Cube map is surjective endomorphism implies abelian

Verbal statement
If the fact about::cube map on a group is an automorphism, or more generally a surjective endomorphism, then the group is an proves property satisfaction of::abelian group.

Statement with symbols
Let $$G$$ be a group such that the map $$\sigma:G \to G$$ defined by $$\sigma(x) = x^3$$ is an automorphism, or more generally, a surjective endomorphism. Then, $$G$$ is an abelian group.

Similar facts for cube map

 * Cube map is endomorphism iff abelian (if order is not a multiple of 3)
 * Cube map is endomorphism implies class four for 2-divisible group

Similar facts for other values

 * Inverse map is automorphism iff abelian
 * Square map is endomorphism iff abelian

Weaker facts for other values

 * nth power map is endomorphism iff abelian (if order is relatively prime to n(n-1))
 * nth power map is automorphism implies (n-1)th power map is endomorphism taking values in the center
 * nth power map is endomorphism implies every nth power and (n-1)th power commute
 * (n-1)th power map is endomorphism taking values in the center implies nth power map is endomorphism

Opposite facts
The statement breaks down if we remove the assumption of surjectivity:

Frattini-in-center odd-order p-group implies p-power map is endomorphism: In particular, for $$p = 3$$, we can obtain non-abelian groups of order $$p^3 = 27$$, such as prime-cube order group:U(3,3) and semidirect product of Z9 and Z3, where the cube map is an endomorphism. In the former case, the cube map is the trivial endomorphism. In the latter, it is a nontrivial endomorphism.

Facts used

 * 1) uses::Group acts as automorphisms by conjugation: For any $$g \in G$$, the map $$c_g = x \mapsto gxg^{-1}$$ is an automorphism of $$G$$.
 * 2) uses::nth power map is surjective endomorphism implies (n-1)th power map is endomorphism taking values in the center
 * 3) uses::Square map is endomorphism iff abelian

Hands-on proof using fact (1)
Given: A group $$G$$ such that the map sending $$x$$ to $$x^3$$ is a surjective endomorphism of $$G$$.

To prove: $$G$$ is abelian.

Proof: We denote by $$c_g$$ the map $$x \mapsto gxg^{-1}$$.

Hands-off proof (using facts (2) and (3))
Given: A group $$G$$ such that the map sending $$x$$ to $$x^3$$ is a surjective endomorphism of $$G$$.

To prove: $$G$$ is abelian.

Proof: By fact (2), we conclude that the square map must be an endomorphism of $$G$$. By fact (3), we conclude that therefore $$G$$ must be abelian.

Difference from the corresponding statement for the square map
In the case of the square map, we can in fact prove something much stronger:

$$(xy)^2 = x^2y^2 \iff xy = yx$$

In the case of the cube map, this is no longer true. That is, it may so happen that $$(xy)^3 = x^3y^3$$ although $$xy \ne yx$$. Thus, to show that $$xy = yx$$ we need to not only use that $$(xy)^3 = x^3y^3$$ but also use that this identity is valid for other elements picked from $$G$$ (specifically, that it is valid for their cuberoots).

Textbook references

 * , Page 48, Exercise 24

Links to related riders

 * Math Stackexchange question on this precise result
 * Math Stackexchange question on this precise result