Idempotent Semigroup/Examples/Relation induced by Inverse Element/Properties/4
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 $a$.
$\RR$ is an equivalence relation.
Proof
Checking in turn each of the criteria for equivalence:
Reflexivity
\(\ds \forall a \in S: \, \) | \(\ds a \circ a \circ a\) | \(=\) | \(\ds a \circ a\) | Definition of Idempotent Semigroup | ||||||||||
\(\ds \) | \(=\) | \(\ds a\) | Definition of Idempotent Semigroup | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds a\) | \(\RR\) | \(\ds a\) | Definition of $\RR$ |
Thus $\RR$ is seen to be reflexive.
$\Box$
Symmetry
\(\ds a\) | \(\RR\) | \(\ds b\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds a \circ b \circ a\) | \(=\) | \(\ds a\) | Definition of $\RR$ | ||||||||||
\(\, \ds \land \, \) | \(\ds b \circ a \circ b\) | \(=\) | \(\ds b\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds b \circ a \circ b\) | \(=\) | \(\ds b\) | Conjunction is Commutative | ||||||||||
\(\, \ds \land \, \) | \(\ds a \circ b \circ a\) | \(=\) | \(\ds a\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds b\) | \(\RR\) | \(\ds a\) | Definition of $\RR$ |
Thus $\RR$ is seen to be symmetric.
$\Box$
Transitivity
Let $a \mathrel \RR b$ and $b \mathrel \RR c$.
Thus we have:
\(\ds a \circ b \circ a\) | \(=\) | \(\ds a\) | ||||||||||||
\(\ds b \circ a \circ b\) | \(=\) | \(\ds b\) |
and:
\(\ds b \circ c \circ b\) | \(=\) | \(\ds b\) | ||||||||||||
\(\ds c \circ b \circ c\) | \(=\) | \(\ds c\) |
Let:
\(\ds x\) | \(=\) | \(\ds c \circ b \circ c\) | ||||||||||||
\(\ds y\) | \(=\) | \(\ds c \circ b \circ a\) | ||||||||||||
\(\ds z\) | \(=\) | \(\ds a \circ b \circ a\) |
We have:
\(\ds b \circ a \circ b\) | \(=\) | \(\ds b\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {b \circ a} \circ \paren {b \circ c}\) | \(=\) | \(\ds b \circ c\) |
and:
\(\ds b \circ c \circ b\) | \(=\) | \(\ds b\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {b \circ c} \circ \paren {b \circ a}\) | \(=\) | \(\ds b \circ a\) |
Hence using Properties of Idempotent Semigroup: $1$:
\(\ds c \circ \paren {b \circ a} \circ c \circ \paren {b \circ c}\) | \(=\) | \(\ds c \circ \paren {b \circ c}\) | ||||||||||||
\(\ds c \circ \paren {b \circ c} \circ c \circ \paren {b \circ a}\) | \(=\) | \(\ds c \circ \paren {b \circ a}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds x \circ y\) | \(=\) | \(\ds y\) | |||||||||||
\(\ds y \circ x\) | \(=\) | \(\ds x\) |
Then we have:
\(\ds b \circ c \circ b\) | \(=\) | \(\ds b\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {a \circ b} \circ \paren {c \circ b}\) | \(=\) | \(\ds a \circ b\) |
and:
\(\ds b \circ a \circ b\) | \(=\) | \(\ds b\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {c \circ b} \circ \paren {a \circ b}\) | \(=\) | \(\ds c \circ b\) |
Hence using Properties of Idempotent Semigroup: $2$:
\(\ds \paren {a \circ b} \circ a \circ \paren {c \circ b} \circ a\) | \(=\) | \(\ds \paren {a \circ b} \circ a\) | ||||||||||||
\(\ds \paren {c \circ b} \circ a \circ \paren {a \circ b} \circ a\) | \(=\) | \(\ds \paren {c \circ b} \circ a\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds z \circ y\) | \(=\) | \(\ds z\) | |||||||||||
\(\ds y \circ z\) | \(=\) | \(\ds y\) |
Thus we have:
\(\ds x \circ y\) | \(=\) | \(\ds y\) | ||||||||||||
\(\ds y \circ x\) | \(=\) | \(\ds x\) | ||||||||||||
\(\ds y \circ z\) | \(=\) | \(\ds y\) | ||||||||||||
\(\ds z \circ y\) | \(=\) | \(\ds z\) |
Hence by Idempotent Semigroup: Relation induced by Inverse Element: $3$:
- $x \mathrel \RR z$
That is:
\(\ds c \circ b \circ c\) | \(\RR\) | \(\ds a \circ b \circ a\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds c\) | \(\RR\) | \(\ds a\) |
Thus $\RR$ is seen to be transitive.
$\Box$
$\RR$ has been shown to be reflexive, symmetric and transitive.
Hence by definition $\RR$ is an equivalence relation.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Exercise $11.19 \ \text {(f)}$