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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 $p!$?

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Isn'it normal since it's index two? –  Alex Youcis Jan 28 '12 at 16:50
    
Reflections. See wikipedia The rotations are a subgroup of index 2, hence normal, so every other symmetry of the p-gon wil normalize it, in particular the reflections and the unity. –  Myself Jan 28 '12 at 16:50
    
@AlexYoucis: Thanks, sorry, my mistake, I misunderstood $S_p$ to be the symmetric group of a p-gon (which should be $D_{2\times p}$, right?) I just mean the cyclic group $\langle(1,2,...,p)\rangle$ in the symmetric group $S_p$. –  Mark Jan 28 '12 at 17:06
    
@Myself: Thanks, sorry for my mistake, but I meant $S_p$. Please c.f. my comment to Alex. –  Mark Jan 28 '12 at 17:07
    
Okay, here's a 'hint': try to determine an element that maps $x = (1,2,\dots,p)$ to $x^2 = (1,3,\dots,p,2,\dots,p-1)$. Because $x^2$ generates that subgroup, such an element will normalize the group. (And similarly for $x^3$, ...) –  Myself Jan 28 '12 at 17:13

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up vote 2 down vote accepted

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.

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Thanks, Tobias, I as confused, I thought $S_p$ was the symmetry group of a p-gon, when it is actually the group of permutations of p elements. Could you suggest what the m=normalizer is in that case? –  Mark Jan 28 '12 at 17:04

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