# Power of Product of Commutative Elements in Monoid

## Theorem

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

Let $a, b \in S$ be invertible elements for $\circ$ that also commute.

Then:

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

## Proof

From Power of Product of Commutative Elements in Semigroup, this result holds if $n \ge 0$.

Since $a$ and $b$ commute, then so do $a^{-1}$ and $b^{-1}$ by Commutation of Inverses in Monoid.

Hence, if $n > 0$:

 $\displaystyle \paren {a \circ b}^{-n}$ $=$ $\displaystyle \paren {\paren {a \circ b}^{-1} }^n$ $\displaystyle$ $=$ $\displaystyle \paren {b^{-1} \circ a^{-1} }^n$ Inverse of Product $\displaystyle$ $=$ $\displaystyle \paren {a^{-1} \circ b^{-1} }^n$ Commutation of Inverses in Monoid $\displaystyle$ $=$ $\displaystyle \paren {a^{-1} }^n \circ \paren {b^{-1} }^n$ $\displaystyle$ $=$ $\displaystyle a^{-n} \circ b^{-n}$

The result follows.

$\blacksquare$