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

Proof

As $G$ is a group, it is a fortiori a monoid.

Hence Product of Commuting Elements with Inverses applies:

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

$\blacksquare$

Sources

• 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $15$