- Hall subgroup
In
mathematics , a Hall subgroup of a finite group "G" is a subgroup whose order iscoprime to its index. They are named after the group theoristPhilip Hall .Definitions
A Hall divisor of an integer "n" is a divisor "d" of "n" such that"d" and "n"/"d" are coprime. The easiest way to find the Hall divisors is to write the prime factorization for the number in question and take any product of the multiplicative terms (the full power of any of the prime factors), including 0 of them for a product of 1 or all of them for a product equal to the original number. For example, to find the Hall divisors of 60, show the prime factorization is 22·3·5 and take any product of {3,4,5}. 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.*If "G" = "A"5, 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 (though they are conjugate under an outer automorphism of "G").
*The simple group of order 660 has two Hall subgroups of order 12 that are not isomorphic.
Hall's theorem
Hall proved that if "G" is a finite
solvable group and "π"is any set of primes, then "G" has a Hall "π"-subgroup, and anytwo Hall "π"-subgroups are conjugate. Moreover any subgroup whose order isa product of primes in "π" is contained in some Hall "π"-subgroup. 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.Hall's theorem can be proved by induction on the order of "G", using the fact that every finite solvable group has a normal elementary abelian subgroup.
A converse to Hall's theorem
Any finite group that has a Hall "π"-subgroup for every set of primes "π" is solvable. This is a generalization of
Burnside's theorem that any group whose order is of the form "p aq b" for primes "p" and "q" is solvable, becauseSylow's theorem implies that all Hall subgroups exist. This does not give another proof of Burnside's theorem, because Burnside's theorem is needed to prove this converse.Sylow systems
A Sylow system is a set of Sylow "p"-subgroups "Sp" for each prime "p" such that "SpSq" = "SqSp" for all "p" and "q". If we have a Sylow system, then the subgroup generated by the groups "Sp" for "p" in "π" is a Hall "π"-subgroup. A more precise version of Hall's theorem says that any solvable group has a Sylow system, and any two Sylow systems are conjugate.
References
*citation|title=Finite groups|first=Daniel|last=Gorenstein|authorlink=Daniel Gorenstein| ISBN =0828403015|year=1980.
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