TITLE: What is the size of the normalizer of a subgroup generated by a $p$-cycle in a symmetric group?
QUESTION [7 upvotes]: 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 $n \geq 2p$.

REPLY [2 votes]: 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
$$
\#N(P) = \frac{\#S_n}{\#X} = p(p-1)(n-p)!.
$$