Let be a group. Suppose that
-----(1)for all .
(i)Show that for all , thenand .(ii)If satisfies (1) and the condition that the identity element in , is the only element in satisfying , show that for all ,and conclude that is abelian.
Hint 1. By condition (1), the map given by is a homomorphism from the group to itself.
Hint 2. If the kernel of consists solely of the identity, then is injective. In particular, if then .
Hint 3. Use the properties in (i) to show that .
To conclude that G is abelian, notice that , from which , and it easily follows that .