Hall subgroup
In mathematics, specifically group theory, a Hall subgroup of a finite group G is a subgroup whose order is coprime to its index. They were introduced by the group theorist.
Definitions
A Hall divisor of an integer n is a divisor d of n such thatd and n/''d are coprime. The easiest way to find the Hall divisors is to write the prime power factorization of the number in question and take any subset of the factors. For example, to find the Hall divisors of 60, its prime power factorization is 22 × 3 × 5, so one takes any product of 3, 22 = 4, and 5. Thus, the Hall divisors of 60 are 1, 3, 4, 5, 12, 15, 20, and 60.
A Hall subgroup of G'' is a subgroup whose order is a Hall divisor of the order of G. In other words, it is a subgroup whose order is coprime to its index.
If π is a set of primes, then a Hall π-subgroup is a subgroup whose order is a product of primes in π, and whose index is not divisible by any primes in π.
Examples
- Any Sylow subgroup of a group is a Hall subgroup.
- The alternating group A4 of order 12 is solvable but has no subgroups of order 6 even though 6 divides 12, showing that Hall's theorem cannot be extended to all divisors of the order of a solvable group.
- If G = A5, the only simple group of order 60, then 15 and 20 are Hall divisors of the order of G, but G has no subgroups of these orders.
- The simple group of order 168 has two different conjugacy classes of Hall subgroups of order 24.
- The simple group of order 660 has two Hall subgroups of order 12 that are not even isomorphic. The normalizer of a Sylow of order 4 is isomorphic to the alternating group A4 of order 12, while the normalizer of a subgroup of order 2 or 3 is isomorphic to the dihedral group of order 12.
Hall's theorem
is any set of primes, then G has a Hall π-subgroup, and any two Hall are conjugate. Moreover, any subgroup whose order is
a product of primes in π is contained in some Hall. This result can be thought of as a generalization of Sylow's Theorem to Hall subgroups, but the examples above show that such a generalization is false when the group is not solvable.
The existence of Hall subgroups can be proved by induction on the order of G, using the fact that every finite solvable group has a normal elementary abelian subgroup. More precisely, fix a minimal normal subgroup A, which is either a or a as G is. By induction there is a subgroup H of G containing A such that H/''A is a Hall of G''/A. If A is a then H is a Hall of G. On the other hand, if A is a, then by the Schur–Zassenhaus theorem A has a complement in H, which is a Hall of G.