# Inverse of the Grothendieck group construction?

Is there an "inverse" of the Grothendieck group construction that would generate a (somehow "simplest") Abelian semigroup given some (suitably qualified, if necessary) Abelian group? (I realize that any Abelian group is already an Abelian semigroup, but this is not very satisfying... I'm trying to get at the "simplest" semigroup that will generate a given Abelian group when one applies the Grothendieck construction to it. So this "reverse construction" should, e.g., yield $\mathbf{N}^+$ (i.e. the positive integers) when applied to $\mathbf{Z}$, not simply produce $\mathbf{Z}$ back.)

Thx

EDIT2: Please read the comments on this post by Theo and Arturo before spending any time on it; it looks like the wording of my question is not right, but I'll need to do some more research to fix it...

EDIT: As I wrote in a comment below, this question was motivated by a passing remark I came across about a "well-known" one-to-one correspondence between cancellative Abelian semigroups and ordered Abelian groups. Judging from the brief, informal account I read of this correspondence, it looked to me like the semigroup $\to$ group half was just the construction of the Grothendieck group from the semigroup, although it was not explicitly named as such. And since this correspondence was described as being bijective, I figured that there may be an inverse construction, which, at least in this special case (partially ordered Abelian group $\to$ cancellative Abelian semigroup), could be construed as a sort of inverse of the Grothendieck group construction. The citation given for all this is a 1940 paper by Clifford, which I don't have access to (beyond the first page), and even if I did, it may not be of much help to me, since its abstract states that the paper gives no proofs, and instead refers the reader to another paper (which I think is this one), even further out of my reach, and in German, a language I can read only with a lot of help from the dictionary...

• How do you expect to make this functorial? The Grothendieck construction is the left adjoint to the inclusion of the category of abelian groups into the category of abelian semi-groups, but your "reverse construction" would need additional data, wouldn't it? – t.b. Aug 22 '11 at 23:04
• @Theo: you may well be right... I haven't even gotten to functors yet, let alone adjoints... – kjo Aug 22 '11 at 23:19
• @Theo: I once read something to the effect that there is a well-established correspondence between Abelian semigroups with a cancellation law and abelian partially ordered groups, and the application of the Grothendieck construction to such a semigroup produces a an Abelian group that may be regarded as partially ordered, with the elements of the form $(2a, a)$ as the positive elements... But I don't know the details, and how they would answer your question about functoriality – kjo Aug 22 '11 at 23:26
• Well, then you should start learning about functors! (this was addressing your previous comment; I'll have to think about your second comment). – t.b. Aug 22 '11 at 23:27
• @kjo: In particular, no finite nontrivial abelian group can be partially ordered (if $a\neq 0$, then we must have $a\lt 0$ or $0\lt a$. If $a$ is of order $n$, then adding $(n-1)a$ to both sides gives the opposite strict inequality). But if you view $C_n$ as a cancellative abelian semigroup (which it certainly is), and apply Grothendieck's construction, you get $C_n$ again. So that bijection is certainly not obtained merely via the enveloping group construction. – Arturo Magidin Aug 23 '11 at 4:01

$$\mathbb{N}_k=\{s\in\mathbb{N}\ |\ s \geq k\}\cup\{0\}$$
It's obviously a submonoid of $\mathbb{N}$ and it can be easily seen that the Grothendieck construction for each $\mathbb{N}_k$ yields $\mathbb{Z}$. A little more difficult is to show that $\mathbb{N}_k\not\simeq\mathbb{N}_t$ if $k\neq t$.