Automorphism iff G is abelian

Question

Let $G$ be a group. Prove the mapping $\alpha(g)=g^{-1}\forall g \in G$ is an automorphism iff $G$ is abelian.

Proof (forwards): Assume $G$ is an automorphism. Show $ab=ba$. How would I even go about it this way?

Proof (backwards): Assume abelian. Then show automorphism, so 1-1, onto, and homomorphism. I have the 1-1 part. For the onto part, is it: if $x \in G$, then there exists a $g \in G$ such that $\alpha(g)=g^{-1}=x$. What next? Lastly for homomorphism, $\alpha(mn)=(mn)^{-1}=m^{-1}n^{-1}=\alpha(m)\alpha(n)$.

Answer 1

Assume that $\alpha$ is an automorphism. Then for each $a,b\in G$ we have:

$ab=\alpha(a^{-1})\alpha(b^{-1})=\alpha(a^{-1}b^{-1})=\alpha((ba)^{-1})=ba$,

hence $G$ is abelian.

Answer 2

If it's automorphism then $\alpha(ab) = \alpha(a) \alpha(b)$ for every pair. but then $$(ab)^{-1} = a^{-1}b^{-1}$$and then $$((ab)^{-1})^{-1} = (a^{-1}b^{-1})^{-1}$$So$$ab = ba$$ And every pair of elements commute

Answer 3

If $G$ is not abelian group we have that $(mn)^{-1}=n^{-1}m^{-1}$ then $\alpha(mn)=n^{-1}m^{-1}\neq\alpha(nm)$

Answer 4

"Forward" proof:

$$1=ab(ab)^{-1}=ab\alpha(ab)=ab\alpha(a)\alpha(b)=aba^{-1}b^{-1}$$

So:

$$ab=ba$$

"Backward" proof, onto part:

If $x \in G$, then exists $g \in G$ that $\alpha(g)=x$. You can put $g=x^{-1}$.