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 ( ).