If it’s trivial, then for all x, y we have x * y * x^-1 = y, or xy=yx, so group is abelian. For abelian groups x -> x^-1 is an automorphism. If it’s nontrivial, we’re done again.

Otherwise we have: x=x^-1, xy=yx. As this group is abelian, we know it’s isomorphic with a product of group of subgroup generated by {x,y} and corresponding quotient group (all subgroups of an abelian group are normal, so the quotient group must exist).

Now {e->e, x->y, y->x, xy->xy} is a nontrivial automorphism of the subgroup, and product of it and identity over the quotient group is a nontrivial automorphism of the entire group.

There’s probably a nicer way that doesn’t require quotient groups anywhere, but I cannot think of it.

]]>