Hi playa,

Let , . We first show that .

Since is finite, map from to is a bijection for every .

Thus, because , we have and and .

It follows that .

Further, since , we have .

So there must exist at least one such that .

We see this has many desirable properties, among others, it commutes with and with .

So we get , thus .

In fact we've proved that is a subgroup of .

But since is too fatty, .

Since all elements of are involutions, must be abelian ( ).