4
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Reposted at mathoverflow. The question has been answered there with 'no', giving a counterexample based on the reference to "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)". I won't repost this as an answer here for the moment, because I have not worked through the reference yet, and so can't claim understand $\endgroup$ – Thomas Klimpel Nov 2 '12 at 10:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.