3-Engel and 2-torsion-free implies 2-local class three for Lie rings

Statement

Suppose $L$ is a 3-Engel Lie ring, i.e., the following identity holds:

$[x,[x,[x,y]]]=0\ \forall \ x,y\in L$

Suppose further that $L$ has no 2-torsion, i.e., $2x=0$ implies $x=0$ for $x\in L$ . Then, $L$ is a Lie ring of 2-local nilpotency class three, i.e., the 2-local nilpotency class of $L$ is at most three.

Facts used

Polarization trick: We use the polarization trick in three variables.
Higgins' lemma on Engel conditions

Proof

Preliminary observations

In order to establish the result, we need to show that all Lie products of length four that involve only two variables must take the value zero. We know that it suffices to restrict attention to right normed expressions because of the Jacobi identity. Up to interchange of $x$ and $y$ and using skew symmetry in $x,y$ , we see that there are only two types of expressions: $[x,[x,[x,y]]]$ and $[x,[y,[x,y]]]$ . Thus, it suffices to prove that $[x,[y,[x,y]]]=0$ for all $x,y\in L$ .

Proof details

Given: A Lie ring $L$ . We have $[x,[x,[x,y]]]=0$ for all $x,y\in L$ . We also have that $L$ has no 2-torsion: $2x=0\implies x=0$ for $x\in L$ .

To prove: $[x,[y,[x,y]]]=0$ for all $x,y\in L$ .

Proof: For notational convenience, we will use the string $abcd$ to denote the right-normed Lie product $[a,[b,[c,d]]]$ .

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	$x_{1}x_{2}x_{3}y+x_{1}x_{3}x_{2}y+x_{2}x_{1}x_{3}y+x_{2}x_{3}x_{1}y+x_{3}x_{1}x_{2}y+x_{3}x_{2}x_{1}y=0$ for all $x_{1},x_{2},x_{3},y\in L$	Fact (1)	3-Engel condition		Apply Fact (1) to the 3-Engel condition
2	$xxyy+xyxy+xxyy+xyxy+yxxy+yxxy=0$ for all $x,y\in L$			Step (1)	Set $x_{1}=x_{2}=x$ and set $x_{3}=y$ in Step (1)
3	$2(xyxy+yxxy)=0$ for all $x,y\in L$			Step (2)	Simplifying Step (2), noting that all products ending with $yy$ must be zero
4	$xyxy+yxxy=0$ . Back to explicit notation, we have $[x,[y,[x,y]]]+[y,[x,[x,y]]]=0$ for all $x,y\in L$		$L$ is 2-torsion-free	Step (3)	direct (note: there's actually a way of reaching this step without using the 2-torsion-free condition , using incomplete polarization, but we are avoiding that here because it's unnecessary).
5	$[x,[y,[x,y]]]+[y,[[x,y],x]]+[[x,y],[x,y]]=0$ for all $x,y\in L$				Direct application of Jacobi identity in $x,y,[x,y]$ .
6	$[x,[y,[x,y]]]=[y,[x,[x,y]]]$ for all $x,y\in L$			Step (5)	In Step (5), the third term on the left side is zero by alternation. Move the second term to the right side and interchange the order of terms on the inside using skew symmetry.
7	$2[x,[y,[x,y]]]=0$ for all $x,y\in L$			Steps (4), (6)	Step-combination direct
8	$[x,[y,[x,y]]]=0$ for all $x,y\in L$		$L$ is 2-torsion-free	Step (7)	Step-given combination direct.