Power of Product of Commutative Elements in Monoid
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Theorem
Let $\struct {S, \circ}$ be a monoid whose identity is $e_S$.
Let $a, b \in S$ be invertible elements for $\circ$ that also commute.
Then:
- $\forall n \in \Z: \paren {a \circ b}^n = a^n \circ b^n$
Proof
From Power of Product of Commutative Elements in Semigroup, this result holds if $n \ge 0$.
Since $a$ and $b$ commute, then so do $a^{-1}$ and $b^{-1}$ by Commutation of Inverses in Monoid.
Hence, if $n > 0$:
\(\ds \paren {a \circ b}^{-n}\) | \(=\) | \(\ds \paren {\paren {a \circ b}^{-1} }^n\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {b^{-1} \circ a^{-1} }^n\) | Inverse of Product | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {a^{-1} \circ b^{-1} }^n\) | Commutation of Inverses in Monoid | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {a^{-1} }^n \circ \paren {b^{-1} }^n\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds a^{-n} \circ b^{-n}\) |
The result follows.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {IV}$: Rings and Fields: $20$. The Integers: Theorem $20.11 \ (4)$
- 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): $\S 3.1$: Monoids