Exponent three implies 2-Engel for groups

Statement

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

Related facts

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 $ba{b}^{-1}$ and the commutator $[a,b]$ is defined as $ab{a}^{-1}{b}^{-1}$.

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

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

Proof: We have:

$[x,yx{y}^{-1}]=xyx{y}^{-1}{x}^{-1}(yx{y}^{-1}{)}^{-1}$

This simplifies to:

$xyx{y}^{-1}{x}^{-1}y{x}^{-1}{y}^{-1}$

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

$xyx{y}^{-1}{x}^{-1}y{x}^{-1}yy$

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:

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

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.