Inverse of Group Commutator

Theorem

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

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

Then $\sqbrk {g, h}$ is the inverse of $\sqbrk {h, g}$:

$\sqbrk {g, h} = \sqbrk {h, g}^{-1}$

Proof

\(\ds \sqbrk {g, h} \circ \sqbrk {h, g}\)

\(=\)

\(\ds \paren {g^{-1} \circ h^{-1} \circ g \circ h} \circ \paren {h^{-1} \circ g^{-1} \circ h \circ g}\)

Definition of Commutator of Group Elements

\(\ds \)

\(=\)

\(\ds \paren {g^{-1} \circ h^{-1} \circ g \circ h} \circ \paren {g^{-1} \circ h^{-1} \circ g \circ h}^{-1}\)

Inverse of Group Product: General Result

\(\ds \)

\(=\)

\(\ds e\)

Group Axiom $\text G 3$: Existence of Inverse Element

\(\ds \)

\(=\)

\(\ds \paren {g^{-1} \circ h^{-1} \circ g \circ h}^{-1} \circ \paren {g^{-1} \circ h^{-1} \circ g \circ h}\)

Group Axiom $\text G 3$: Existence of Inverse Element

\(\ds \)

\(=\)

\(\ds \paren {h^{-1} \circ g^{-1} \circ h \circ g} \circ \paren {g^{-1} \circ h^{-1} \circ g \circ h}\)

Inverse of Group Product: General Result

\(\ds \)

\(=\)

\(\ds \sqbrk {h, g} \circ \sqbrk {g, h}\)

Definition of Commutator of Group Elements

$\blacksquare$