Question

Prove that the mapping Θ: g->g-1 of G into its self is always a one to one and onto and is automophism if and only if G is abelian.

Answer 1

What have you tried so far?

Answer 2

Consider the direction, "If G is abelian, then the map \(\Theta(g)=g^{-1}\) is an automorphism".

You need to show two things:

1) The map \(\Theta\) is a bijection, and
2) The map \(\Theta\) is a group homomorphism.

Answer 3

For 1), the bijectivity of \(\Theta\) comes from the fact that \(\left(g^{-1}\right)^{-1}=g\). When trying to formally prove injectivity and surjectivity, there will be spots to use this identity about inverses.

Answer 4

For 2), you need to show for any \(g,h\in G\), that: $$\Theta(gh)=\Theta(g)\Theta(h).$$This is where G being abelian will have to come into play. If G is not abelian, this is not true.

Answer 5

For the converse, "If \(\Theta\) is an automorphism, then G is abelian.", look at the term: $$\Theta(gh)=(gh)^{-1}$$in two ways; one as an automorphism, and one as the inverse of a product of two elements.
Some books refer to the inverse of a product as the "sock-shoes" principle. That would be really useful here.