Schur-Zassenhaus Theorem

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

Let $N$ be 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
The proof proceeds by induction.

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

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

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

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

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

Let $\operatorname{Syl}_p \left({N}\right)$ be the set of Sylow $p$-subgroups of $N$.

By the First Sylow Theorem:
 * $\operatorname{Syl}_p \left({N}\right) \ne \O$

Let:
 * $P \in \operatorname{Syl}_p \left({N}\right)$
 * $G_0$ 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 for Groups and thence Lagrange's Theorem (Group Theory), it follows that:
 * $N_0$ is a Hall subgroup of $G_0$
 * $\left[{G_0 : N_0}\right] = \left[{G : H}\right]$.

Suppose $G_0 < G$.

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

We have that:
 * $\left\vert{H}\right\vert = \left[{G_0 : N_0}\right]$

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 $P$.

By:
 * Center is Characteristic Subgroup
 * $P$ is normal in $G$
 * Characteristic Subgroup of Normal Subgroup is Normal

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

Let $Z \left({P}\right) = N$.

Then there exists a long exact sequence of cohomology groups:
 * $0 \to H^1 \left({G / N, P^N}\right) \to H^1 \left({G, P}\right) \to H^1 \left({N, P}\right) \to H^2 \left({G / N, P}\right) \to H^2 \left({G, P}\right)$

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 equivalence relation).

Then:
 * $\left\vert{G_1}\right\vert = \left\vert{K / Z\left({P}\right)}\right\vert \times \left\vert{Z \left({P}\right)}\right\vert = \left\vert{G / N}\right\vert \times \left\vert{Z \left({P}\right)}\right\vert$

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 $G$.

Also known as
Some sources refer to this theorem as Schur's theorem, but that name is also used for an unrelated result in Ramsey theory.