Second Sylow Theorem

Let be a Group and the set of Sylow Subgroups.
Then all elements of are Conjugate.

Proof

We prove a stronger statement (DONT NEED TO).
Suppose and is a -subgroup.
Then for some .
Consider the Group Action of
on left cosets of by left multiplication.
Note that ,
so there is at least one Orbit not divisible by .
But all orbits divide , so there is an orbit of size .
So for some , and every
we have
i.e. .
Hence for some we have .