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