# Proof of Wilson’s Theorem using Sylow’s Theorem

Wilson’s theorem $(p-1)!\equiv -1 \pmod p$ is a useful theorem in Number Theory, and may be proved in several different ways. One of the interesting proofs is to prove it using Sylow’s Third Theorem.

Let $G=S_p$, the symmetric group on p elements, where p is a prime.

$|G|=p!=p(p-1)!$

By Sylow’s Third Theorem, we have $n_p\equiv 1\pmod p$. The Sylow p-subgroups of $S_p$ have $p-1$ p-cycles each.

There are a total of $(p-1)!$ different p-cycles (cyclic permutations of p elements).

Thus, we have $n_p (p-1)=(p-1)!$, which implies that $n_p=(p-2)!$

Thus $(p-2)!\equiv 1\pmod p$, and multiplying by p-1 gives us $(p-1)!\equiv p-1\equiv -1\pmod p$ which is precisely Wilson’s Theorem. 🙂

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## Author: mathtuition88

https://mathtuition88.com/

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