Commutator of Group Element with Identity is Identity
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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:
- $\sqbrk {g, e} = e = \sqbrk {e, g}$
Proof
| \(\ds \forall g \in G: \, \) | \(\ds \sqbrk {g, e}\) | \(=\) | \(\ds g^{-1} \circ e^{-1} \circ g \circ e\) | Definition of Commutator of Group Elements | ||||||||||
| \(\ds \) | \(=\) | \(\ds g^{-1} \circ e \circ g \circ e\) | Identity is Self-Inverse | |||||||||||
| \(\ds \) | \(=\) | \(\ds g^{-1} \circ g\) | Definition of Identity Element | |||||||||||
| \(\ds \) | \(=\) | \(\ds e\) | Definition of Inverse Element |
| \(\ds \forall g \in G: \, \) | \(\ds \sqbrk {e, g}\) | \(=\) | \(\ds e^{-1} \circ g^{-1} \circ e \circ g\) | Definition of Commutator of Group Elements | ||||||||||
| \(\ds \) | \(=\) | \(\ds e^ \circ g^{-1} \circ e \circ g\) | Identity is Self-Inverse | |||||||||||
| \(\ds \) | \(=\) | \(\ds g \circ g^{-1}\) | Definition of Identity Element | |||||||||||
| \(\ds \) | \(=\) | \(\ds e\) | Definition of Inverse Element |
$\blacksquare$