Theorem states-
Let $G$ be a finite group of order $mn$ and $N$ be a normal subgroup of order $n$, then schur zassenhaus states that there exist a complement of $N$ in $G$ of order $m$ and all such are conjugate if more than one.
The proof i have gone through are very long, but every proof is reduction upto case where $N$ is proved abelian and then some long transversals based continuation of proof(as in Isaacs & also Robinson , too long and confusing), but in a paper by Conrad he states than rest of proof can be done by cohomology theory(the vanishing of $H^2(G/N,N$), he says.) I did self read homology and cohomology from Rotman a few months back but will need a quick revision though. So if somebody can tell me the main steps how the proof proceed via cohomology argument, it will be a huge help. I hope proof via cohomology will be much shorter and elegant , is it? Thanks