Idempotent Semigroup/Examples/Relation induced by Inverse Element/Properties/6

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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 $a$.


The quotient structure $\struct {S / \RR, \circ_\RR}$ is a commutative idempotent semigroup.

The equivalence classes under $\RR$ are anticommutative subsemigroups of $\struct {S, \circ}$.


Proof

From Idempotent Semigroup: Relation induced by Inverse Element: $5$:

$\RR$ is a congruence relation on $\struct {S, \circ}$.

Hence $\struct {S / \RR, \circ_\RR}$ is indeed a quotient structure, and Quotient Structure is Well-Defined applies.


From Quotient Structure of Semigroup is Semigroup, we have that $\struct {S / \RR, \circ_\RR}$ is a semigroup.

Then we have:

\(\ds \forall \eqclass a \RR \in S / \RR: \, \) \(\ds \) \(\) \(\ds \eqclass a \RR \circ_\RR \eqclass a \RR\)
\(\ds \) \(=\) \(\ds \eqclass {a \circ a} \RR\) Definition of Operation Induced on Quotient Set
\(\ds \) \(=\) \(\ds \eqclass a \RR\) Definition of Idempotent Operation

Thus $\struct {S / \RR, \circ_\RR}$ is an idempotent semigroup.


To prove commutativity, it is sufficient to prove that $\paren {a \circ b} \mathrel \RR \paren {b \circ a}$.

We have:

\(\ds \forall a, b \in S: \, \) \(\ds \) \(\) \(\ds \paren {a \circ b} \circ \paren {b \circ a} \circ \paren {a \circ b}\)
\(\ds \) \(=\) \(\ds a \circ \paren {b \circ b} \circ \paren {a \circ a} \circ b\) Semigroup Axiom $\text S 1$: Associativity
\(\ds \) \(=\) \(\ds a \circ b \circ a \circ b\) Definition of Idempotent Operation
\(\ds \) \(=\) \(\ds \paren {a \circ b} \circ \paren {a \circ b}\) Semigroup Axiom $\text S 1$: Associativity
\(\ds \) \(=\) \(\ds a \circ b\) Definition of Idempotent Operation


and:

\(\ds \forall a, b \in S: \, \) \(\ds \) \(\) \(\ds \paren {b \circ a} \circ \paren {a \circ b} \circ \paren {b \circ a}\)
\(\ds \) \(=\) \(\ds b \circ \paren {a \circ a} \circ \paren {b \circ b} \circ a\) Semigroup Axiom $\text S 1$: Associativity
\(\ds \) \(=\) \(\ds b \circ a \circ b \circ a\) Definition of Idempotent Operation
\(\ds \) \(=\) \(\ds \paren {b \circ a} \circ \paren {b \circ a}\) Semigroup Axiom $\text S 1$: Associativity
\(\ds \) \(=\) \(\ds b \circ a\) Definition of Idempotent Operation

That is:

$\paren {a \circ b} \mathrel \RR \paren {b \circ a}$

Hence:

\(\ds \forall \eqclass a \RR, \eqclass b \RR \in S / \RR: \, \) \(\ds \) \(\) \(\ds \eqclass a \RR \circ_\RR \eqclass b \RR\)
\(\ds \) \(=\) \(\ds \eqclass {a \circ b} \RR\)
\(\ds \) \(=\) \(\ds \eqclass {b \circ a} \RR\)
\(\ds \) \(=\) \(\ds \eqclass b \RR \circ_\RR \eqclass a \RR\)

Thus we have shown that $\struct {S / \RR, \circ_\RR}$ is a commutative idempotent semigroup.

$\Box$


Let $\eqclass x \RR$ be an arbitrary element of $S / \RR$.

We have that $\eqclass x \RR$ is a subset of $\struct {S, \circ}$.


We now show that $\struct {\eqclass x \RR, \circ}$ is a subsemigroup of $\struct {S, \circ}$.

From Subsemigroup Closure Test, it is sufficient to show that $\struct {\eqclass x \RR, \circ}$ is closed.

Hence, let $a, b \in \eqclass x \RR$.

Then:

$a \mathrel \RR b$

and so:

\(\ds a\) \(=\) \(\ds a \circ b \circ a\)
\(\ds b\) \(=\) \(\ds b \circ a \circ b\)


We have:

\(\ds a \circ b\) \(=\) \(\ds \paren {a \circ b} \circ \paren {a \circ b}\) Definition of Idempotence
\(\ds \) \(=\) \(\ds \paren {a \circ b} \circ a \circ \paren {a \circ b}\) Definition of Idempotence

and:

\(\ds a\) \(=\) \(\ds a \circ b \circ a\) Definition of $\RR$
\(\ds \) \(=\) \(\ds a \circ a \circ b \circ a\) Definition of Idempotence
\(\ds \) \(=\) \(\ds a \circ \paren {a \circ b} \circ a\) Semigroup Axiom $\text S 1$: Associativity

That is:

$a \mathrel \RR a \circ b$

and so:

$a, b \in \eqclass x \RR \implies a \circ b \in \eqclass x \RR$

Thus Semigroup Axiom $\text S 0$: Closure is satisfied.

Hence from Subsemigroup Closure Test $\eqclass x \RR$ is a subsemigroup of $\struct {S, \circ}$


It remains to be shown that $\struct {\eqclass x \RR, \circ}$ is anticommutative.

Suppose $a, b \in \eqclass x \RR$ such that $a \circ b = b \circ a$.

We have:

\(\ds a \circ b\) \(=\) \(\ds b \circ a\)
\(\ds \leadsto \ \ \) \(\ds a \circ b \circ b\) \(=\) \(\ds b \circ a \circ b\) Definition of Idempotent Operation, Definition of $\RR$
\(\ds \leadsto \ \ \) \(\ds a \circ b\) \(=\) \(\ds b\)


Then:

\(\ds a \circ b\) \(=\) \(\ds b \circ a\)
\(\ds \leadsto \ \ \) \(\ds a \circ b \circ a\) \(=\) \(\ds b \circ a \circ a\)
\(\ds \leadsto \ \ \) \(\ds a\) \(=\) \(\ds b \circ a\) Definition of Idempotent Operation, Definition of $\RR$
\(\ds \) \(=\) \(\ds a \circ b\) by hypothesis
\(\ds \) \(=\) \(\ds b\) a priori


That is:

$a \circ b = b \circ a \implies a = b$

and so $\struct {\eqclass x \RR, \circ}$ is an anticommutative subsemigroup of $\struct {S, \circ}$.

$\blacksquare$


Sources

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