Number of Elements of order $p$ in $S_{p}$

Question

An exercise from Herstein asks to prove that the number of elements of order $p$, $p$ a prime in $S_{p}$, is $(p-1)!+1$. I would like somebody to help me out on this, and also I would like to know whether we can prove Wilson's theorem which says $(p-1)! \equiv -1 \ (\text{mod} \ p)$ using this result.

Answer 1

Maybe you mean the number of elements of order dividing $p$ (so that you are including the identity)? (Think about the case $p = 3$ --- there are two three cycles, not three of them.) For the general question, think about the possible cycle structure of an element of order $p$ in $S_p$.

You can go from the formula in your question to Wilson's theorem by counting the number of $p$-Sylow subgroups (each contains $p-1$ elements of order $p$), and then appealing to Sylow's theorem. (You will find that there are $(p-2)!$ $p$-Sylow subgroups, and by Sylow's theorem this number is congruent to $1$ mod $p$. Multiplying by $p-1$, we find that $(p-1)!$ is congruent to $-1$ mod $p$.)

Answer 2

Every element of order $p$ in $S_p$ is a $p$-cycle. The symmetric group $S_{p-1}$ acts transitively on these $p$ cycles.