If $g^3 = e$ for all $g \in G$, then $hgh^{-1}$ and $g$ commute for all $h,g \in G$.

Question

This is an exercise from a lecture on Introduction to group theory and it is given in the introductory part so I am not sure what methods to use to prove this. It is given right after an exercise that if $g^2=e$ for all $g \in G$ then $G$ is abelian, so I assume this problem can be solved with similar basic algebraic manipulations but I have got stuck on it.

Suppose that $g^3 = e$ for all $g \in G$ for a group $G$. Then show that $hgh^{-1}$ and $g$ commute for all $h,g \in G$.

I've been thinking about this problem for a while but I can't figure out a way to show this. I would greatly appreciate any help.

Answer 1

It suffices to show

$$\begin{align} [hgh^{-1},g]&=\color{purple}{hgh^{-1}}\color{orange}g(\color{purple}{hgh^{-1}})^{-1}\color{orange}{g}^{-1}\\ &=(hg\underbrace{h)(h}_{h^{-1}}gh)g^{-1}h^{-1}g^{-1}\\ &=(hgh)^2g^{-1}h^{-1}g^{-1}\\ &=(hgh)^{-1}g^{-1}h^{-1}g^{-1}\\ &=h^{-1}g^{-1}h^{-1}g^{-1}h^{-1}g^{-1}\\ &=(h^{-1}g^{-1})^3\\ &=e. \end{align}$$

Answer 2

The question can be read in a somewhat different way: if a group $G$ has exponent $3$, then all the conjugacy classes of $G$ are commutative sets. The latter is equivalent to all the centralisers of elements of $G$ being normal subgroups. In fact, this only holds for so-called $2$-Engel groups ($[[x,g],g]=1$ for all $x,g \in G$), for details see here. Groups of exponent $3$ are $2$-Engel, which is easy to prove. The converse is not true in general, as the quaternion group of order $8$ shows.