Idempotent Semigroup/Examples/Relation induced by Inverse Element/Properties/1
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Example of Idempotent Semigroup
Let $\struct {S, \circ}$ be an idempotent semigroup.
Let $\RR$ be the relation on $S$ defined as:
- $\forall a, b \in S: a \mathrel \RR b \iff \paren {a \circ b \circ a = a \land b \circ a \circ b = b}$
That is, such that $a$ is the inverse of $b$ and $b$ is the inverse of $a$.
Let $x \circ y = y$ and $y \circ x = x$.
Then for all $z \in S$:
- $\paren {z \circ x} \mathrel \RR \paren {z \circ y}$
and:
- $\paren {x \circ z} \mathrel \RR \paren {y \circ z}$
Proof
From Semigroup Axiom $\text S 0$: Closure we take it for granted that $\struct {S, \circ}$ is closed under $\circ$.
From Semigroup Axiom $\text S 1$: Associativity we take it for granted that $\circ$ is associative.
Hence parentheses will be used whenever it makes groupings of operations more clear.
We have:
\(\ds \forall z \in S: \, \) | \(\ds z \circ x\) | \(=\) | \(\ds \paren {z \circ y} \circ \paren {z \circ x}\) | Properties of Idempotent Semigroup: $1$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {z \circ x} \circ \paren {z \circ x}\) | \(=\) | \(\ds \paren {z \circ x} \circ \paren {z \circ y} \circ \paren {z \circ x}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {z \circ x}\) | \(=\) | \(\ds \paren {z \circ x} \circ \paren {z \circ y} \circ \paren {z \circ x}\) | Definition of Idempotent Operation |
Similarly:
\(\ds \forall z \in S: \, \) | \(\ds z \circ y\) | \(=\) | \(\ds \paren {z \circ x} \circ \paren {z \circ y}\) | Properties of Idempotent Semigroup: $1$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {z \circ y} \circ \paren {z \circ y}\) | \(=\) | \(\ds \paren {z \circ y} \circ \paren {z \circ x} \circ \paren {z \circ y}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {z \circ y}\) | \(=\) | \(\ds \paren {z \circ y} \circ \paren {z \circ x} \circ \paren {z \circ y}\) | Definition of Idempotent Operation |
Hence by definition of $\RR$:
- $\forall z \in S: \paren {z \circ x} \mathrel \RR \paren {z \circ y}$
$\Box$
Then:
\(\ds \forall z \in S: \, \) | \(\ds x \circ z\) | \(=\) | \(\ds x \circ \paren {z \circ x} \circ z\) | Definition of Idempotent Operation | ||||||||||
\(\ds \) | \(=\) | \(\ds x \circ \paren {z \circ y} \circ \paren {z \circ x} \circ z\) | Properties of Idempotent Semigroup: $1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {x \circ z} \circ \paren {y \circ z} \circ \paren {x \circ z}\) | Semigroup Axiom $\text S 1$: Associativity |
and:
\(\ds \forall z \in S: \, \) | \(\ds y \circ z\) | \(=\) | \(\ds y \circ \paren {z \circ y} \circ z\) | Definition of Idempotent Operation | ||||||||||
\(\ds \) | \(=\) | \(\ds y \circ \paren {z \circ x} \circ \paren {z \circ y} \circ z\) | Properties of Idempotent Semigroup: $1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {y \circ z} \circ \paren {x \circ z} \circ \paren {y \circ z}\) | Semigroup Axiom $\text S 1$: Associativity |
Hence by definition of $\RR$:
- $\forall z \in S: \paren {x \circ z} \mathrel \RR \paren {y \circ z}$
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Exercise $11.19 \ \text {(b)}$