Hall's theorem
From Groupprops
Contents
Statement
Suppose is a finite group such that for any subset
of the set of prime divisors of
,
has a
-Hall subgroup. Then,
is a Solvable group (?) (specifically,
is a Finite solvable group (?)).
Related facts
- Existence of pi-subgroups for all prime sets pi is equivalent to existence of p-complements for all primes p
- ECD condition for pi-subgroups in solvable groups is a converse to this result
Facts used
- Order has only two prime factors implies solvable (this result is popularly called Burnside's
theorem).
- Three solvable subgroups of pairwise coprime indices implies solvable
Proof
Case that the order has one or two prime factors
If the order has two or fewer prime factors, fact (1) tells us that the group is solvable. (If the order has only one prime factor, the group is in fact nilpotent).
Case that the order has three or more prime factor
We prove this claim by induction on the number of prime factors of the order. Note that the cases of one or two prime factors have already been dealt with.
Suppose the order of is
where
are distinct primes.
- There exist
-complements for each prime
: In other words, for each
, there exists a subgroup
whose index is
. This follows from the assumption that there exist Hall subgroups of all possible orders.
- For any subset
of
, the intersection of the subgroups
with
, is a
-Hall subgroup: PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]
- Each
satisfies the hypothesis of containing Hall subgroups of all possible orders: This follows from step (2).
-
is solvable for each
: This follows from the induction hypothesis.
-
is solvable: This follows from fact (2), and the observation that when
, we have a collection of at least three solvable subgroups of pairwise coprime indices.