Schur-Zassenhaus Theorem

Schur's 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 a complement of N, H such that G is the semidirect product of N and H.

Note: N is a Hall subgroup iff the index and order of N in G are relatively prime numbers. Remark: This proof uses many other theorems and we try to cite them where it is believed to be necessary. It is also not from first principles.

Proof:

We induct on $$|G|$$.

We may assume that $$N \neq \{1 \}$$.

Let $$p$$ be a prime number dividing $$|N|$$ and let $$P \in Syl_p(N)$$ (where $$Syl_p$$ is the set of Sylow p subgroups) which exists by Sylow's Theorem. Let $$G_0$$ be the normalizer in $$G$$ of $$P$$ and $$N_0 = N \cap G_0$$.

By Frattini's Argument(Theorem) $$G = G_0N$$. By the Second Isomorphism Theorem, $$N_0$$ is a Hall subgroup of $$G_0$$ and $$|G_0:N_0|=|G:N|$$.

Now, if $$G_0 <G $$ then by induction applied to $$N_0$$ in $$G_0$$ we get that $$G_0$$ contains a complement $$H \in N_0$$. Since $$|H|=|G_0:N_0|$$, $$H$$ is also a complement to $$N$$ in $$G$$, thus we may assume that $$P$$ is normal in $$G$$ (i.e. $$G_0 <G $$).

Since $$Z(P)$$ is characteristic in $$P$$, it is also normal in $$G$$.

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

Otherwise, $$Z(P) \neq N$$. In this case $$N/Z(P)$$ is a normal (Hall) subgroup of $$G/Z(P)$$. By induction, $$N/Z(P)$$ has a complement $$H/Z(P)$$ in $$E//Z(P)$$.

Let $$G_1$$ be the preimage of $$H//Z(P)$$ in $$G$$ (under the equiv. relation). Then $$|G_1|=|K/Z(P)||Z(P)|=|G/N||Z(P)|$$.

Therefore, $$Z(P)$$ is normal Hall subgroup of $$G_1$$.

By induction $$Z(P)$$ has a complement in $$G_1$$ and is also a complement of $$N$$ in $$G$$.

QED