If $p$ is a prime, then $S_{p-1}$ has no element with order $kp$, $k \in \mathbb{N}$.

Question

How can I prove the statement:

If $p$ is a prime, then $S_{p-1}$ has no element with order $kp$, $k \in \mathbb{N}$.

I don't know what is the best approach, I've tried to use permutations, but I can't proceed. Any help would be appreciated! Thanks in advance!

Answer 1

The order of $S_n $ is $n!$. By Lagrange the order of any element would divide $(p-1)!$. But $kp\nmid (p-1)!$.