Wilson’s theorem and further thought…

In any course on Elementary Number Theory you meet Wilson’s Theorem. This says that for any prime p we have that (p-1) \equiv -1 \,mod \,p.

How do we prove this? Well I have reshaped the usual proof in order to generalise.

My claim: For any finite Abelian group, the product of all elements is equal to the product of 2-torsion elements (i.e. the self-inverse ones).

This is easy to see…every element in the product has to have an inverse (we are in a group) and either a given element is self-inverse or not. Thus when taking the product as a whole everything that is not self-inverse gets inverted (if we jumble up the product)…leaving only those elements that are self-inverse.

Now we may prove Wilson’s theorem. The non-zero integers mod p form a finite group under multiplication. The product of all elements is the same as (p-1)!. By the above this product is the same as the product of 2-torsion elements. These ones will correspond to solutions of a^2 \equiv 1 \,mod \,p. Solving we find that 1, p-1 are the only 2-torsion elements…hence (p-1)! \equiv p-1 \equiv -1 \,mod \,p. QED

In a similar vein we find that for any n>0 and odd prime p then discarding the multiples of p from (p^n - 1)! gives us something which is also -1 \,mod \,p^n.

Maybe other finite groups will provide other Wilson’s Theorems?


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