Proving the normalizer of a prime order subgroup in S(p)

[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.