# 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 $p(p-1)(n-p)!$.
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 the number of $$p$$-cycles in every $$P\in X$$ is $$p-1$$, and every $$p$$-cycle is contained in a unique $$P\in S_n$$ (namely $$P = \langle p\rangle$$), 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 $$\#N(P) = \frac{\#S_n}{\#X} = p(p-1)(n-p)!.$$