Condition for Group given Semigroup with Idempotent Element
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Theorem
Let $\struct {S, \circ}$ be a semigroup.
Let there exist an idempotent element $e$ of $S$ such that for all $a \in S$:
- there exists at least one element $x$ of $S$ satisfying $x \circ a = e$
- there exists at most one element $y$ of $S$ satisfying $a \circ y = e$.
Then $\struct {S, \circ}$ is a group.
Proof
Let $a$ be arbitrary.
We have:
\(\ds \exists x \in S: \, \) | \(\ds x \circ a\) | \(=\) | \(\ds e\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {x \circ a} \circ \paren {x \circ a}\) | \(=\) | \(\ds e \circ e\) | |||||||||||
\(\ds \) | \(=\) | \(\ds e\) | by hypothesis: $e$ is idempotent | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds x \circ \paren {a \circ x \circ a}\) | \(=\) | \(\ds e\) | Semigroup Axiom $\text S 1$: Associativity | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds a \circ x \circ a\) | \(=\) | \(\ds a\) | both $a \circ x \circ a$ and $a$ are $y \in S$ such that $x \circ y = e$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds a \circ e\) | \(=\) | \(\ds a\) | as $x \circ a = e$ |
So $e$ is a right identity.
Again, let $a$ be arbitrary.
Let $x \in S$ be such that $x \circ a = e$.
We have:
\(\ds e\) | \(=\) | \(\ds x \circ a\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {x \circ e} \circ a\) | Definition of Right Identity | |||||||||||
\(\ds \) | \(=\) | \(\ds x \circ \paren {e \circ a}\) | Semigroup Axiom $\text S 1$: Associativity | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds e \circ a\) | \(=\) | \(\ds a\) | both $e \circ a$ and $a$ are $y \in S$ such that $x \circ y = e$ |
So $e$ is a left identity.
We note the meaning of the criterion:
- there exists at least one element $x$ of $S$ satisfying $x \circ a = e$
As we now know that $e$ is a left identity, the above means that $x$ is a left inverse for $a$ in $S$.
To summarise, we have an algebraic structure $\struct {S, \circ}$:
- $(1): \quad$ which is closed, from Semigroup Axiom $\text S 0$: Closure
- $(2): \quad$ which is associative, from Semigroup Axiom $\text S 1$: Associativity
- $(3): \quad$ which has a left identity
- $(4): \quad$ for which every element has a left inverse.
That is, $\struct {S, \circ}$ fulfils all the left group axioms.
Hence the result.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.15$
- Arturo Magidin (https://math.stackexchange.com/users/742/arturo-magidin), Conditions for Group given Semigroup with Idempotent Element, URL (version: 2022-03-02): https://math.stackexchange.com/q/4394150