Commutation Property in Group

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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          

$\blacksquare$


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