Inverse map is automorphism iff abelian

This article describes an easy-to-prove fact about basic notions in group theory, that is not very well-known or important in itself
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This article gives a result about how information about the structure of the automorphism group of a group (abstractly, or in action) can control the structure of the group
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Statement

The following are equivalent for a group:

1. The map sending every element to its inverse, is an endomorphism
2. The map sending every element to its inverse, is an automorphism
3. The group is abelian

The equivalence of (1) and (2) is direct from the fact that the inverse map is bijective.

Related facts

• Square map is endomorphism iff abelian
• Cube map is endomorphism iff abelian (if order is not a multiple of 3)
• Fixed-point-free involution on finite group is inverse map

Proof

Given: A group $G$

To prove: $G$ is Abelian iff the map $x\mapsto x^{-1}$ is an automorphism.

Proof: The following fact is true:

$(xy{)}^{-1}=y^{-1}{x}^{-1}$

Thus, we see that:

$(xy{)}^{-1}=x^{-1}{y}^{-1}⟺x^{-1},y^{-1}$ commute

Since the inverse map is a bijection, this tells us that the above is a homomorphism iff any two elements commute.

Textbook references

• Algebra by Michael Artin, ISBN 0130047635, 13-digit ISBN 978-0130047632, ^{More info}, Page 71, Exercise 12(b) of Section 3 (Isomorphisms)