Cube map is endomorphism iff abelian (if order is not a multiple of 3)

Verbal statement
Consider a finite group whose order is not a multiple of 3. Then, the fact about::cube map (viz the map sending each element of the group to its cube) is an endomorphism if and only if the group is abelian.

Statement with symbols
Let $$G$$ be a finite group whose order is not a multiple of 3. Then, the map $$\sigma$$ defined as $$\sigma(x) = x^3$$ is an endomorphism if and only if $$G$$ is abelian.

Related facts
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.

Facts used

 * 1) uses::Abelian implies universal power map is endomorphism: In an abelian group, the $$n^{th}$$ power map is an endomorphism for all $$n$$.
 * 2) uses::Cube map is surjective endomorphism implies abelian
 * 3) uses::kth power map is bijective iff k is relatively prime to the order

Abelian implies cube map is endomorphism
This is a direct consequence of fact (1).

Cube map is endomorphism implies abelian
Given: A finite group $$G$$ whose order is relatively prime to $$3$$, and such that $$x \mapsto x^3$$ is an endomorphism of $$G$$.

To prove: $$G$$ is abelian.

Proof:

Difference from the corresponding statement for the square map
In the case of the square map, we 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