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

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