The question of [existence of a universal inverse semigroup](https://math.stackexchange.com/q/104893/12490) 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](https://math.stackexchange.com/q/162765/12490) at math.stackexchange.com.



Is the universal inverse semigroup of a commutative semigroup an embedding?
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. 


