What Determines the Normalizer of a Sylow p-Subgroup in Sym(p)?

[MarkovMarakov]

Homework Statement

What is the normalizer of the Sylow p-subgroup in the symmetric group Sym(p) generated by the element (1,2,...,p) where p is a prime number?
Thanks

Homework Equations

na

The Attempt at a Solution

I know that the normalizer has order p(p-1). And I know that it has to include the group generated by (1,2,...,p). I know there must be elements outside <(1,2,...,p)> that conjugates (1,2,...,p) to (1,2,...,p)^n but what are their forms?
Your help will be greatly appreciated.

 

[jbunniii]

Try writing out the p elements of the subgroup in question:

(1 2 3 ... p)
(1 3 5 ... )
etc.
p-1 of these are p-cycles and the final element is the identity map. The identity map is normalized by any element of Sym(P), so we need only consider which elements of Sym(P) map one of the p-cycles to another.

Each p-cycle can be written in one of p equivalent ways, by choosing where to write 1 in the cycle.

Furthermore, each cycle is completely characterized by, for example, the distance between 1 and 2 in the cycle.

Conjugation of a p-cycle by any element of Sym(p) simply relabels the elements and results in another p-cycle. How many of these relabelings will result in one of the powers of (1 2 ... p)?

It's clear that there are p choices for where to place 1, and p-1 choices for where to place 2. These two choices completely characterize the cycle if it is to be a power of (1 2 ... p). This gives you your p*(p-1) for the size of the normalizer.

This analysis should give you enough info to work out what the elements of the normalizer must look like.