Proving |N(P)| for Prime P in S(p) - Dummit & Foote

[hermanni]

I'm reading my textbook (Dummit & Foote) and having trouble at conjugacy section , here's the question:

Prove that if p is a prime and P is a subgroup of S(p) of order p, then
| N (P) | = p(p-1). (Argue that every conjugate of P contains exactly p-1 p-cycles and use the formula for the number of p-cycles to compute the index of
N(P) in S(p) .
N(P) : normalizer of P in S(p).

I reaaly have no clue , can someone help??

 

[ystael]

Suppose, as you say, that P is a subgroup of \mathfrak{S}_p of order p. What can the elements of P be, in concrete terms? There aren't that many choices.

Also, given a permutation \sigma \in \mathfrak{S}_p, there is a general way to write down the elements of its conjugacy class \{ \pi\sigma\pi^{-1} \mid \pi \in \mathfrak{S}_p \}. (It helps to write out \sigma in cycle decomposition first.)

 

[hermanni]

I think that P must be cyclic (since it's of prime order ) so it's generated by an element of order p , which must be an p-cycle.Am I right?

 

[ystael]

Right. So if \kappa is a p-cycle, and P = \langle\kappa\rangle is the subgroup generated by \kappa, what can the conjugates \pi P \pi^{-1} look like?

 

[hermanni]

Well, conjugates have the same cycle structure.Conjugate of P has also p elements , but I still can't see why they must be all p-cycles.