Inversion Mapping is Automorphism iff Group is Abelian

Theorem
Let $\struct {G, \circ}$ be a group.

Let $\iota: G \to G$ be the inversion mapping on $G$, defined as:
 * $\forall g \in G: \map \iota g = g^{-1}$

Then $\iota$ is an automorphism $G$ is abelian.

Proof
From Inversion Mapping is Permutation, $\iota$ is a permutation.

It remains to be shown that $\iota$ has the morphism property $G$ is abelian.

Sufficient Condition
Suppose $\iota$ is an automorphism.

Then:

Thus from Inverse of Commuting Pair, $x$ commutes with $y$.

This holds for all $x, y \in G$.

So $\struct {G, \circ}$ is abelian by definition.

Necessary Condition
Let $\struct {G, \circ}$ be abelian.

Thus $\iota$ has the morphism property and is therefore an automorphism.