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Question: What is the size of the normalizer of a $p$-cycle (prime $p$) in the symmetric group $S_n$ ($n \geq p$)?

If $n<2p$, we can actually find the size $N:=|N_{S_n}(\langle (12\cdots p) \rangle)|$ using Sylow III. Since $p$ exactly divides $n$, subgroups of order $p$ are Sylow $p$-subgroups. Thus Sylow III implies the number of Sylow $p$-subgroups $n_p$ satisfies $$n_p=\frac{|G|}{N}.$$ By combinatorial arguments, we find $$n_p=\frac{1}{p(p-1)}\binom{n}{p}p!.$$ (There are $\binom{n}{p}p!$ ways of writing a $p$-cycle; $p$ equal copies of each cycle are written; they belong to subgroups consisting of $p-1$ distinct $p$-cycles.) So the normaliser has size $$N=\frac{p!\ (n-p)!}{(p-2)!}.$$

This argument, however, doesn't work when $n \geq 2p$.

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It's the direct product of $N_{S_p}(\langle (1,2,\ldots,p) \rangle)$ and $S_{n-p}$, so it has order $p(p-1)(n-p)!$.

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  • $\begingroup$ Thank you; I can't believe I missed this. $\endgroup$ – Rebecca J. Stones Sep 5 '13 at 11:16
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Here is an alternative solution which is similar to the Sylow approach in the Question.

Let $X$ be the set of all subgroups of $S_n$ generated by a $p$-cycle.

By the standard counting method, the number of $p$-cycles in $S_n$ is $\frac{n!}{p\cdot (n-p)!}$. Since every $P\in S_n$ contains $6$ $p$-cyles, and every $p$-cycle is contained in an unique $P\in S_n$, we get $$ \#X = \frac{n!}{p(p-1)(n-p)!}. $$

$S_n$ acts on $X$ by conjugation. This group action is transitive (since in the $S_n$ two elements are conjugated iff they have the same cycle type), and for any $P\in X$, the stabilizer of $P$ is the normalizer $N(P)$ of $P$ in $S_n$. Now by the orbit-stabilizer formula $$ \#N(P) = \frac{\#S_n}{\#X} = p(p-1)(n-p)!. $$

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