Powers of Commutative Elements in Semigroups

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Let $\left ({S, \circ}\right)$ be a semigroup.

Let $a, b \in S$ both be cancellable elements of $S$.

Then the following results hold:

Commutativity of Powers

$\forall m, n \in \N_{>0}: a \circ b = b \circ a \implies a^m \circ b^n = b^n \circ a^m$

but it is not necessarily the case that:

$\forall m, n \in \N_{>0}: a^m \circ b^n = b^n \circ a^m \implies a \circ b = b \circ a$

Product of Commutative Elements

$\forall n \in \N_{>1}: \paren {x \circ y}^n = x^n \circ y^n \iff x \circ y = y \circ x$

Also see