Idempotent Semigroup/Properties/1
Jump to navigation
Jump to search
Property of Idempotent Semigroup
Let $\struct {S, \circ}$ be an idempotent semigroup.
Let $x \circ y = y$ and $y \circ x = x$.
Then for all $z \in S$:
- $z \circ x \circ z \circ y = z \circ y$
and:
- $z \circ y \circ z \circ x = z \circ x$
Proof
From Semigroup Axiom $\text S 1$: Associativity we take it for granted that $\circ$ is associative.
\(\ds \forall z \in S: \, \) | \(\ds z \circ x \circ z \circ y\) | \(=\) | \(\ds z \circ x \circ z \circ x \circ y\) | as $x \circ y = y$ by hypothesis | ||||||||||
\(\ds \) | \(=\) | \(\ds z \circ x \circ y\) | Definition of Idempotent Operation: $\paren {z \circ x} \circ \paren {z \circ x} = z \circ x$ | |||||||||||
\(\ds \) | \(=\) | \(\ds z \circ y\) | as $x \circ y = y$ by hypothesis |
\(\ds \forall z \in S: \, \) | \(\ds z \circ y \circ z \circ x\) | \(=\) | \(\ds z \circ y \circ z \circ y \circ x\) | as $y \circ x = x$ by hypothesis | ||||||||||
\(\ds \) | \(=\) | \(\ds z \circ y \circ x\) | Definition of Idempotent Operation: $\paren {z \circ y} \circ \paren {z \circ y} = z \circ y$ | |||||||||||
\(\ds \) | \(=\) | \(\ds z \circ x\) | as $y \circ x = x$ by hypothesis |
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Exercise $11.19 \ \text {(a)}$