What is the size of the normalizer of a subgroup generated by a $p$-cycle in a symmetric group? 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 $n \geq 2p$.

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