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 by Induction
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

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