## 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?