# Hall subgroup

Let $G$ be a finite group^{}. A subgroup^{} $H$ of $G$ is said to
be a *Hall subgroup* if

$$\mathrm{gcd}(|H|,|G/H|)=1.$$ |

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 ${p}^{k}m$, $\mathrm{gcd}(p,m)=1$,
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 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.

###### Theorem 1 (Hall (1928)).

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

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.

###### Example 2.

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

###### 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 (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 $6$, 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 ${A}_{5}$.

Title | Hall subgroup |
---|---|

Canonical name | HallSubgroup |

Date of creation | 2013-03-22 14:02:02 |

Last modified on | 2013-03-22 14:02:02 |

Owner | Algeboy (12884) |

Last modified by | Algeboy (12884) |

Numerical id | 22 |

Author | Algeboy (12884) |

Entry type | Definition |

Classification | msc 20D20 |

Related topic | VeeCuhininsTheorem |

Related topic | SylowTheorems |

Defines | Hall’s theorem |

Defines | Hall $\pi $-subgroup |