Schur-Zassenhaus Theorem

Let $$G$$ be a finite group and $$N$$ be a normal subgroup in $$G$$.

Theorem
If $$N$$ is a Hall subgroup of $$G$$, then there exists $$H$$, a complement of $$N$$, such that $$G$$ is the semidirect product of $$N$$ and $$H$$.

Proof by Induction
By definition, $$N$$ is a Hall subgroup iff the index and order of $$N$$ in $$G$$ are relatively prime numbers.

Let $$G$$ be a group whose identity is $$e$$.

We induct on $$\left|{G}\right|$$, where $$\left|{G}\right|$$ is the order of $$G$$.

We may assume that $$N \ne \left\{{e}\right\}$$.

Let $$p$$ be a prime number dividing $$\left|{N}\right|$$.

Let $$Syl_p \left({N}\right)$$ be the set of Sylow $p$subgroups of $$N$$.

By the First Sylow Theorem, $$Syl_p \left({N}\right) \ne \varnothing$$.

Let:
 * $$P \in Syl_p \left({N}\right)$$;
 * $$G_0$$ be the normalizer in $$G$$ of $$P$$:
 * $$N_0 = N \cap G_0$$.

By Frattini's Argument $$G = G_0 N$$.

By the Second Isomorphism Theorem and thence Lagrange's Theorem, it follows that:
 * $$N_0$$ is a Hall subgroup of $$G_0$$;
 * $$\left[{G_0:N_0}\right] = \left[{G : H}\right]$$.

Now suppose $$G_0 < G$$.

Then by induction applied to $$N_0$$ in $$G_0$$, we find that $$G_0$$ contains a complement $$H \in N_0$$.

Now $$|H| = \left[{G_0:N_0}\right]$$, and so $$H$$ is also a complement to $$N$$ in $$G$$.

So we may assume that $$P$$ is normal in $$G$$ (i.e. $$G_0 < G$$).

Let $$Z \left({P}\right)$$ be the center of $$P$$.

Since $$Z \left({P}\right)$$ is characteristic in $$P$$, it is also normal in $$G$$.

If $$Z \left({P}\right) = N$$ then there is a long exact sequence of cohomology groups:
 * $$0 \to H^1(G/N, P^N) \to H^1(G,P) \to H^1(N,P)\to H^2(G/N,P) \to H^2(G,P)$$ which splits as desired.

Otherwise, $$Z \left({P}\right) \ne N$$.

In this case $$N / Z \left({P}\right)$$ is a normal (Hall) subgroup of $$G / Z \left({P}\right)$$.

By induction, $$N / Z \left({P}\right)$$ has a complement $$H / Z \left({P}\right)$$ in $$E // Z \left({P}\right)$$.

Let $$G_1$$ be the preimage of $$H // Z \left({P}\right)$$ in $$G$$ (under the equiv. relation).

Then $$\left|{G_1}\right| = \left|{K / Z\left({P}\right)}\right| \times \left|{Z \left({P}\right)}\right| = \left|{G / N}\right| \times \left|{Z \left({P}\right)}\right|$$.

Therefore, $$Z \left({P}\right)$$ is normal Hall subgroup of $$G_1$$.

By induction, $$Z \left({P}\right)$$ has a complement in $$G_1$$ and is also a complement of $$N$$ in $$G$$.