Idempotent Semigroup/Examples/Relation induced by Inverse Element/Properties/3

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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:

\(\ds x \circ y\) \(=\) \(\ds y\)
\(\ds y \circ x\) \(=\) \(\ds x\)
\(\ds y \circ z\) \(=\) \(\ds y\)
\(\ds z \circ y\) \(=\) \(\ds z\)


Then:

$x \mathrel \RR z$


Proof

From Semigroup Axiom $\text S 1$: Associativity we take it for granted that $\circ$ is associative.

Hence parentheses will be employed whenever it makes groupings of operations more clear.


Recall:

From $x \circ y = y$ and $y \circ x = x$

$(1): \quad \forall z \in S: z \circ x \circ z \circ y = z \circ y$

using Properties of Idempotent Semigroup: $1$

From $y \circ z = y$ and $z \circ y = z$

$(2): \quad \forall x \in S: y \circ x \circ z \circ x = y \circ x$

using Properties of Idempotent Semigroup: $2$, exchanging $z$ and $x$


We have:

\(\ds z \circ x \circ z\) \(=\) \(\ds z \circ x \circ z \circ y\) by hypothesis: $z \circ y = z$
\(\ds \) \(=\) \(\ds z \circ y\) from $(1)$
\(\ds \) \(=\) \(\ds z\) by hypothesis: $z \circ y = z$

and:

\(\ds x \circ z \circ x\) \(=\) \(\ds y \circ x \circ z \circ x\) by hypothesis: $y \circ x = x$
\(\ds \) \(=\) \(\ds y \circ x\) from $(2)$
\(\ds \) \(=\) \(\ds x\) by hypothesis: $y \circ x = x$

Hence the result, by definition of $\RR$.

$\blacksquare$


Sources

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