Commutation with Group Elements implies Commuation with Product with Inverse

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Theorem

Let $G$ be a group.

Let $a, b, c \in G$ such that $a$ commutes with both $b$ and $c$.


Then $a$ commutes with $b c^{-1}$.


Proof

\(\displaystyle a b c^{-1}\) \(=\) \(\displaystyle b a c^{-1}\) as $a$ commutes with $b$
\(\displaystyle \) \(=\) \(\displaystyle b c^{-1} a\) Commutation with Inverse in Monoid

$\blacksquare$


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