Inverse map is automorphism iff abelian

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.

Similar facts for other power maps
We say that a group is a n-abelian group if the $$n^{th}$$ power map is an endomorphism. Here are some related facts about $$n$$-abelian groups.

Applications

 * Fixed-point-free involution on finite group is inverse map
 * Automorphism sends more than three-fourths of elements to inverses implies abelian

Related facts for other algebraic structures

 * Multiplication by n map is an endomorphism iff derived subring has exponent dividing n(n-1) for Lie rings. In particular, we can construct examples of non-abelian Lie rings where the negation map is an automorphism.
 * Inverse map is automorphism not implies abelian for loop

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:

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

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

Textbook references

 * , Page 71, Exercise 12(b) of Section 3 (Isomorphisms)