Group Commutators are Commuting Elements

Theorem
Let $\struct {G, \circ}$ be a group whose identity is $e$.

Let $\sqbrk {g, h}$ denote the commutator of $g$ and $h$.

Then $\sqbrk {g, h}$ commutes with $\sqbrk {h, g}$, in the sense that:
 * $\sqbrk {g, h} \circ \sqbrk {h, g} = \sqbrk {h, g} \circ \sqbrk {g, h}$

Proof
From Inverse of Group Commutator:


 * $\forall g, h \in G: \sqbrk {g, h} = \sqbrk {h, g}^{-1}$

The result follows from Group Element Commutes with Inverse.