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:
\(\ds \forall x, y \in G: \, \) | \(\ds \map \iota {x \circ y}\) | \(=\) | \(\ds \map \iota x \circ \map \iota y\) | Definition of Morphism Property | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {x \circ y}^{-1}\) | \(=\) | \(\ds x^{-1} \circ y^{-1}\) | 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.
\(\ds \forall x, y \in G: \, \) | \(\ds \paren {x \circ y}^{-1}\) | \(=\) | \(\ds x^{-1} \circ y^{-1}\) | Inverse of Commuting Pair | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map \iota {x \circ y}\) | \(=\) | \(\ds \map \iota x \circ \map \iota y\) | Definition of $\iota$ |
Thus $\iota$ has the morphism property and is therefore an automorphism.
$\blacksquare$
Sources
- 1964: Walter Ledermann: Introduction to the Theory of Finite Groups (5th ed.) ... (previous) ... (next): Chapter $\text {I}$: The Group Concept: Examples: $(9)$
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: Group Homomorphism and Isomorphism: $\S 60 \gamma$
- except this source requests only that the morphism property is demonstrated, and not the bijectivity.
- 1974: Thomas W. Hungerford: Algebra ... (previous) ... (next): $\S 1.2$
- except this source proves only the necessary condition.