Third Sylow Theorem

Let be a Group and the set of Sylow Subgroups. The number satisfies and .

Proof

By the Second Sylow Theorem, acts Transitively on , so so it suffices to show . Let be a Sylow -subgroup. Consider the Conjugation Group action of on . Note that all orbits of this action are divisors of . Also . Suppose another orbit has order 1, i.e. orbit of is of size 1 for some Then , so and are Sylow -subgroups of . So for some we have . Hence there is exactly one orbit of size , but is the sum of orbit sizes so it has to be .

Corollary

If then there is a Normal Sylow -subgroup of .

Proof

is closed under Conjugation by Second Sylow Theorem