When a normal subgroup $N$ admits a complament? Let $G$ be a finite group and let $N$ be a normal subgroup. I am looking for conditions on $N$ (and maybe also on $G$) such that there exist a subgroup $H$ of $G$ such that 
$$G=N\rtimes H.$$
Clearly, this is not always the case.
The easiest example for that is $C_2$ as a subgroup of $C_4$.
On the other hand, if $G$ is solvable and $N$ is a Hall normal subgroup then it admit a complament $H$ where $H$ is a Hall subgroup with order $|\frac{G}{N}|$.
I am looking for other restrictions on $N$, under such restrictions it admits a complament. Also I would be happy to see a strictly group-theoretical proof (without cohomology) of the fact that a normal Hall subgroup always admits a complament. 
 A: There is  proof of Schur-Zassenhaus, due to Wielandt, which does not make (explicit) use of group cohomology, but uses group actions instead. As has been noted in comments, it is easy to reduce to the case that $N$ is an elementary $p$-group for some  prime $p.$ Let $[G:N] = h.$
We define an equivalence relation $\sim$ of transversals to $S$ by 
$\{s_{1},\ldots,s_{h} \} \sim \{t_{1},\ldots,t_{h} \}$ if and only if $\prod_{i=1}^{h}(s_{i}t_{i}^{-1}) = 1$ (note that the order of the product does not matter, since $s_{i}t_{i}^{-1} \in N$ for each and $N$ is Abelian). We denote the equivalence class of $T$
by $[T]$. Note that $G$ acts on the equivalence classes via $T \to xT$ for $x \in G.$
The action is transitive, for given transversals $S$ and $T$ with $\prod_{i=1}^{h}(s_{i}t_{i}^{-1}) = n \in N,$ we may choose $u \in N$ with $u^{h} =n^{-1}.$ Then $T \sim uS$
since $\prod_{i=1}^{h} us_{i}t_{i}^{-1} = u^{h}n =1.$ Thus $[T] = [uS].$ In fact, we have shown that $N$ is transitive, so there are at most $|N|$ equivalence classes. Furthermore, given a transversal $S$ and $y \in N$ we have $[yS] = [S]$ if and only if $y^{h} = 1$, if and only if $y = 1$, since $h$ and $|N|$ are coprime. Thus there are exactly $|N|$ equivalence classes, and the stablizer of an equivalence class has order $[G:N]$, and is a complement to $N.$
