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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 can we prove Wilson's theorem which says $(p-1)! \equiv -1 \ (\text{mod} \ p)$ using this result.

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up vote 6 down vote accepted

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 $p$.)

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

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So the number sought for is a divisor of $(p-1)!$ :) (You probably want to say that action is simply transitive or regular) – Mariano Suárez-Alvarez Sep 29 '10 at 8:22
Yes, but I don't want to spoil everything. – Martin Brandenburg Sep 29 '10 at 8:31

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