# Power of Product of Commutative Elements in Semigroup

## Theorem

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

Let $x, y \in S$ both be cancellable elements of $S$.

Then:

$\forall n \in \N_{>1}: \paren {x \circ y}^n = x^n \circ y^n \iff x \circ y = y \circ x$

## Proof

### Necessary Condition

Let $x \circ y = y \circ x$.

$\forall n \in \N_{>1}: \paren {x \circ y}^n = x^n \circ y^n$

$\Box$

### Sufficient Condition

Suppose $\forall n \in \N_{>1}: \paren {x \circ y}^n = x^n \circ y^n$.

In particular, when $n = 2$,

 $\displaystyle \paren {x \circ y}^2$ $=$ $\displaystyle x^n \circ y^2$ $\displaystyle x \circ y \circ x \circ y$ $=$ $\displaystyle x \circ x \circ y \circ y$ Definition of Semigroup: $\circ$ is Associative $\displaystyle y \circ x$ $=$ $\displaystyle x \circ y$ $x$ and $y$ are assumed to be cancellable

$\blacksquare$

## Examples

### Elements of $3$rd Symmetric Group

Let $S = \set {1, 2, 3}$.

Let $S_3$ denote the symmetric group on $3$ letters.

Let $\rho, \sigma \in S_3$ defined in two-row notation as:

 $\displaystyle \rho$ $=$ $\displaystyle \dbinom {1 \ 2 \ 3} {2 \ 3 \ 1}$ $\displaystyle \sigma$ $=$ $\displaystyle \dbinom {1 \ 2 \ 3} {1 \ 3 \ 2}$

Then:

$\rho^2 \sigma^2 \ne \paren {\rho \sigma}^2$