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A congruence on a semigroup $S$ is an equivalence relation $\sigma\subseteq S\times S$ that respect to the multiplication. In other words $$(a,b), (c,d)\in\sigma \implies (ac, bd)\in\sigma. $$ Given a such congruence the quotient $S/\sigma$ has a well defined semigroup structure $[a][b]=[ab]$ and a quotient map $S\to S/\sigma.$

Now suppose $S$ is an inverse semigroup. Then $[a^{\star}]$ act as an inverse of $[a],$ but I do not see a reason for it to be unique. First I tried to show that if $b\in S$ has the property that$(a, aba), (b, bab)\in\sigma$ then $(a^{\star}, b)\in\sigma,$ but this doesn't seems to work.

  • On the other hand, this seems bit strange. What am I doing wrong here?
  • If $S/\sigma$ is merely a regular semigroup, what conditions on $\sigma$ would force it to be an inverse semigroup?
Bumblebee
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    You can require that $\sigma$ be closed under ${}^{\star}$ (i.e., $(a,b)\in\sigma\implies (a^{\star},b^{\star})\in\sigma$). This will mean $\sigma$ is a congruence of "inverse semigroups" viewed as semigroups with an additional unary operation. Since inverse semigroups form a variety under those conditions, the quotient will certainly be an inverse semigroup. But I do not know if this is necessary or merely sufficient. – Arturo Magidin Mar 26 '21 at 17:10

1 Answers1

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Any quotient of an inverse semigroup is an inverse semigroup. An easy way to prove this is to use the following characterisation:

Theorem. A semigroup is an inverse semigroup if and only if it is regular and its idempotents commute.

You have already shown that if $S$ is inverse, then $S/\sigma$ is regular. It remains to show that the idempotents of $S/\sigma$ commute. One needs the following lemma (see Lemma 2.4.3 in [1])

Lemma. Let $[a]$ be an idempotent in $S/\sigma$. Then there is an idempotent $e$ in $S$ such that $[e] = [a]$.

Proof. Since $[a]$ is idempotent, $(a, a^2) \in \sigma$. Let $x$ be an inverse of $a^2$ and let $e = axa$. Then $$ e^2 = a(xa^2x)a = axa = e $$ and, modulo $\sigma$ $$ e = axa \equiv a^2xa^2 = a^2 \equiv a $$ so $(e,a) \in \sigma$. $\quad\blacksquare$

It is now easy to finish the proof of the theorem. Let $[a]$ and $[b]$ be idempotents in $S/\sigma$. By the lemma, there exist idempotents $e$ in $f$ in $S$ such that $(e,a) \in \sigma$ and $(f,b) \in \sigma$. It follows that, modulo $\sigma$, $$ ab \equiv ef = fe \equiv ba $$ and thus $[a][b] = [b][a]$.

[1] John M. Howie, Fundamentals of Semigroup Theory, Oxford, Oxford University Press, coll. « London Mathematical Society Monographs. New Series » (no 12), 1995, x+351 p. (ISBN 0-19-851194-9, Math Reviews 1455373).

J.-E. Pin
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  • Thank you very much. I think this is it. I will accept this answer after I went through this argument one more time. – Bumblebee Mar 27 '21 at 18:36