Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I recently stumbled upon the Mathieu groupoid and I found them fascinating.

It appears as a subset of $S_{13}$ which is not closed under multiplication, but it turns out to be a groupoid with 13 objects. The "first" sporadic finite simple group $M_{12}$ appear as the Automorphism of a point. One can further find $M_{11}$ from this.

Now that people classified finite simple groups with decades of effort, I'm wondering if there has been any attempt to classify finite groupoids.

(Maybe some results like this is already implied by the classification of finite simple groups, but the groupoid $M_{13}$ seems so amazing, so I got curious.)

share|improve this question
    
I guess all I'm saying in my answer is that there has been an "attempt to classify finite groupoids" to the same exact extent that there has been an attempt to classify finite groups. I honestly don't know what exactly has been achieved in this regard, but last I heard, we are extremely far away from that. –  Todd Trimble Oct 6 '12 at 5:51
3  
Any groupoid is a disjoint union of groups, so classifying finitie groups or finite groupoids are the same problem. –  Fernando Muro Oct 6 '12 at 9:56
3  
@Fernado: I think you mean: any groupoid is a disjoint union of connected groupoids. A;ex Heller once remarked to me that the classification of vector spaces is easy, but the classification of vector spaces with one endomorphism is both interesting and non trivial; with two endomorphisms is hard; and with three is unsolved. So I think in this example we are given more than just a groupoid-see my comment below. –  Ronnie Brown Oct 6 '12 at 13:52
3  
@Ronnie: a connected groupoid is the same as a group if we look through the glasses of equivalences of categories, which I think of as the right notion of 'bein the same' in this context. –  Fernando Muro Oct 6 '12 at 19:00
    
Silly question: what is $\pi_0(M_{13})$ (the set of isomorphism classes)? Note that the 2008 article The transitivity of Conway's $M_{13}$ linked to on the Wikipedia page is not about the concept of groupoid transitivity. –  David Roberts Nov 29 '12 at 1:33
show 3 more comments

4 Answers 4

The problem of classifying finite groupoids is essentially of the same order of difficulty as classifying finite groups, which as far as I am aware we are very, very far away from doing. (There is a classification theorem for finite simple groups. I don't know what is meant by classifying "finite subgroups".)

The basic idea is that groupoids are disjoint unions of connected groupoids, and connected groupoids are equivalent (in the technical sense of categorical equivalence) to groups as 1-object categories. Specifically, if you have a connected groupoid $G$ and choose an object $x$, then $G$ is equivalent to the group of automorphisms $\hom_G(x, x)$ (which I will abbreviate to $G(x, x)$.

So for example, I claim that a finite connected groupoid $G$ is classified by the cardinality of its object set $G_0$ together with the isomorphism type of a typical automorphism group $G(x, x)$. In other words, if $G$, $H$ are finite connected groupoids $G$, $H$ and there exists a bijection $F_0: G_0 \to H_0$ between their objects sets and a group isomorphism $\phi: G(x, x) \to H(y, y)$ between typical automorphism groups (supposing WLOG that $y = F_0(x)$), then $G$ and $H$ are isomorphic as groupoids.

The proof is easy. Let $x_0 = x$, $x_1, \ldots, x_n$ be the objects of $G$. For each $j > 0$, choose at random a morphism $g_j: x_0 \to x_j$, and let $g_0 = 1_{x}$; similarly choose at random a morphism $h_j: F_0(x_0) \to F_0(x_j)$ (but again with $h_0 = 1_{F_0(x)}$. Define a functor $F: G \to H$ to be $F_0$ at the object level. To define $F$ at the morphism level, notice that any morphism $f: x_i \to x_j$ is of the form $g_j \circ g \circ g_{i}^{-1}$ for some unique $g \in G(x, x)$. Then define $F(f)$ to be $h_j \circ \phi(g) \circ h_{i}^{-1}$. Then check that this defines a functor and indeed an isomorphism between $G$ and $H$; the details are straightforward.

share|improve this answer
    
"finite subgroups" is a funny typo. I just fixed. –  temp Oct 6 '12 at 5:27
    
I can see this part (equivalence of categories). It is just that the Mathieu groupoid seems so natural that makes me wonder how one can come up with other definition of M_12 before discover M_13... –  temp Oct 6 '12 at 5:32
3  
Interesting to see this! From the wiki article it seems that this groupoid is given with an action on sets, so it may be better seen in the context of covering morphisms of groupoids, which are equivalent to actions: see the book by Philip Higgins on "Categories and groupoids" downloadable from TAC Reprints No. 7 (2005) pp 1--195. –  Ronnie Brown Oct 6 '12 at 11:32
    
That's a very pertinent remark, Ronnie. –  Todd Trimble Oct 6 '12 at 13:01
add comment

Asking to classify finite groups in general is essentially a pie in the sky question. We know that every finite group is 'built up' of finite simple groups, but even with those classified there are still many different ways a given set of groups can be combined to produce new ones. Worse still, enumeration of finite groups seems to suggest what one would intuitively expect to be true: the ones with less structure can be glued together in far more ways than those with rich and complicated structure.

