Cube map is surjective endomorphism implies 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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Statement

Verbal statement

If the Cube map (?) on a group is an automorphism, or more generally a surjective endomorphism, then the group is an 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.

Related facts

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. 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. nth power map is surjective endomorphism implies (n-1)th power map is endomorphism taking values in the center
3. Square map is endomorphism iff abelian

Proof

Hands-on proof using fact (1)

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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 $x \mapsto gxg^{-1}$.

Step no.	Assertion	Facts used	Given data used	Previous steps used	Explanation
1	$g^2h^3 = h^3g^2$ for all $g,h \in G$, i.e., every square commutes with every cube	Fact (1)	Cube map is an endomorphism	--	[SHOW MORE] Consider $g,h \in G$. Then, by fact (1), $c_g$ is an automorphism, so we have: $\! c_g(h^3) = c_g(h)^3$, giving $gh^3g^{-1} = (ghg^{-1})^3$. On the other hand, since the cube map is an endomorphism, we have $(ghg^{-1})^3 = g^3h^3g^{-3}$. Combining these, we get $gh^3g^{-1} = g^3h^3g^{-3}$. Canceling the left-most $g$ and the right-most $g^{-1}$ and rearranging yields that $g^2h^3 = h^3g^2$.
2	$g^2x = xg^2$ for all $g,x \in G$		Cube map is surjective	Step (1)	[SHOW MORE] Since the cube map is surjective, every element of $G$ is a cube. Combining this with step (1) yields that $g^2x = xg^2$ for every $g,x \in G$.
3	$g^2x^2 = xgxg$ for all $g,x \in G$		Cube map is an endomorphism	--	[SHOW MORE] Since the cube map is an endomorphism, we get $(gx)^3 = g^3x^3$, so expanding and canceling the left-most and right-most terms yields $xgxg = g^2x^2$.
4	We get $gx = xg$ for all $g,x \in G$.			Steps (2), (3)	[SHOW MORE] Using step (2), we can rewrite $g^2x^2$ as $xg^2x$. Combining with step (3) yields that $xgxg = xg^2x$. Canceling $xg$ from the left, we get $gx = xg$, which is the goal.

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).

References

Textbook references

• Topics in Algebra by I. N. Herstein, ^{More info}, Page 48, Exercise 24

External links

Links to related riders

• Mathlinks discussion forum page for Power maps being homomorphisms implies Abelianness
• Math Stackexchange question on this precise result