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. 🙂

If you are interested in reading some Math textbooks, do check out our recommended list of Math texts for undergraduates.

You may also want to check out Match Wits With Mensa: The Complete Quiz Book, which is our most popular recommended book on this website.

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