Inversion Mapping is Automorphism iff Group is Abelian

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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 if and only if $G$ is abelian.


Proof

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

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


Sufficient Condition

Suppose $\iota$ is an automorphism.

Then:

\(\displaystyle \forall x, y \in G: \map \iota {x \circ y}\) \(=\) \(\displaystyle \map \iota x \circ \map \iota y\) Definition of Morphism Property
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {x \circ y}^{-1}\) \(=\) \(\displaystyle x^{-1} \circ y^{-1}\) Definition of $\iota$


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.

$\Box$


Necessary Condition

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

\(\displaystyle \forall x, y \in G: \paren {x \circ y}^{-1}\) \(=\) \(\displaystyle x^{-1} \circ y^{-1}\) Inverse of Commuting Pair
\(\displaystyle \leadsto \ \ \) \(\displaystyle \map \iota {x \circ y}\) \(=\) \(\displaystyle \map \iota x \circ \map \iota y\) Definition of $\iota$


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

$\blacksquare$


Sources

except this source requests only that the morphism property is demonstrated, and not the bijectivity.
except this source proves only the necessary condition.