# From monoids to groups

I was looking at the case when you go from the monoid of natural numbers to the group of integers by means of a suitable equivalence relation. The key here was to find inverses for each natural and I was wondering how it could be generalized. I mean:

Given any monoid, when can I find inverses for your elements?

In other words, can I always go from a monoid to a group? If not, when is it? When do not?

If you could tell me where I can find information about this (a book or paper) it would be very helpful. Thank you.

• See here, and here. For commutative monoids, the construction is often called the “Grothendieck group of the (commutative) monoid”. For not-necessarily commutative monoids, it is called the “universal enveloping group” of the monoid. – Arturo Magidin Nov 28 '20 at 0:30

Yes, you can always obtain a group from a monoid, through a process that is called group completion.
In general, the completion is a process very similar to the one you cited: going from $$(\mathbb N,+)$$ to $$(\mathbb Z,+)$$ is the standard, classic example; another one is going from the monoid $$(\mathbb N^\ast,\cdot)$$ to its group completion $$(\mathbb Q_{>0},\cdot)$$.

The "easy" case is when your monoid $$(M,+)$$ is commutative; its group completion, sometimes called Grothendieck group (for its ties with algebraic K-theory, if I'm not mistaken), can be defined as a quotient of $$M\times M$$ (with coordinate-wise operation, $$(m,m')+(n,n'):=(m+n,m'+n')$$) by the equivalence relation described as follows: $$(m,m')\sim(n,n') \iff \exists\;k\in M:\;m+n'+k=m'+n+k$$ Sometimes (i.e. when the cancellation law holds in $$M$$) the term $$k$$ is not necessary (you can just choose the identity of your monoid as $$k$$). In your original example, you have that the group completion of $$(\mathbb N,+)$$ is made of equivalence classes of $$(m,m')\in\mathbb N^2$$, where $$(m,m')\sim(n,n')\iff m+n'=m'+n$$, hence you can see that "$$(m,m')$$ basically means $$m-m'$$".

Notice that this procedure is very reminiscent of the construction of rational numbers! In fact, as we mentioned before, if you take as $$M$$ the monoid of positive integers with multiplication $$(\mathbb N^\ast,\cdot)$$, you get as its group completion elements of $$\mathbb N^*\times \mathbb N^*$$ modulo the relation: $$(m,m')\sim(n,n') \iff m\cdot n'=m'\cdot n,$$ so $$(m,m')$$ is basically what we're used to denote as $$\frac{m}{m'}$$.

The most important application of this construction I know of is, as I mentioned early, K-groups in K-theory; for instance, real/complex vector bundles over a "nice" (e.g., compact Hausdorff) topological space, and finitely generated projective modules over a ring, up to isomorphisms, form a monoid $$(M,\oplus)$$ with the direct sum, and then $$K_0$$ is defined as the group completion of $$M$$.

• Not to say that "group completion" is a completely absurd nomenclature, but "total group of fractions (differences, in the additive case)" would perhaps be even more natural, since it intimately relates to the analogous construction of localising rings and modules and, furthermore, the object obtained in general by applying this construction is a monoid which does not necessarily embed the original monoid and only "renders" part of the elements of this original monoid (the ones in the multiplicative system chosen) as invertible elements in the newly created structure. – ΑΘΩ Nov 28 '20 at 1:50
• But then on second thought I better understand what you are referring to by "group completion". The idea would be that "group completions" and "monoids of fractions" are distinct constructions to begin with and only in the case of commutative monoids with total multiplicative system do they produce canonically isomorphic objects (groups). – ΑΘΩ Nov 28 '20 at 2:03
• The more common name is “universal enveloping group”, as far as I am aware. But it is important to note is that the original monoid need not be embedded in the constructed group, and I don’t think you even hint at that. – Arturo Magidin Nov 28 '20 at 2:07

If $$M$$ is a semigroup, the group presentation $$\langle (x_m)_{m\in M}\mid (x_mx_n=x_{mn})_{m,n\in M}\rangle$$ defines a group $$\hat{M}$$ and the canonical map $$M\to\hat{M}$$, $$m\mapsto x_m$$ is a semigroup homomorphism (monoid homomorphism if $$M$$ is a monoid), and this satisfies the universal property. As mentioned by Arturo Magidin it is not injective in general (by the universal property, it's injective iff $$M$$ embeds into a group; a necessary condition is that $$M$$ is left and right cancelative).

While this construction is theoretically obvious, it is practically not very useful and not "practically explicit" (since it's a huge set modulo a huge equivalence relation). If $$M$$ is commutative, a much more practical construction is provided in Ottavio's answer; in particular if $$M$$ is commutative and cancelative then $$M\to\hat{M}$$ is injective.