Commutation with Group Elements implies Commuation with Product with Inverse

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

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

$\blacksquare$