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By general nonsense the forgetful functor from groups to monoids has a left adjoint. It maps a monoid $(X,\cdot,1)$ to the free group on $\{\underline{x} : x \in X\}$ modulo the relations $\underline{1}=1$, $\underline{x \cdot y}=\underline{x} \cdot \underline{y}$. Notice that the elements of this group have the form $\underline{x_1} \cdot \underline{x_2}^{-1} \cdot \underline{x_3} \cdot \underline{x_4}^{-1} \cdot \dotsc$.

Question 1: Does this group have a name? It is analogous to the Grothendieck group, which is the left adjoint of the forgetful functor from abelian groups to commutative monoids, so perhaps we may call it the non-abelian Grothendieck group?

Question 2: Is there any criterion when an element of this group vanishes?

Martin Brandenburg
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  • The group still seems to be free, since the relations just say that some generator is equal to a product of other relators. – HJRW Jan 07 '15 at 19:59
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    No, this argument would also apply to the Grothendieck group, which is not always free abelian. Notice that every relation $x \cdot y = a \cdot b$ implies the relation $\underline{x} \cdot \underline{y} = \underline{a} \cdot \underline{b}$ in the group. – Martin Brandenburg Jan 07 '15 at 20:02
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    Wait, my mistake - apologies. – HJRW Jan 07 '15 at 20:08
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    I think semigroups theorists most often say maximal group image. Universal group and group completion are fairly common. The second volume of Clifford and Preston gives a detailed account of Malcev and Lambek results. – Benjamin Steinberg Jan 07 '15 at 20:30
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    For an alternative approach to the terminology, I can't help noticing that this is the 1-object version of localizing a category at all its morphisms. Since this localization corresponds (correct me if I'm wrong) to taking the 1-truncation of the classifying space of the category, there might be some topological terminology that would be suitable... – Tim Campion Jan 07 '15 at 20:50
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    It is the fundamental group of the classifying space of M. – Benjamin Steinberg Jan 07 '15 at 20:56
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    Of course undecidability of the word problem for groups means that it not possible to give an effective characterisation of triviality of an element of the group. – Benjamin Steinberg Jan 07 '15 at 20:58
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    But if the monoid's word problem is decidable, perhaps the word problem of the group is decidable? – Tim Campion Jan 07 '15 at 21:48
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    @TimCampion, I know of finitely presented inverse monoids with decidable word problem whose universal group has undecidable word problem. I don't think they are finitely presented in the category of monoids but in any event such finitely presented examples must exist. – Benjamin Steinberg Jan 08 '15 at 02:52
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    This group is also called the group completion of the monoid. – Omar Antolín-Camarena Jan 08 '15 at 19:49

3 Answers3

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George Bergman refers to it as the universal enveloping group of the monoid. See Section 3.11, page 81, of his universal algebra notes/forthcoming book: http://math.berkeley.edu/~gbergman/245/

He also references P.M. Cohn's Universal Algebra book, and two papers of Mal'cev in which he establishes conditions for the universal map to be an embedding.

P.S. As far as I know, the construction predates Grothendieck's work; but because of the applications the latter found for the abelian case, his work is better known. In fact, this is not the first time I've seen the general case described as "the nonabelian version of the Grothendieck group"...

Arturo Magidin
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I have seen "(universal) enveloping group of the monoid" used for this construction. Mal'cev has found necessary and sufficient conditions for injectivity of the comparison map.

Anatoliy I.Mal’cev, Über die Einbettung von assoziativen Systemen in Gruppen, Mat. Sb. N.S. 6 (1939) 331-336.

Anatoliy I. Mal’cev, Über die Einbettung von assoziativen Systemen in Gruppen, II, Mat. Sb. N.S. 8 (1940) 251-264.

See also Chapter VII in P. M. Cohn, Universal Algebra, second edition, Reidel, 1981.

Andreas Thom
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    It should be added that the quasivariety of monoids embeddable in a group has no finite basis of quasi-identities and there is no algorithm to decide if a finitely presented monoid embeds in a group. – Benjamin Steinberg Jan 07 '15 at 20:28
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    Thanks, the result about the finite basis is in the second paper of Mal'cev. – Andreas Thom Jan 07 '15 at 20:29
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I'd just like to point out that in the many-object case this has been called the fundamental groupoid of a category by Paré, which would suggest that it could be called the fundamental group of a monoid, a possibility strengthened by the fact that, as Benjamin Steinberg points out in the comments, it's the fundamental group of the classifying space of the monoid. Another possibility might be classifying group, but that's probably ill-advised since it's not clear (to me) what it classifies.

Tim Campion
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  • But if I had to vote, in the single-object case I'd go with group completion as in Omar Antolín-Camarena's comment above. Even if "universal enveloping group" is more standard in the general algebra literature, it just seems overly heavy-handed to me. – Tim Campion Jan 08 '15 at 23:52