Recovering a group from its quotient group 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?
 A: 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.
A: 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.
