# Question about monoid (Grothendieck's) group-completion

1. Jun 16, 2010

### mnb96

Hello,
in these pages (http://en.wikipedia.org/wiki/Grothendieck_group" [Broken]) I found the description of a particular construction that makes a Group from a cancellative-abelian-monoid.
The group embeds the original monoid, and it is called Grothendieck's group.

I can't find more sources about this topic and I would like to know if the group constructed with this method has minimal cardinality.

Last edited by a moderator: May 4, 2017
2. Jun 22, 2010

### ohubrismine

First, note that a monoid fails to be a group precisely because the set lacks an inverse for each of its elements.
The construction above basically adds an inverse for each element.
So for a finite set, the group will contain roughly double the number of elements in the monoid. For a monoid on a transfinite but countable set, the group will also be countable.
In this sense, the group does have minimal cardinality because it will contain precisely the elements necessary to form a group.

Last edited by a moderator: May 4, 2017
3. Jun 22, 2010

### Martin Rattigan

A finite abelian cancellation semigroup (a fortiori monoid) is already a group. The construction will produce an isomorphic copy of the same group in this case, giving exactly single the number of elements (so only roughly double with a rather free interpretation of the phrase).

The construction works equally well for a (nonempty) abelian cancellation semigroup and gives a group with minimal cardinality in either case. This cardinality is always the same as the semigroup or monoid you start with (assuming of course these are abelian cancellation versions).

4. Jun 22, 2010

### ohubrismine

I was really addressing the non-group abelian monoid (the second of the sources given above).
By the way, did you know Alexandre Grothendieck won the Field's medal?

5. Jun 22, 2010

### Martin Rattigan

There are no finite cancellative examples of "non group abelian monoids", so there would seem to be nothing to address in terms of the original question in the finite case.

If you meant to address finite "non group abelian monoids" that don't satisfy the cancellation law then the Grothendieck construction will indeed give a group, but the original monoid will, of course, never embed in it. The number of elements in the group would generally be even further adrift from "roughly double" in this case.

See e.g. last example in https://www.physicsforums.com/showthread.php?t=405363&highlight=Grothendieck".

Last edited by a moderator: Apr 25, 2017
6. Jun 22, 2010

### Martin Rattigan

I think I read that Grothendieck won a Fields medal, but probably not for this construction (which I think any average algebra student would probably invent without too much difficulty). I believe the construction had in any case been in use well before Grothendieck, so it's not too clear why it's got his name attached in the first place.

Last edited: Jun 22, 2010
7. Jun 22, 2010

### mnb96

Wait a minute.
So you are basically saying that if an abelian semigroup S is cancellative, then S is a group? Then what are the inverse elements?

8. Jun 22, 2010

### Martin Rattigan

No, I'm saying a nonempty finite cancellation semigroup is a group. Obviously $\mathbb{N}$ is a cancellation semigroup under $+$ but not a group.

If $S$ is a finite cancellation semigroup and $s\in S$, then by the pigeonhole principle $Ss=sS=S$.

Then $(\exists x\in S)sx=s$ and if $t\in S$, $(\exists y\in S)ys=t$. Then $ysx=ys$, so $tx=t$. That is $x$ is a right identity. Since $sS=S$, each element $s\in S$ also has a right inverse, thus $S$ is a group.

This holds for any nonempty finite cancellation semigroup $S$ and in particular in the abelian case we are interested in for the Grothendieck construction.

For finite abelian cancellation semigroups (or monoids) the Grothendieck construction will result in a copy of the semigroup (which is actually a group). For infinite abelian cancellation semigroups or monoids the semigroup or monoid may embed as a proper subsemigroup, but since $\mathfrak{m}^2=\mathfrak{m}$ for any infinite cardinal $\mathfrak{m}$, the cardinality of the Grothendieck group will still be the same as the original semigroup or monoid - hence minimal.

Last edited: Jun 22, 2010
9. Jun 23, 2010

### mnb96

I see!
That's very clear. thanks a lot!
I might ask you something more for the infinite case...

10. Jun 24, 2010

### mnb96

1) Is this true also when the original semigroup (or monoid) is "continuous"? For example, consider the monoid $(\mathbb{R}^{+} \cup \{ 0 \} , +)$, does its group-completion have the same cardinality?

2) If the cardinality is always the same, does this all mean that any finite or infinite cancellative abelian semigroup is a group?

11. Jun 24, 2010

### Martin Rattigan

1) Yes.
2) No. (The example you gave is obviously not a group.)

I assumed the axiom of choice - some people might like to add that as a proviso.

Last edited: Jun 24, 2010
12. Jun 24, 2010

### mnb96

This sounds counter-intuitive to me.
The positive reals (together with 0) is a monoid and not a group.
On the other hand, its group-completion has the same cardinality...???

If there is a bijection between the monoid and its group completion, why shouldn´t they be the same thing?
Perhaps when one start to deal with infinities, common sense and intuition just do harm.

Last edited: Jun 24, 2010
13. Jun 24, 2010

Perhaps.