Let $G$ be a finite group. A subgroup $H$ of $G$ is said to be a Hall subgroup if $$\gcd (|H|,|G/H|)=1.$$ In other words, $H$ is a Hall subgroup if the order of $H$ and its index in $G$ are coprime. These subgroups are name after Philip Hall who used them to characterize solvable groups.

Hall subgroups are a generalization of Sylow subgroups. Indeed, every Sylow subgroup is a Hall subgroup. According to Sylow's theorem, this means that any group of order $p^k m$ , $\gcd(p,m)=1$ , has a Hall subgroup (of order $p^k$ ).

A common notation used with Hall subgroups is to use the notion of $\pi$ -groups. Here $\pi$ is a set of primes and a Hall $\pi$ -subgroup of a group is a subgroup which is also a $\pi$ -group, and maximal with this property.

The sets of primes $\pi$ in Hall's theorem can be restricted to the subsets of primes which divide $|G|$ . However, this result fails for non-solvable groups.

Proof. Suppose that $A_5$ has a Hall $\{2,5\}$ -subgroup $H$ . As $|A_5|=60$ , it follows that $|H|=20$ . Thus, there are three cosets of $H$ . Since a group always acts on the cosets of a subgroup, it follows that $A_5$ acts on the three member set $C$ of cosets of $H$ . This induces a non-trivial homomorphism from $A_5$ to $S_C\cong S_3$ (here, $S_C$ is the symmetric group on $C$ , see this for more detail). Since $A_5$ is simple, this homomorphism must be one-to-one, implying that its image must have order at most $6$ , an impossibility.

This example can also be proved by direct inspection of the subgroups of $A_5$ . In any case, $A_5$ is non-abelian simple and therefore it is not a solvable group. Thus, Hall's theorem does not apply to $A_5$ .