Schur-Zassenhaus existence theorem

We argue by induction on $\mathrm{\mid }G\mathrm{\mid }$. If $N=1$, then $G$ is a complement to $N$. If $N=G$, then $G\mathrm{/}N=1$, so the trivial subgroup is a complement. Assume from now on that $1<N<G$ and that the statement holds for all smaller finite groups.

Suppose $M$ is a normal subgroup of $G$ with $1<M<N$. Then $N\mathrm{/}M$ is a normal Hall subgroup of $G\mathrm{/}M$, so the induction hypothesis gives a subgroup $K\mathrm{/}M\le G\mathrm{/}M$ complementary to $N\mathrm{/}M$. Thus $K<G$, $G=NK$, and $N\cap K=M$. Inside the proper group $K$, the subgroup $M$ is normal Hall because $\mathrm{\mid }K\mathrm{/}M\mathrm{\mid }=\mathrm{\mid }G\mathrm{/}N\mathrm{\mid }$ is coprime to $\mathrm{\mid }M\mathrm{\mid }$. The induction hypothesis applied to $K$ gives a complement $H$ to $M$ in $K$, so $K=MH$. Since $H\le K$, we have $H\cap N=H\cap (N\cap K)=H\cap M=1$, while $NH=N(MH)=NK=G$. Thus $H$ complements $N$ in $G$. For the remainder we may therefore assume that $N$ contains no nontrivial proper subgroup normal in $G$.

Choose a prime $p$ dividing $\mathrm{\mid }N\mathrm{\mid }$ and let $P$ be a Sylow $p$-subgroup of $N$, whose existence is supplied by [L4]; in particular $P\mathrm{\ne }1$. Because $N⊴G$, every $G$-conjugate of $P$ is again a Sylow $p$-subgroup of $N$. Hence [L4] gives, for each $g\in G$, an $n\in N$ with $gP{g}^{-1}=nP{n}^{-1}$. Thus $n^{-1}g\in N_G(P)$ and $G=N{N}_G(P)$. If $P<N$ and $N_G(P)=G$, then $P$ is a nontrivial proper normal subgroup of $G$ contained in $N$, contrary to step 2.1. Thus $P<N$ forces $R:=N_G(P)<G$. Put $A=R\cap N$. The subgroup $A$ is normal in $R$, and $G=NR$ gives $R\mathrm{/}A\cong G\mathrm{/}N$; hence $A$ is a normal Hall subgroup of $R$. Induction gives a complement $H$ to $A$ in $R$, so $R=AH$. Consequently $G=NR=NH$ and $N\cap H=A\cap H=1$, proving the theorem when $P<N$. The only remaining case is $P=N$, so $N$ is a $p$-group.

By [L5], the finite $p$-group $N$ has nontrivial center. The center is characteristic in $N$, so [L6] makes it normal in $G$; step 2.1 therefore forces $Z(N)=N$, and $N$ is abelian. Now ${\mathrm{\Omega }}_1(N)=\{x\in N:x^p=1\}$ is a subgroup, is nontrivial by [L3], and is characteristic because automorphisms preserve $p$th powers. Hence [L6] makes it normal in $G$, and step 2.1 again forces ${\mathrm{\Omega }}_1(N)=N$. Therefore every nonidentity element of $N$ has order $p$, so $N$ is elementary abelian.

Write $Q=G\mathrm{/}N$, choose a set-theoretic section $s:Q\to G$ with $s(1)=1$, and write the abelian group $N$ additively. Because $N$ is abelian, the formula $q\cdot n=s(q)ns(q{)}^{-1}$ does not depend on the chosen lift of $q$, and [L7] makes it an action of $Q$ on $N$ by automorphisms. Define $f:Q\times Q\to N$ by $s(q)s(r)=f(q,r)s(qr)$. This is well defined because $\pi (s(q)s(r))=qr$, so $s(q)s(r)s(qr{)}^{-1}\in N$.

Associativity gives $f(q,r)+f(qr,t)=q\cdot f(r,t)+f(q,rt)$ for all $q,r,t\in Q$. Let $m=\mathrm{\mid }Q\mathrm{\mid }$, choose an integer $b$ with $bm\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}p)$, define $\sigma (q)={\sum }_{x\in Q}f(q,x)$ and $c(q)=b\sigma (q)$, and sum the cocycle identity over $x\in Q$. Since $x\mapsto rx$ is a permutation of $Q$, this yields $mf(q,r)+\sigma (qr)=q\cdot \sigma (r)+\sigma (q)$. Because every element of the elementary abelian $p$-group $N$ has order dividing $p$, multiplication by $bm$ is the identity on $N$. Hence $f(q,r)=c(q)+q\cdot c(r)-c(qr)$.

Define $s^{\mathrm{\prime }}:Q\to G$ by $s^{\mathrm{\prime }}(q)=-c(q)+s(q)$, meaning $s^{\mathrm{\prime }}(q)=i(-c(q))s(q)$ inside the extension. Since $i(-c(q))\in N$, we have $\pi (s^{\mathrm{\prime }}(q))=q$. Using the formula from step 6.1, $s^{\mathrm{\prime }}(q)s^{\mathrm{\prime }}(r)=(-c(q)-q\cdot c(r)+f(q,r))s(qr)=-c(qr)s(qr)=s^{\mathrm{\prime }}(qr)$. So $s^{\mathrm{\prime }}$ is a homomorphic section of $G\to Q$. By [L2], the extension splits, equivalently $N$ has a complement in $G$. This completes the induction.