# 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

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

$\blacksquare$