group theoretic proof of Wilson’s theorem

Here we present a group theoretic proof of it.
Clearly, it is enough to show that $(p-2)!\equiv 1\pmod{p}$ since $p-1\equiv-1\pmod{p}$ . By Sylow theorems, we have that $p$ -Sylow subgroups of $S_{p}$ , the symmetric group on $p$ elements, have order $p$ , and the number $n_{p}$ of Sylow subgroups is congruent to 1 modulo $p$ . Let $P$ be a Sylow subgroup of $S_{p}$ . Note that $P$ is generated by a $p$ -cycle. There are $(p-1)!$ cycles of length $p$ in $S_{p}$ . Each $p$ -Sylow subgroup contains $p-1$ cycles of length $p$ , hence there are $\frac{(p-1)!}{p-1}=(p-2)!$ different $p$ -Sylow subgrups in $S_{p}$ , i.e. $n_{P}=(p-2)!$ . From Sylow’s Second Theorem, it follows that $(p-2)!\equiv 1\pmod{p}$ ,so $(p-1)!\equiv-1\pmod{p}$ .

Title	group theoretic proof of Wilson’s theorem
Canonical name	GroupTheoreticProofOfWilsonsTheorem
Date of creation	2013-03-22 13:35:27
Last modified on	2013-03-22 13:35:27
Owner	ottocolori (1519)
Last modified by	ottocolori (1519)
Numerical id	10
Author	ottocolori (1519)
Entry type	Proof
Classification	msc 11-00