number of subgroup in $S_n$ which are isomorphic to $\mathbb{Z}_n$

Question

This is a generalization of my previous question in number of subgroups in $S_5$ which are isomorphic to $\mathbb{Z}_5$

I want to find the number of subgroups in $S_n$ that are isomorphic to $\mathbb{Z}_n$.

Inspired by @Brian M. Scott, I write my proposal as follow:

Since $\mathbb{Z}_n$ is cyclic, I guess $p=\{1,2,3,4,5,\cdots, n\}$ and $p, p^2, p^3, \cdots, p^n=e$.

Letting $p=(1,2,3,\cdots, n)$ and since it is cyclic there are $(n-1)!$ case of permutations. So my guess is $\frac{(n-1)!}{n-1} = (n-2)!$

Am I right? Is there any other way to find the number of subgroups?

Answer 1

No. Powers of $n$-cycles $(123\dots n)$ are not $n$-cycles unless the power is prime to the order. And there are more elments of order $n$ than just an $n$-cycle.

Examples:

• $n=4$: There are indeed $3!=6$ 4-cycles, but each $C_4$ only contains 2 of them.
• $n=6$: cycletypes (6) and (2,3). There are $5!=120$ cycletype (6) and another $\binom{6}{3,2,1}$ cycletype (2,3). Each $C_6$ contains $\phi(6)=2$ generators.
• $n=8,9$ similar to $n=4$.
• $n=10$: cycletypes (10), (2,5), (2,2,5) are all order 10. I'll leave you to count it.
• $n=12$: I'll leave you to list the cycletypes and count.