# Power of Product of Commuting Elements in Semigroup equals Product of Powers

## 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:

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

That is:

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

## Proof

The proof proceeds by the Principle of Mathematical Induction:

Let $\map P n$ be the proposition:

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

### Basis of the Induction

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

So $\map P 1$ holds.

This is the basis for the induction.

### Induction Hypothesis

Suppose that $\map P k$ holds:

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

This is the induction hypothesis.

It remains to be shown that:

$\map P k \implies \map P {k + 1}$

That is, that:

$\map {\circ^{k + 1} } {a \circ b} = \paren {\circ^{k + 1} a} \circ \paren {\circ^{k + 1} b}$

### Induction Step

This is the induction step:

 $\ds \map {\circ^{k + 1} } {a \circ b}$ $=$ $\ds \paren {\map {\circ^k} {a \circ b} } \circ \paren {a \circ b}$ Definition of $\circ^{k + 1}$ $\ds$ $=$ $\ds \paren {\paren {\circ^k a} \circ \paren {\circ^k b} } \circ \paren {a \circ b}$ Induction Hypothesis $\ds$ $=$ $\ds \paren {\circ^k a} \circ \paren {\paren {\paren {\circ^k b} \circ a} \circ b}$ Semigroup Axiom $\text S 1$: Associativity $\ds$ $=$ $\ds \paren {\circ^k a} \circ \paren {\paren {a \circ \paren {\circ^k b} } \circ b}$ Powers of Commuting Elements of Semigroup Commute $\ds$ $=$ $\ds \paren {\paren {\circ^k a} \circ a} \circ \paren {\paren {\circ^k b} \circ b}$ Semigroup Axiom $\text S 1$: Associativity $\ds$ $=$ $\ds \paren {\circ^{k + 1} a} \circ \paren {\circ^{k + 1} b}$ Definition of $\circ^{k + 1}$

So $\map P {k + 1}$ holds.

Thus by the Principle of Mathematical Induction, the result holds for all $n \in \N_{>0}$:

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

$\blacksquare$