Hall subgroup

Let $G$ be a finite group. A subgroup $H$ of $G$ is said tobe a Hall subgroup if

\\gcd(|H|,|G/H|)=1.

In otherwords, $H$ is a Hall subgroup if the order of $H$ and its indexin $G$ are coprime. These subgroups arename 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’stheorem, 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 (http://planetmath.org/PiGroupsAndPiGroups). Here $\\pi$ is a setof primes and a Hall $\\pi$ -subgroup of a group is a subgroup which isalso a $\\pi$ -group, and maximal with this property.

Theorem 1 (Hall (1928)).

A finite group $G$ is solvable iff $G$ has a Hall $\\pi$ -subgroup forany set of primes $\\pi$ .

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

Example 2.

The group $A_{5}$ has no Hall $\\{2,5\\}$ -subgroup. That is, $A_{5}$ has no subgroup of order $20$ .

Proof.

Suppose that $A_{5}$ has a Hall $\\{2,5\\}$ -subgroup $H$ .As $|A_{5}|=60$ , it follows that $|H|=20$ . Thus, thereare three cosets of $H$ . Since a group always actson the cosets of a subgroup, it follows that $A_{5}$ actson the three member set $C$ of cosets of $H$ . This inducesa non-trivial homomorphism from $A_{5}$ to $S_{C}\\cong S_{3}$ (here, $S_{C}$ is the symmetric group on $C$ , seethis (http://planetmath.org/GroupActionsAndHomomorphisms) for more detail).Since $A_{5}$ is simple, this homomorphism must be one-to-one,implying that its image must have order at most $6$ , animpossibility.∎

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 asolvable group. Thus, Hall’s theorem does not apply to $A_{5}$ .