Number of elements of order $m$ in the symmetric group $S_m$

Question

What is the number of elements of order $m$ in the symmetric group $S_m$ given that $m$ is a prime.

I believe it is $\left(m-1\right)!$.

Because on cycle form the its possible Length is $m$ only. Only $1$ cycle is possible with length $m$ and I guess it had $\left(m-1\right)!$ possibilities only.

$S_m$ is defined for all positive integers $m$, whether prime or composite. This problem is harder when $m$ is composite. For example, $(12)(345)(6)$ has order $6$ in $S_6$. — C Monsour, Jun 25 '18 at 02:19

Ross Millikan · Answer 1 · 2018-06-25T14:54:18.037

1

For prime $m$ your result is correct. We cannot comment on your reasoning as you did not present it. To get a cycle of length $m$ you can send $1$ to anywhere but $1$, which is $m-1$ choices. Then you send than number anywhere but $1$ or itself, which is $m-2$ choices. This results in $(m-1)!$ elements in $S_m$ with period $m$.

edited Jun 25 '18 at 14:54

answered Jun 25 '18 at 02:40

Ross Millikan

374,822

Given that the OP has not asked the question they wanted to ask I think this would be better as a comment. – Jyrki Lahtonen Jun 25 '18 at 14:30
@JyrkiLahtonen: I don't see where OP wants a different question. i added some detail. – Ross Millikan Jun 25 '18 at 14:54

Number of elements of order $m$ in the symmetric group $S_m$

1 Answers1