Commutation Property in Group
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Theorem
Let $\struct {G, \circ}$ be a group.
Then $x$ and $y$ commute if and only if $x \circ y \circ x^{-1} = y$.
Proof
\(\ds x \circ y\) | \(=\) | \(\ds y \circ x\) | ||||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \paren {x \circ y} \circ x^{-1}\) | \(=\) | \(\ds \paren {y \circ x} \circ x^{-1}\) | Cancellation Laws | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds x \circ y \circ x^{-1}\) | \(=\) | \(\ds y \circ \paren {x \circ x^{-1} }\) | Definition of Associative Operation | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds x \circ y \circ x^{-1}\) | \(=\) | \(\ds y \circ e\) | Definition of Inverse Element | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds x \circ y \circ x^{-1}\) | \(=\) | \(\ds y\) | Definition of Identity Element |
$\blacksquare$
Sources
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $6$: An Introduction to Groups: Exercise $6 \ \text{(i)}$