There is at least one non-identity element in S: to see this, pair up every element in G with its inverse....what do you get?

By the way, the map is a homomorphism on iff is abelian, so you can use it since you assume the group is abelian...but the result still is true if the group isn't abelian: there's always an element (one, at least) of order 2 in any finite group of even order.

Tonio