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 $n > 0$:

\(\displaystyle \paren {a \circ b}^{-n}\) \(=\) \(\displaystyle \paren {\paren {a \circ b}^{-1} }^n\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {b^{-1} \circ a^{-1} }^n\) Inverse of Product
\(\displaystyle \) \(=\) \(\displaystyle \paren {a^{-1} \circ b^{-1} }^n\) Commutation of Inverses in Monoid
\(\displaystyle \) \(=\) \(\displaystyle \paren {a^{-1} }^n \circ \paren {b^{-1} }^n\)
\(\displaystyle \) \(=\) \(\displaystyle a^{-n} \circ b^{-n}\)

The result follows.

$\blacksquare$


Sources