# 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

 $\ds a \circ b$ $=$ $\ds b \circ a$ $\ds \leadsto \ \$ $\ds \paren {\circ^1 a} \circ b$ $=$ $\ds b \circ \paren {\circ^1 a}$ Definition of $\circ^1$

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:

 $\ds \paren {\circ^{k + 1} a} \circ b$ $=$ $\ds \paren {\paren {\circ^k a} \circ a} \circ b$ Definition of $\circ^n a$ $\ds$ $=$ $\ds \paren {\circ^k a} \circ \paren {a \circ b}$ Semigroup Axiom $\text S 1$: Associativity $\ds$ $=$ $\ds \paren {\circ^k a} \circ \paren {b \circ a}$ $b$ commutes with $a$ under $\circ$ $\ds$ $=$ $\ds \paren {\paren {\circ^k a} \circ b} \circ a$ Semigroup Axiom $\text S 1$: Associativity $\ds$ $=$ $\ds \paren {b \circ \paren {\circ^k a} } \circ a$ Induction Hypothesis $\ds$ $=$ $\ds b \circ \paren {\paren {\circ^k a} \circ a}$ Semigroup Axiom $\text S 1$: Associativity $\ds$ $=$ $\ds b \circ \paren {\circ^{k + 1} a}$ Definition of $\circ^n a$

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

Thus:

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

$\Box$

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

 $\ds b \circ \paren {\circ^m a}$ $=$ $\ds \paren {\circ^m a} \circ b$ $\ds \leadsto \ \$ $\ds \paren {\circ^n b} \circ \paren {\circ^m a}$ $=$ $\ds \paren {\circ^m a} \circ \paren {\circ^n b}$

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:

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

$\blacksquare$