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

## Theorem

Let $\struct {S, \circ}$ be a monoid whose identity element is $e$.

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:

$\forall n \in \N: \circ^n \paren {a \circ b} = \paren {\circ^n a} \circ \paren {\circ^n b}$

That is:

$\forall n \in \N: \paren {a \circ b}^n = a^n \circ b^n$

## Proof

Because $\struct {S, \circ}$ is a monoid, it is a fortiori also a semigroup.

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

That is:

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

It remains to be shown that the result holds for the cases where $n = 0$.

Thus:

 $\ds \paren {a \circ b}^0$ $=$ $\ds \circ^0 \paren {a \circ b}$ Definition of $\circ$ $\ds$ $=$ $\ds e$ Definition of $\circ^0$ $\ds$ $=$ $\ds e \circ e$ Definition of Identity Element $\ds$ $=$ $\ds \paren {\circ^0 a} \circ \paren {\circ^0 b}$ Definition of $\circ^0$ $\ds$ $=$ $\ds a^0 \circ b^0$ Definition of $\circ$

Thus:

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

holds for $n = 0$.

Thus:

$\forall n \in \N: \paren {a \circ b}^n = a^n \circ b^n$

$\blacksquare$