Commutation with Group Elements implies Commuation with Product with Inverse
Jump to navigation
Jump to search
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 $(11)$