Powers of Commuting Elements of Semigroup Commute

Theorem
Let $\struct {S, \circ}$ be a semigroup.

For $a \in S$, let $\circ^n a = a^n$ denote the $n$th power of $a$.

Let $a, b \in S$ such that $a$ commutes with $b$:
 * $a \circ b = b \circ a$

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

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

Proof
The proof proceeds by the Principle of Mathematical Induction:

Let $\map P n$ be the proposition:


 * $\paren {\circ^n a} \circ b = b \circ \paren {\circ^n a}$

Basis of the Induction
demonstrating that $\map P 1$ holds.

This is the basis for the induction.

Induction Hypothesis
Suppose that $\map P k$ holds:


 * $\paren {\circ^k a} \circ b = b \circ \paren {\circ^k a}$

This is the induction hypothesis.

It remains to be shown that:
 * $\map P k \implies \map P {k + 1}$

That is, that:
 * $\paren {\circ^{k + 1} a} \circ b = b \circ \paren {\circ^{k + 1} a}$

Induction Step
This is the induction step:

Thus:

So $\map P {k + 1}$ is true.

Thus:
 * $\forall m \in \N_{>0}: \paren {\circ^m a} \circ b = b \circ \paren {\circ^m a}$

By repeating the argument above, replacing $a$ with $b$ and $b$ with $\circ^m a$, we have:

Hence the result:
 * $\forall m, n \in \N_{>0}: \paren {\circ^m a} \circ \paren {\circ^n b} = \paren {\circ^n b} \circ \paren {\circ^m a}$

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