Cube map is endomorphism implies class three

Statement

Suppose $G$ is a group such that the cube map $x\mapsto x^{3}$ is an endomorphism of $G$. Further, suppose that $G$ is 2-divisible, i.e., every element of $G$ is a square element.

Then, $G$ is a nilpotent group and its nilpotency class is at most four. It is not yet clear whether the nilpotency class should be at most three.

Note that the condition of 2-divisibility is true for any odd-order group, and more generally, any group in which every element has finite odd order. It is also true for any rationally powered group and in fact for any group powered over the prime 2.

Related facts

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

Facts used

1. nth power map is endomorphism implies every nth power and (n-1)th power commute
2. Exponent three implies class three for groups

Proof

Given: A group $G$ such that the map $\sigma =x\mapsto x^{3}$ is an endomorphism of $G$. Further, every element of $G$ has finite odd order.

To prove: $G$ is nilpotent and its nilpotency class is at most four.

Proof:

Fact no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	$\sigma (G)$ is a characteristic subgroup of $G$ and $G/\sigma (G)$, if nontrivial, is a group of exponent three		$\sigma =x\mapsto x^{3}$ is an endomorphism		PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE] One of the people editing this page intended to fill in this information at a later stage, but hasn't gotten around to doing it yet. If you see this placeholder for a long time, file an error report at the error reporting page. -- direct reasoning
2	$\sigma (G)$ is in the center of $G$	Fact (1)	$\sigma =x\mapsto x^{3}$ is an endomorphism $G$ is 2-divisible		By Fact (1), every cube commutes with every square. Thus, every element of $\sigma (G)$ commutes with every square. By 2-divisibility, every element of $G$ is a square, so we obtain that every element of $\sigma (G)$ commutes with every element of $G$. Thus, $\sigma (G)$ is in the center of $G$.
3	$G/\sigma (G)$ is a nilpotent group of nilpotency class at most three.	Fact (2)		Step (1)	Step-fact combination direct
4	$G$ is a nilpotent group of nilpotency class at most four.			Steps (2), (3)	Step-combination direct, plus the observation that if the quotient by a subgroup in the center has class at most $c$, then the group itself has class at most $c+1$.