Prove: A group $G$ is abelian if and only if the map $G\rightarrow G$ given by $x\mapsto x^{-1}$ is an automorphism.

Question

Problem: A group $G$ is abelian if and only if the map $G\rightarrow G$ given by $x\mapsto x^{-1}$ is an automorphism. Prove it.

Solution: $(\Rightarrow)$ If $f(x_1)=f(x_2)$, then $x_1^{-1}=x_2^{-1}\Rightarrow x_2=x_1$, by multiplying both sides by $x_1x_2$ and the assumption that $G$ is commutative. Thus $f$ is injective. Also, since $G$ is a group, $\forall x\in G, \exists x^{-1}\in G; xx^{-1}=e_G$ which proves that $f$ is surjective.
$(\Leftarrow)$ $\forall x,y\in G; xy=\Big((xy)^{-1}\Big)^{-1}=f\Big(f(xy)\Big)=f(y^{-1}x^{-1})$, but since $G$ is a automorphism (and therefore a homomorphism), $xy=f(y^{-1})f(x^{-1})=yx \square$

Anything wrong with the solution?

Answer 1

If $G$ is abelian then the map $x\to x^{-1}$ is a homomorphism: $$(ab)^{-1}=(ba)^{-1}=a^{-1}b^{-1}$$ If $x\to x^{-1}$ is a homomorphism, $G$ is abelian: $$ab=(b^{-1}a^{-1})^{-1}=((ba)^{-1})^{-1}=ba$$

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$x\to x^{-1}$ is a bijection as proved in the question. So if it is a homomorphism, it is an automorphism.