Inverse of Group Commutator
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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 |
$\blacksquare$