Commutation Property in Group
From ProofWiki
Theorem
Let $\left({G, \circ}\right)$ be a group.
Then $x$ and $y$ commute iff $x \circ y \circ x^{-1} = y$.
Proof
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle x \circ y\) | \(=\) | \(\displaystyle y \circ x\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \iff\) | \(\displaystyle \) | \(\displaystyle \left({x \circ y}\right) \circ x^{-1}\) | \(=\) | \(\displaystyle \left({y \circ x}\right) \circ x^{-1}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Cancellation Laws | ||
| \(\displaystyle \) | \(\displaystyle \iff\) | \(\displaystyle \) | \(\displaystyle x \circ y \circ x^{-1}\) | \(=\) | \(\displaystyle y \circ \left({x \circ x^{-1} }\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Associativity | ||
| \(\displaystyle \) | \(\displaystyle \iff\) | \(\displaystyle \) | \(\displaystyle x \circ y \circ x^{-1}\) | \(=\) | \(\displaystyle y \circ e\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Behaviour of Inverse | ||
| \(\displaystyle \) | \(\displaystyle \iff\) | \(\displaystyle \) | \(\displaystyle x \circ y \circ x^{-1}\) | \(=\) | \(\displaystyle y\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Behaviour of Identity |
$\blacksquare$
Sources
- Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978)... (previous)... (next): Exercise $6.6 \ \text{(i)}$