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.

Answer 3

I will prove that the number of elements of order $p$ is $(p-1)!$. Recall that if $\sigma$ is a permutation, we have a decomposition in terms of independent cycles so:

$$\sigma=\sigma_1\circ … \circ \sigma_k$$

Because these cycles are independent, we must have:

$$\sigma^n=\sigma_1^n\circ … \circ \sigma_k^n$$

We will have this being the identity if and only if $\sigma_j^n =1$ for every $j$. Equivalently $|\sigma_j| $ must divide $n$. The smallest such $n$ is $|\sigma|=n=LCM(|\sigma_1|,…,|\sigma_k|)$.

We want $p= LCM(|\sigma_1|,…,|\sigma_k|)$. This will happen if and only if every $|\sigma_j|=p$ (none of these cycles is the identity). If they have any chance to be independent, there must be only one such cycle (there are $p$ elements in these cycles).

So we need to count how many cycles with $p$ elements there are. There are $p!$ lists of p elements. We can arrange these lists in subsets with $p$ lists all of which are associated with the same permutation $(i_1,…,i_{p-1},i_p)=(i_p,i_1,…,i_{p-1})=…$, so there are only $p!/p$ cycles we must count.