Inverse map is automorphism iff abelian

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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

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.

Value of $n$ (note that the condition for $n$ is the same as the condition for $1-n$) Characterization of $n$-abelian groups Proof Other related facts
0 all groups obvious
1 all groups obvious
2 abelian groups only 2-abelian iff abelian endomorphism sends more than three-fourths of elements to squares implies abelian
-1 abelian groups only -1-abelian iff abelian
3 3-abelian group means: 2-Engel group and derived subgroup has exponent dividing three Levi's characterization of 3-abelian groups cube map is surjective endomorphism implies abelian, cube map is endomorphism iff abelian (if order is not a multiple of 3), cube map is endomorphism implies class three
-2 same as for 3-abelian (based on n-abelian iff (1-n)-abelian)

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

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