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.)

  • $\begingroup$ 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. $\endgroup$ – Todd Trimble Oct 6 '12 at 5:51
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    $\begingroup$ Any groupoid is a disjoint union of groups, so classifying finitie groups or finite groupoids are the same problem. $\endgroup$ – Fernando Muro Oct 6 '12 at 9:56
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    $\begingroup$ @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. $\endgroup$ – Ronnie Brown Oct 6 '12 at 13:52
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    $\begingroup$ @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. $\endgroup$ – Fernando Muro Oct 6 '12 at 19:00
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    $\begingroup$ Indeed, it is more correct to say that groups are equivalent to pointed connected groupoids. See the appendix in Baez and Shulman's article: arxiv.org/abs/math/0608420, p. 50 and following. $\endgroup$ – Todd Trimble Feb 27 '13 at 18:19

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 Besche, Eick and O'Brien:

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!

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    $\begingroup$ 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. $\endgroup$ – Nick Gill Nov 27 '12 at 16:13
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    $\begingroup$ Wow, thanks for the link and the factoid, Ben! $\endgroup$ – Todd Trimble Nov 27 '12 at 16:22
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    $\begingroup$ I was so struck by this calculation that I had to spread the (old) news: golem.ph.utexas.edu/category/2012/11/… $\endgroup$ – Tom Leinster Nov 28 '12 at 21:09
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    $\begingroup$ I absolutely don't understand how this address the question: this answer doesn't even mention the word "groupoid". -1 $\endgroup$ – Joël Feb 27 '13 at 15:33
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    $\begingroup$ @Joel, this sounds like a good reason to downvote the question (which it seems you did) but not the answer. This answer explains why classifying even finite groups is hopeless in a very precise way and so implicitly answers the question. $\endgroup$ – Benjamin Steinberg Feb 27 '13 at 17:42

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.

  • $\begingroup$ "finite subgroups" is a funny typo. I just fixed. $\endgroup$ – temp Oct 6 '12 at 5:27
  • $\begingroup$ 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... $\endgroup$ – temp Oct 6 '12 at 5:32
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    $\begingroup$ 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. $\endgroup$ – Ronnie Brown Oct 6 '12 at 11:32
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    $\begingroup$ That's a very pertinent remark, Ronnie. $\endgroup$ – Todd Trimble Oct 6 '12 at 13:01

Everything that's been written so far about the classification of finite groupoids reducing to the classification of finite groups is true but, I think, misleading. In order to actually produce a list of finite groups from a finite groupoid $X$ you need to choose a basepoint in each connected component of $X$. For example, to produce $M_{12}$ from $M_{13}$ you need to choose one of $13$ points.

There are various reasons it's undesirable to make such choices, with maybe the most practical one being that they are often unavailable once you introduce extra structure. For example, given a group $G$ you might want to study groupoids equipped with a $G$-action. These are strictly more interesting than disjoint unions of groups equipped with a $G$-action, and the reason is precisely that a $G$-groupoid need not have any $G$-invariant basepoints.

This is in fact an issue in the present example:

The Mathieu groupoid $M_{13}$ is equipped with an action of $\text{SL}_3(\mathbb{F}_3)$ which does not fix any points. When you pick a basepoint and get $M_{12}$ back you lose this action.

Another example of this phenomenon is the following: the configuration space $\text{Conf}_k(\mathbb{R}^2)$ of $k$ ordered points in the plane is naturally an Eilenberg-MacLane space $K(P_k, 1)$, where $P_k$ is the pure braid group. On the other hand, $\text{Conf}_k(\mathbb{R}^2)$ clearly has an action of $S_k$ on it given by permuting points. This action does not fix any basepoints, and so it does not induce an action on $P_k$.

It does induce an action on a groupoid equivalent to $P_k$ with $k!$ points given by taking the fundamental groupoid of $\text{Conf}_k(\mathbb{R}^2)$ at a set of basepoints which are setwise fixed under the action of $S_k$. A more complicated choice of basepoints setwise fixed under the action of $S_k$ can be used to give a model of the little disks operad as an operad in groupoids. This operad cannot be simplified to an operad in groups, despite the fact that all of the groupoids involved are connected, precisely because of basepoint issues (the basepoints must be chosen not only compatibly with the $S_k$-actions but with operadic composition).

Edit: One way to measure the interestingness of the $S_k$ action on $\text{Conf}_k(\mathbb{R}^2)$ is that it does not even fix a basepoint in the homotopy coherent sense: equivalently, there is a corresponding short exact sequence

$$1 \to P_k \to B_k \to S_k \to 1$$

that does not split. On the other hand, as pointed out in the comments, any extension of $\text{SL}_3(\mathbb{F}_3)$ by $M_{12}$ must be trivial.

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    $\begingroup$ Extensions of $G$ by a nonabelian $M$ are classified by cohomology of $G$ with coefficients in what the ancients called a crossed module; i.e. by homotopy classes of maps from $BG$ to the delooping of $\mathrm{Aut}(BM)$. Since $M_{12}$ has no center, $\mathrm{Aut}(BM_{12}) \cong \mathrm{Out}(BM_{12}) \cong \mathbf{Z}/2$. That means there can be no nontrivial extensions of a group like $\mathrm{SL}_3(\mathbf{F}_3)$ or $\mathrm{PGL}_3(\mathbf{F}_3)$ by $M_{12}$, right? $\endgroup$ – David Treumann Mar 29 '15 at 16:07
  • $\begingroup$ One can prove by elementary methods that if $N$ is a simple normal subgroup of the finite group $G$ with $G/N$ simple and some prime $p$ divides $|G/N|$ but does not divide $|Aut(N)|$, then $G$ is isomorphic with $N \times G/N$. In the case at hand, $p=13$ works. $\endgroup$ – John Shareshian Mar 29 '15 at 16:33
  • $\begingroup$ @David: hmm, you're right. My bad. $\endgroup$ – Qiaochu Yuan Mar 29 '15 at 17:44
  • $\begingroup$ @Qiaochu Yuan: Groupoids equipped with a $G$-action where $G$ is a group occur in the study of orbit spaces, where the groupoid is the fundamental groupoid $\pi_1(X)$. See Chapter 11 of Topology and Groupoids, where the "orbit groupoid" is related to the fundamental groupoid of the orbit space! $\endgroup$ – Ronnie Brown Mar 30 '15 at 14:04

@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 (source)

Now what the objects of a groupoid add to a group is a kind of "spatial" character. This allows between different groupoids all sorts of new possible interactions, 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.

March 27, 2015: I discussed this matter in the 1980s with Alex Heller and he remarked:"We have long passed the days when the classification of objects up to isomorphism was the sole object of mathematics. Thus the classification of vector spaces is trivial; the classification of vector spaces with one morphism is interesting, and is the rational canonical form; the classification of vector spaces with two endomorphisms is difficult; and with three endomorphisms is unknown."

Groupoids internal to a given category are of much interest, partly because groupoids generalise equivalence relations, and so the idea of quotienting.

A discussion on mathoverflow on many base points seems relevant.

  • $\begingroup$ 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.) $\endgroup$ – Todd Trimble Feb 27 '13 at 15:50
  • $\begingroup$ @Ronnie, what about inverse semigroups? This is almost the same subject as etale groupoids. $\endgroup$ – Benjamin Steinberg Feb 27 '13 at 17:08
  • $\begingroup$ @BenjaminSteinberg: 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? $\endgroup$ – Ronnie Brown Feb 28 '13 at 11:13

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.


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