3 Let G be a group and define G G via g g1 a Prove that is

3. Let G be a group and define : G G via (g) = g^-1 (a) Prove that is a bijection. (b) Prove that is an isomorphism if and only if G is Abelain.

Solution

This is not a homomorphism. If a, b G, then (ab) = (ab) 1 = b 1a 1 , which is not equal to (a)(b) = a 1 b 1 in general.

When G is abelian, this map is a homomorphism. From the computation we did in part (b), we see that (ab) = b 1 a 1 = a 1 b 1 = (a)(b).

Note that if a ker , then e = (a) = a 1 , which means that a = e, and ker = {e}. This implies that is one-to-one. It is also onto: if a G, then (a 1 ) = (a 1 ) 1 = a