More explicitly, groups of order at most 2000 or so have been classified (see for instance this ten year old paper of Breach, Eick and O'Brian:

http://www.math.auckland.ac.nz/~obrien/research/2000-announce.pdf )

The conclusion? Out of the 49 910 529 484 groups of order at most 2000 a staggering 49 487 365 422 of them have order 1024 - almost every group is not just nilpotent but is in fact a 2-group!

share|improve this answer
6  
Corrections: Breach -> Besche, O'Brian -> O'Brien. Also it might be worth mentioning that the number of ways of combining groups to get new ones is counted using second cohomology. The c-word should be enough to convince people that this is a highly non-trivial process. –  Nick Gill Nov 27 '12 at 16:13
2  
Wow, thanks for the link and the factoid, Ben! –  Todd Trimble Nov 27 '12 at 16:22
2  
I was so struck by this calculation that I had to spread the (old) news: golem.ph.utexas.edu/category/2012/11/… –  Tom Leinster Nov 28 '12 at 21:09
    
This is similar to the phenomenon that 99% of all finite semigroups have a zero element and satisfy xyz=0. –  Benjamin Steinberg Nov 29 '12 at 2:03
1  
I absolutely don't understand how this address the question: this answer doesn't even mention the word "groupoid". -1 –  Joël Feb 27 '13 at 15:33
show 4 more comments

Todd Trimble basically answered the question you literally asked, but it sounds like you may be thinking about a slightly different question, e.g., is there a classification of objects like the Mathieu groupoid, where it may show up as an exceptional example?

To elaborate, the Mathieu groupoid is not only a groupoid, but it is equipped with a distinguished representation on a finite state machine. The set of states is the set of reachable labelings of vertices, and the transition operations are described by the generators of the groupoid. In other words, you may be seeking not a classification of finite groupoids (which, as Todd Trimble mentioned, is equivalent to a classification of finite groups), but a classification of reversible finite state machines.

That said, it seems unlikely that anyone has made a concerted attempt, even on the level of Hölder's 2-step program, due to a lack of structure.

share|improve this answer
add comment

@Fernando: @Todd: I'd just like to add to Todd's remark on the classification of groupoids up to isomorphism. It was early realised that any groupoids is the disjoint union of its connected components; and that given any $cx \in Ob(G)$ for a connected groupoid $G$ ithen $G$ s isomorphic to $G(x) * T$ where $G(x)$ is the vertex, or object group, at $x$ and $T$ is a "tree groupoid", i.e. $T(y,z)$ is a singleton for all $y,z \in Ob(G)$. However this determination depends on first choosing the object $x$ and then for each $ y \ne x$ in $Ob(G)$, choosing an element in $G(x,y)$. So there are lots of choices. As Fernando remarks, a single connected groupoid is up to homotopy "the same as" a group.

However the relation of groupoids to other areas of mathematics is interesting.

diagram

Now what the objects of a groupoid add to a group is a kind of "spatial" character. This allows all sorts of new possible interactions beteeen different grouopids, quite unlike those of groups. This is especially relevant to van Kampen type situations. Further, the choices involved in the above determination imply that the classification of diagrams of groupoids does not reduce to the classification of diagrams of groups.

Further, morphisms of groupoids have much more variety than do those for groups: for groupoids we have equivalences, fibrations, covering morphisms (related to actions on sets), quotient morphisms (factor by a normal subgroupoid), universal morphisms (identify objects in some way), orbit morphisms, .... So it is often in the relations between groupoids rather than the classification of single groupoids that we should see the benefit of their use. This reflects the categorical viewpoint.

share|improve this answer
    
I absolutely agree; +1. Related is the fact that equivalences between groupoids (not only isomorphisms) are important, and this would not be apparent if one sought to "reduce" groupoids to groups. (My answer should not be taken in such a reductionist spirit; it was only saying that finite groupoids are at least as hard to classify as finite groups, and that's very hard indeed.) –  Todd Trimble Feb 27 '13 at 15:50
    
@Ronnie, what about inverse semigroups? This is almost the same subject as etale groupoids. –  Benjamin Steinberg Feb 27 '13 at 17:08
    
@Benjamion: the relation with inverse semigroups, in terms of ordered groupoids, pioneered by c. Ehresmann and developed by Mark Lawson et al, is fascinating, very related to local-global ideas which are a background to why Ehresmann's approach to category theory is so different from that of other schools. Are $C^*$-algebras relevant, as they are for groupoids? Other ideas: atlases of inverse semigoups (see Bak/Brown/Minian/Porter JHRS 1 (2006) 101-167))? atlases of $C^*$ algebras? –  Ronnie Brown Feb 28 '13 at 11:13
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.