# Condition for Group given Semigroup with Idempotent Element

## 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.

$\blacksquare$