Let be a finite group. A subgroup of is said to be a Hall subgroup if

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 , , has a Hall subgroup (of order ).

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

Theorem 1 (Hall (1928))   A finite group is solvable iff has a Hall -subgroup for any set of primes .

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

Example 2   The group has no Hall -subgroup. That is, has no subgroup of order .

Proof. Suppose that has a Hall -subgroup . As , it follows that . Thus, there are three cosets of . Since a group always acts on the cosets of a subgroup, it follows that acts on the three member set of cosets of . This induces a non-trivial homomorphism from to (here, is the symmetric group on , see this for more detail). Since is simple, this homomorphism must be one-to-one, implying that its image must have order at most , an impossibility.

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