Commutativity of Powers in Monoid

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Theorem

Let $\left ({S, \circ}\right)$ be a monoid whose identity is $e_S$.

Let $a, b \in S$ be invertible elements for $\circ$ that also commute.

Then:

$\forall m, n \in \Z: a^m \circ b^n = b^n \circ a^m$


Proof

By Commutativity of Powers in Semigroup, if $m > 0$ and $n > 0$ then $a^m$ commutes with $b^n$.

By Commutation with Inverse in Monoid, again if $m > 0$ and $n > 0$ then $a^m$ commutes with $\left({b^n}\right)^{-1} = b^{-n}$.

Similarly $b^n$ commutes with $a^{-m}$.

But as $a^{-m}$ commutes with $b^n$, it also commutes with $\left({b^n}\right)^{-1} = b^{-n}$, again by Commutation with Inverse in Monoid.

Hence the result.

$\blacksquare$


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