Symmetric group and commuting elements What is the normalizer of the subgroup $\langle(1,2,...,p)\rangle$ in $S_p$? Clearly it will have to contain the subgroup $\langle(1,2,...,p)\rangle$, but also some additional $p-1$ elements that commute with the elements within $\langle(1,2,...,p)\rangle$. But what are they?
EDIT: Sorry, I have misread the question, it should be the subgroup $\langle(1,2,...,p)\rangle$ in the permutation group $S_p$, which I believe has order $p!$?
 A: The symmetry group of a $p$-gon has order $2p$, so the subgroup generated by the given element has index 2 and is thus normal, ie the normalizer is the entire symmetry group.
Edit: As it was intended to be in the entire symmetric group $S_p$, we need a bit more.
First, we note that the normalizer consists of exactly those elements $\sigma$ such that $\sigma(12\dots p)\sigma^{-1}$ is some power of $(12\dots p)$. So we need to know how powers of $(12\dots p)$ look, and what we get if we conjugate $(12\dots p)$ by some permutation.The last question is easy, as we have $\sigma(12\dots p)\sigma^{-1} = (\sigma(1)\sigma(2)\dots\sigma(p))$. As for the powers of $(12\dots p)$, we can easily calculate that for $n < p$ we have $(12\dots p)^n = (1(1+n)(1+2n)\dots (1+pn))$ with all entries calculated mod $p$.
This means that if $\sigma$ nomalizes the given subgroup, then $\sigma$ has the form $\sigma(k) = ak + b\, (\textrm{mod}\ p)$ with $a\neq 0\, (\textrm{mod}\ p)$. Thus there are a total of $p(p-1)$ elements in the normalizer. (This is all assuming that $p$ is a prime by the way. If not, one needs to modify some of these things).
Edit: I got called away while writing the first edit, and it ended up being quite wrong actually. I have edited the last paragraph, and hopefully, it should now be correct.
