# Commutativity of Powers in Semigroup

## 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.

$\blacksquare$