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I am new in semigroup theory and I have an problem related to quasi ideals. I am trying to solve the following problem

Problem: For a semigroup $S$, the following are equivalent

  1. $S$ is regular

  2. $A=ASA$ for every quasi ideal $A$

My approach for $1\implies 2$

Consider $a\in A$ since $S$ is regular so there exist $x\in S$ such that $a=axa$ so $A\subseteq ASA$

But I am not able to prove $ASA\subseteq A$

Any help in this regard

EDIT. According to Clifford and Preston (The algebraic theory of semigroups, Volume I, p. 85), a subset $A$ of $S$ is a quasi-ideal if $AS \cap SA \subseteq A$.

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    $\begingroup$ Well, for starters, you are supposed to prove that $ASA\subseteq A$, not that $SAS\subseteq A$. $\endgroup$
    – Arturo Magidin
    Commented Jan 30, 2020 at 17:40
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    $\begingroup$ Note that $asa=(as)a = a(sa)\in SA\cap AS$. $\endgroup$
    – Arturo Magidin
    Commented Jan 30, 2020 at 17:42
  • $\begingroup$ oh yes,my mistake $\endgroup$
    – kam
    Commented Jan 30, 2020 at 17:42
  • $\begingroup$ @ArturoMagidin Thanks! i get it $\endgroup$
    – kam
    Commented Jan 30, 2020 at 17:46
  • $\begingroup$ Could you please give the definition of a quasi ideal of a semigroup? $\endgroup$
    – J.-E. Pin
    Commented Feb 1, 2020 at 6:50

1 Answer 1

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Let $A$ be a quasi-ideal of $S$. Since $SA \subseteq S$ and $AS \subseteq S$, one gets $ASA \subseteq AS \cap SA \subseteq A$.

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