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 $n \geq 2p$.