Commutativity of Powers in Semigroup

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Theorem

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

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


Then:

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


Proof

Necessary Condition

Let $a, b \in S: a \circ b = b \circ a$.

Then from Powers of Commuting Elements of Semigroup Commute:

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

$\Box$


Sufficient Condition

For the above relationships and equalities to hold, it follows that $a$ and $b$ must commute.


The result follows.

$\blacksquare$