Powers of Commutative Elements in Semigroups

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

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

Then $$\forall m, n \in \mathbb{N}^*: a^m \circ b^n = b^n \circ a^m \iff a \circ b = b \circ a$$.

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

Because $$\left({S, \circ}\right)$$ is a semigroup, $$\circ$$ is associative on $$S$$.

Let $$T$$ be the set of all $$n \in \mathbb{N}^*$$ such that:

$$a^n \circ b = b \circ a^n$$

We have $$a \circ b = b \circ a \Longrightarrow a^1 \circ b = b \circ a^1$$.

So $$1 \in T$$.

Now suppose $$n \in T$$. Then we have:

$$ $$ $$ $$ $$ $$ $$

So $$n + 1 \in T$$.

Thus by the Principle of Finite Induction, $$T = \mathbb{N}^*$$. Thus:

$$\forall m \in \mathbb{N}^*: a^m \circ b = b \circ a^m$$

Thus, from the preceding: $$\forall m, n \in \mathbb{N}^*: a^m$$ and $$b^n$$ also commute with each other.


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

The result follows.