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Let $S$ be a semigroup that satisfies the property \begin{align*} \forall a \in S, \quad aS=S \wedge Sa=S. \end{align*}

I want to show that $S$ is a group, ie, that $S$ satisfies

(1) $\exists e \in S \quad \forall a \in S : \quad ea=ae=a$;

<p>(2) $\forall a \in S \quad \exists b \in S : \quad ab=ba=e$.</p>

However, I can only show that

(1') $\exists e \in S \quad \forall a \in S : \quad ea=a$;

(2') $\forall a \in S \quad \exists b \in S : \quad ba=e$.

With what I've shown, is it possible to state that $S$ is a group?

Leafar
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1 Answers1

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Take $a$ in $S$. We have $e\in S$ so that $ae=a$ (Because $aS=S$).

Now take $b\in S$. we have $s\in S$ so that $b=sa$ (Because $Sa=S$). Therefore $be=(sa)e=s(ae)=sa=b$. So $e$ is a right identity.

Let $a$ be as above,we have $e'\in S$ so that $ea=a$ (Because $Sa=S$).

Now take $b\in S$. we have $s'\in S$ so that $b=as'$ (Because $aS=S$). Therefore $e'b=e'(as')=(ea)s'=as'=b$. So $e'$ is left identity.

Now notice $e=e'e$ because $e'$ is left identity and $e'e=e'$ because $e$ is right identity. So $e=e'e=e'$ and $e$ is identity.

Take $b\in S$, we have $c$ so .that $bc=e$ and $d$ so that $db=e$ because $bS=S$ and $Sb=s$.

Now notice $c=ec=(db)c=d(bc)=de=d$ and so $c=d$ is the inverse of $b$.

Asinomás
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