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.

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  • 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!

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