When is a group $G$ isomorphic to $(G/N) \times N$?

Solution 1:

A trivial answer to your question that the normal subgroup $N$ splits off as a direct factor of $G$: $$ G=N\times Q$$ for some subgroup $Q< G$. (I.e. there exists another normal subgroup $Q< G$ such that $Q\cap N=\{1\}$, $[N,Q]=\{1\}$ and $G=NQ$.) Whether you are looking for anything beyond this statement is unclear to me.

Edit. It is clear that if $G=G_1\times G_2$ then $G/G_1\cong G_2$, the isomorphism is given by the map $G_1\times g_2 \mapsto g_2$.

Solution 2:

As Arthur mention in a comment, whenever there is a group homomorphism $p:G\to N$ whose restriction to $N$ is the identity, then $G$ is isomorphic to $N\ltimes \ker(p)$, where the action of $N$ on $\ker(p)$ is given by conjugation in $G$. But if $N$ is assumed to be normal, then this action is actually trivial, and thus $G$ is actually isomorphic to $N\times\ker(p)$. Indeed, in that case, for every $n\in N$ and $k\in \ker(p)$ we have $nkn^{-1}k^{-1}\in N$, and thus $$nkn^{-1}k^{-1}=p(nkn^{-1}k^{-1})=nn^{-1}=1,$$ which means that $nkn^{-1}=k$.

So a possible answer to your question is that there is an isomorphism if and only if $N$ is a retract of $G$.