{Z})$ and $H_2=(\mathbb{Z}/2\mathbb{Z})\times\{\overline{0}\}$. We have $$H_1\cong H_2\cong\mathbb{Z}/2\mathbb{Z},\qquad G_1/H_1\cong G_2/H_2\cong\mathbb{Z}/2\mathbb{Z}$$ but $G_1\not\cong G_2$. The general problem of determining what groups $G$ could arise, given a specified normal subgroup and the corresponding quotient, is called the group extension problem.

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    $\begingroup$ Dear Zev, It is good to remember though that Schur--Zassenhaus applies if $G/H$ and $H$ are of coprime order. Cheers, $\endgroup$ – Matt E May 30 '13 at 2:21
  • $\begingroup$ @MattE But it is also important to remember that even then, we cannot in general uniquely recover $G$. $\endgroup$ – Tobias Kildetoft May 30 '13 at 15:19
  • $\begingroup$ @Tobias: Dear Tobias, As I wrote in my answer, we have to also remember the action of $G/H$ on $H$ (obviously). Regards, $\endgroup$ – Matt E May 30 '13 at 22:59

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