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

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:

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:

and:

That is:
 * $\paren {a \circ b} \mathrel \RR \paren {b \circ a}$

Hence:

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

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:

We have:

and:

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 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:

Then:

That is:
 * $a \circ b = b \circ a \implies a = b$

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