3
$\begingroup$

The question of existence of a universal inverse semigroup of an arbitrary semigroup has been answered before (this is a construction similar to the Grothendieck group). Let's refer to the universal inverse semigroup of a semigroup $S$ as $G_I[S]$ (for this question). It was noted that $G_I[S]$ for a commutative semigroup $S$ is not necessarily commutative (because of nilpotent elements). For a general non-commutative semigroup $S$, it's also clear that $G_I[S]$ is in general not an embedding, i.e. $S \not\subset G_I[S]$. However, it seems to me as if $G_I[S]$ will be an embedding for a commutative semigroup $S$, i.e. $S \subset G_I[S]$ (more precisely, $G_I[S]$ contains a sub-semigroup isomorphic to $S$). Is this true?

Note This question is identical to this question at math.stackexchange.com.

Improve this question
$\endgroup$
5
  • $\begingroup$ Why does it seems to you? This question is indeed more appropriate for math.stackexchange. Take $S=[0,+\infty]$ for a counterexample. $\endgroup$
    – Fernando Muro
    Jul 2, 2012 at 13:52
  • $\begingroup$ @Fernando: This semigroup (with addition, I assume) embeds into an inverse semigroup. $\endgroup$
    – user6976
    Jul 2, 2012 at 14:40
  • $\begingroup$ Sorry, I was thinking of the associated group. $\endgroup$
    – Fernando Muro
    Jul 2, 2012 at 14:42
  • $\begingroup$ In fact if you read Schein's paper (see my answer, the paper is available online), you will see that the question of whether all commutative semigroups embed into inverse semigroups was considered non-trivial. Even an example of a semigroup with commuting idempotents that does not embed into an inverse semigroup was not known for some time. $\endgroup$
    – user6976
    Jul 2, 2012 at 15:54
  • $\begingroup$ @Fernando: It seemed to me, because I had only checked it for upper triangular matrices with at most one non-zero entry per row and column. Now it's obvious to me that these matrices were much too closely related to partial one-to-one transformations. I admit that the question is more appropriate for math.stackexchange, because finding a counter-example was easy once somebody told me that there is one. However, I really like the paper from B. Schein... $\endgroup$
    – Thomas Klimpel
    Jul 2, 2012 at 19:56

1 Answer 1

Reset to default
6
$\begingroup$

B. Schein described all semigroups embeddable into inverse semigroups in Schein, Boris M., Subsemigroups of inverse semigroups, Le Matematiche LI (1996), Supplemento, 205–227 (in fact the paper was written in the 50s). From that paper it easily follows that not every commutative semigroup embeds into an inverse semigroup. Indeed, look at the quasi-identity $\& R\to u=v$ called $A_1$ on page 218 there. Consider the (finite) nilpotent of class 3 commutative semigroup $S$ given by the presentation $R$ in the variaety of commutative nilpotent semigroups of class 3. More concretely, note that all words that appear in $R$, $u,v$ are of length 2. The semigroup $S$ consists of all words of length 1 or 2 in letters that appear in $R$, and 0; the product is the concatenation whenever it is inside that set of words or 0 otherwise. Words that are equal according to $R$ are identified in $S$. The equalities of $R$ hold in $S$ by definition while the equality $u=v$ is not true in that semigroup. Hence this finite commutative semigroup $S$ does not satisfy $A_1$. But by B. Schein $A_1$ is necessary for embeddability into an inverse semigroup. Hence $S$ does not embed into an inverse semigroup. In particular, the map from $S$ into its universal inverse semigroup is not injective.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.