Commutator is Identity iff Elements Commute

Theorem
Let $G$ be a group whose identity is $e$.

Let $x, y \in G$.

Let $\sqbrk {x, y}$ denote the commutator of $x$ and $y$.

Then $\sqbrk {x, y} = e$ $x$ and $y$ commute.

Proof
As $G$ is a group, it is by definition a monoid.

Hence Product of Commuting Elements with Inverses applies:


 * $x y x^{-1} y^{-1} = e = x^{-1} y^{-1} x y$