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.

$\blacksquare$

Sources

• 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): commutator