Power of Product of Commutative Elements in Semigroup

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

Then by Power of Product of Commuting Elements in Semigroup equals Product of Powers:

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

$\Box$


Sufficient Condition

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

In particular, when $n = 2$,

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

$\blacksquare$


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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:

\(\ds \rho\) \(=\) \(\ds \dbinom {1 \ 2 \ 3} {2 \ 3 \ 1}\)
\(\ds \sigma\) \(=\) \(\ds \dbinom {1 \ 2 \ 3} {1 \ 3 \ 2}\)


Then:

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


Sources

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