# 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 GroupTheoreticProofOfWilsonsTheorem 2013-03-22 13:35:27 2013-03-22 13:35:27 ottocolori (1519) ottocolori (1519) 10 ottocolori (1519) Proof msc 11-00