Hall's theorem on solvable groups

Remarks

The proof is by induction on $\abs{G}$. A minimal normal subgroup $N$ of a solvable group is an elementary abelian $p$-group. One applies induction to $G/N$ and takes the preimage $K$ of a Hall $\pi$-subgroup of $G/N$. If $p \in \pi$, then $K$ is a Hall $\pi$-subgroup of $G$. If $p \notin \pi$, then a complement of $N$ in $K$, given by the Schur–Zassenhaus theorem, is one.

Example: $G = S_3 \times \Z/5\Z$ of order $30$ is solvable, and $S_3 \times \set{0}$ is a Hall $\set{2, 3}$-subgroup, of order $6$ and index $5$.

Solvability is necessary. The simple group $A_5$ of order $60$ has no subgroup of order $15$, so it has no Hall $\set{3, 5}$-subgroup. In the simple group $\mathrm{PSL}_2(\F_7)$ of order $168$ there are two conjugacy classes of subgroups isomorphic to $S_4$; these are Hall $\set{2, 3}$-subgroups, of index $7$, that are not conjugate.

Philip Hall also proved a converse. Let $p'$ denote the set of all primes other than $p$. If a finite group has a Hall $p'$-subgroup for every prime $p$, then it is solvable.

Going deeper

For a student

The theorem extends the Sylow theorems from one prime to sets of primes, but only for solvable groups. Together with the converse, it characterises solvable groups by the existence of Hall subgroups. Hall $p'$-subgroups are also called Sylow complements, since a Hall $p'$-subgroup and a Sylow $p$-subgroup together make up the whole order of the group.