Exponent three implies 2-Engel for groups

Statement
Any group of exponent equal to three must be a proves property satisfaction of::2-Engel group (also known as a Levi group).

Not true for Lie rings

 * Exponent three not implies 2-Engel for Lie rings

Converse of sorts
This converse says that although a 2-Engel group need not have exponent three, the extent to which 2-Engel differs from class two is captured by exponent three:


 * 2-Engel group implies third member of lower central series has exponent dividing three
 * 2-Engel Lie ring implies third member of lower central series is in 3-torsion

Applications

 * Exponent three implies class three for groups

Proof
We use the definition that a group is a 2-Engel group if and only if an two conjugates commute.

Proof using left action convention
In this convention, the conjugate of $$a$$ by $$b$$ is denoted $$bab^{-1}$$ and the commutator $$[a,b]$$ is defined as $$aba^{-1}b^{-1}$$.

Given: A group $$G$$ of exponent three, elements $$x,y \in G$$

To prove: $$[x,yxy^{-1}]$$ is the identity element.

Proof: We have:

$$[x,yxy^{-1}] = xyxy^{-1}x^{-1}(yxy^{-1})^{-1}$$

This simplifies to:

$$xyxy^{-1}x^{-1}yx^{-1}y^{-1}$$

Rewrite the right most $$y^{-1}$$ as $$y^2 = yy$$, using that $$y^3$$ is the identity element, and get:

$$xyxy^{-1}x^{-1}yx^{-1}yy$$

We now see two adjacent occurrences of $$x^{-1}y$$, so we have a $$(x^{-1}y)^2$$ in the expression. Using that $$(x^{-1}y)^3$$ is the identity element, we obtain that $$(x^{-1}y)^2 = (x^{-1}y)^{-1} = y^{-1}x$$. We get:

$$xyxy^{-1}(y^{-1}x)y$$

We now see two adjacent occurrences of $$y^{-1}$$, giving $$y^{-2}$$, which simplifies to $$y$$, again using that $$y^3$$ is the identity element. We get:

$$xyxyxy$$

We now simplify this to the identity element using that $$(xy)^3$$ is the identity element, and we are done.