Powers of Commutative Elements in Monoids

Theorem
These results are an extension of the results in Powers of Commutative Elements in Semigroups in which the domain of the indices is extended to include all integers.

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

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

Then the following results hold.

Commutativity of Powers
$$\forall m, n \in \mathbb{Z}: a^m \odot b^n = b^n \odot a^m$$

Product of Commutative Elements
$$\forall n \in \mathbb{Z}: \left({a \odot b}\right)^n = a^n \odot b^n$$

Commutativity of Powers
By the similar result for semigroups, if $$m > 0$$ and $$n > 0$$ then $$a^m$$ commutes with $$b^n$$.

By Commutation with Inverses: Theorem 1, 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 Inverses: Theorem 1.

Hence the result.

Product of Commutative Elements
From the similar result for semigroups, this holds if $$n \ge 0$$.

Since $$a$$ and $$b$$ commute, then so do $$a^{-1}$$ and $$b^{-1}$$ by Commutation with Inverses: Theorem 2.

Hence, if $$n > 0$$:

$$ $$ $$ $$ $$

The result follows.