Prove that if is a prime and is a subgroup of of order , then .

Note that every nonidentity element of is a -cycle, since the order of an element in is the least common multiple of the lengths of the cycles in its decomposition and is prime. Note also that any two conjugates of either intersect trivially or coincide, since is prime. Moreover, every -cycle is contained in some conjugate of because all -cycles are conjugate. Finally, in the previous exercise we computed that the number of -cycles in is . Thus there are distinct conjugates of in , hence via the Orbit-Stabilizer Theorem.

## Comments

I don’t get how you arrived at (p-2)! how do you there are p-1 distinct number of conjugacy classes?

The -cycles are partitioned into sets of order each, corresponding to the nonidentity elements of each Sylow -subgroup. There are total -cycles, hence Sylow -subgroups.

a typo: at the last sentence.e

a direct method: any permutation that normalizes should send 1 to one of and 2 to one of those except the image of 1. This p(p-1) choice completely determines the permutation.

Very nice. Thanks!