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

$\blacksquare$

Sources

• 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): $\S 3.2$: Groups; the axioms: Exercise $(11)$