Disclaimer: I'm not a group theorist, I arrived at the following question from algebraic geometry.
The first half of the Schur-Zassenhaus theorem states that, if $N$ is a normal subgroup of a finite group $G$ and the order of $N$ is prime with its index, then $N$ has a complement in $G$. Usually, the proof starts with the case in which $N$ is abelian: in this case, the theorem follows from the fact that $H^{2}(G/N,N)$ vanishes. The proof then proceeds with some induction argument.
At least in the abelian case, though, a weaker hypothesis is sufficient. Suppose that, for every prime $p\mid |N|$, there exists a $p$-Sylow $P\subset G/N$ with a section $P\to G$. Then $N$ has a complement in $G$. We can prove this again using cohomology: if $g\in H^{2}(G/N,N)$ is the class of the extension, the restriction of $g$ to $H^{2}(P,N)$ is trivial by hypothesis, and if we go back to $H^{2}(G/N,N)$ by applying corestriction we obtain $mg=0$, where $m$ is the index of $P$. Since the order of $g$ divides $|N|$ and $p\nmid m$, by doing this for every prime dividing $N$ we get that $m=0$.
For instance, if we find an intermediate subgroup $N\subset H\subset G$ such that $N$ has a complement in $H$ and the index of $H$ is prime with $|N|$, then $N$ has a complement in $G$ as well.
Question: do you know other assumptions on $N$ which guarantee that this works?
With an induction argument, it is possible to do the case in which $N$ is an iterated semi-direct product of abelian groups with pairwise co-prime order. Write $Q=G/N$. Under the assumption, $N$ has a non-trivial characteristic subgroup $M$ whose order is prime with the index in $N$. By induction, there is a section $Q\to G/M$, its inverse image in $G$ is an extension $E$ of $Q$ by $M$. Let $p$ be a prime dividing $|M|$ and $P\subset Q$ a $p$-Sylow, by hypothesis we have a section $P\to G$. Since $p$ does not divide $|N/M|$, by the second half of Schur-Zassenhaus we can assume that the composition $P\to G\to G/M$ coincides with $P\subset Q\to G/M$. It follows that the image of $P\to G$ is contained in $E\subset G$, hence by induction $E$ is split and we conclude.
I'm interested in how far we can take this, and possibly counterexamples.
