Schur's Theorem (Group Theory)

Theorem

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

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 $\left\vert{G}\right\vert$, 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 $Syl_p \left({N}\right)$ 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 $\left\vert{G_1}\right\vert = \left\vert{K / Z\left({P}\right)}\right\vert \times \left\vert{Z \left({P}\right)}\right\vert = \left\vert{G / N}\right\vert \times \left\vert{Z \left({P}\right)}\right\vert$.

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 $G$.

$\blacksquare$

Source of Name

This entry was named for Issai Schur.