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
- 1965: J.A. Green: Sets and Groups ... (previous) ... (next): Chapter $4$. Groups: Exercise $8$