I am currently working on a masters project in which I prove the Schur-Zassenhaus theorem using the classification of group extensions. This classification relies on homological algebra and to my knowledge this is the "best" proof of it.
My questions : is there a proof of Schur-Zassenhaus without using homological algebra? (i.e. without using the bijection between $H^2(G,A)$ and the equivalence classes of group extensions).
The Schur-Zassenhaus theorem states : If $E$ is any finite group containing a normal subgroup $N$ whose order and index are relatively prime, then $N$ has a complement in $E$.
I have found this arxiv https://arxiv.org/abs/1403.1425 by Sergei O. Ivanov and Nikolay N. Mostovsky but it has since been removed.
There is an elementary proof which goes back to Gaschütz. It can be found in the book "The theory of finite groups" by Kurzweil and Stellmacher (see 3.3.1 and 6.2.1). The idea is an "average trick": Let $N$ be a normal Hall subgroup in $G$. By easy reduction theorems using Sylow, we may assume that $N$ is abelian. Let $R,S$ be set of representatives for the cosets of $N$ in $G$. Now consider $$R|S:=\prod_{\substack{(r,s)\in R\times S\\Nr=Ns}}rs^{-1}\in N$$ (this is well-defined since $N$ is abelian). We call $R$ and $S$ equivalent if $R|S=1$. One shows that $G$ acts by multiplication on the set of equivalence classes of coset rep systems. Finally, a stabilizer of this action is a complement of $N$. Moreover, since $G$ acts transitively, all complements are conjugate. Gaschütz has used this argument to prove his related theorem: Let $N$ be an abelian normal subgroup of $G$, $N\le H\le G$ such that $(|N|,|G:H|)=1$. If $N$ has a complement in $H$, then $N$ has a complement in $G$.