## Compute the order of the normalizer of a Sylow p-subgroup in Sym(p)

Prove that if $p$ is a prime and $P$ is a subgroup of $S_p$ of order $p$, then $|N_{S_p}(P)| = p(p-1)$.

Note that every nonidentity element of $P$ is a $p$-cycle, since the order of an element in $S_n$ is the least common multiple of the lengths of the cycles in its decomposition and $p$ is prime. Note also that any two conjugates of $P$ either intersect trivially or coincide, since $|P|$ is prime. Moreover, every $p$-cycle is contained in some conjugate of $P$ because all $p$-cycles are conjugate. Finally, in the previous exercise we computed that the number of $p$-cycles in $S_p$ is $(p-1)!$. Thus there are $(p-2)!$ distinct conjugates of $P$ in $S_p$, hence $|N_{S_p}(P)| = p(p-1)$ via the Orbit-Stabilizer Theorem.

• neal  On July 29, 2011 at 2:40 am

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

• nbloomf  On August 5, 2011 at 10:00 am

The $p$-cycles are partitioned into sets of order $p-1$ each, corresponding to the nonidentity elements of each Sylow $p$-subgroup. There are $(p-1)!$ total $p$-cycles, hence $(p-1)!/(p-1)$ Sylow $p$-subgroups.

• Gobi Ree  On December 19, 2011 at 12:11 am

a typo: $|N_{S_p}(P)|=p(p-1)$ at the last sentence.e

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

• nbloomf  On January 19, 2012 at 10:30 am

Very nice. Thanks!