$M_{13}$ is the **Mathieu groupoid** defined by Conway in

Conway, J. H. $M_{13}$. Surveys in combinatorics, 1997 (London), 1–11, London Math. Soc. Lecture Note Ser., 241, Cambridge Univ. Press, Cambridge, 1997.

$M_{13}$ has been studied in various places, most notably (for me) here:

Conway, John H.; Elkies, Noam D.; Martin, Jeremy L.

The Mathieu group $M_{12}$ and its pseudogroup extension $M_{13}$. Experiment. Math. 15 (2006), no. 2, 223–236.

The title contains my question, albeit stated a little flippantly. Let me start by stating that I don't want to suggest that $M_{13}$ **isn't** a big deal - in particular, I've spent a deal of time reading the paper by Conway, Elkies and Martin and there is some very cool stuff in it. So...

Descriptions of $M_{13}$ usually start with Conway's anecdote describing the parallel between the structure of $M_{12}$ and the simple group $SL_3(3)$ leading him to wonder if somehow one can extend $M_{12}$ to give a groupoid in a natural way. And this groupoid would be 6-transitive (in some sense of the word). Now $M_{13}$ does the trick.

**Transitivity**: My beef is that one can find any number of copies of $M_{12}$ lying around in $S_{13}$ and, if you take an appropriate set of 13 cosets of $M_{12}$, then you will end up with a groupoid that is (in some sense) 6-transitive. In particular, if you take the product of $M_{12}$ with any transitive subgroup of $S_{13}$ whose stabilizer lies in $M_{12}$, then you will get such a groupoid. Now $M_{13}$ is just the product of $M_{12}$ with (a copy of) $SL_3(3)$, and so there you have it.

I could just as easily do the same with $M_{24}$ and any regular subgroup of $S_{25}$.

**The Game**: Of course $M_{13}$ is defined in a particularly elegant way - via a "game" played on the projective plane of order 3. For me, though, the really amazing thing about this set-up is that the set of *closed moves* gives the group $M_{12}$. The set $M_{13}$ is just a byproduct of this (striking and very intriguing) fact. There are other natural extensions of $M_{12}$ using this game (Some coauthors and I define one such here.)

Having said this, Scott Carnahan answered this MO question by stating that this game gives "a representation of $M_{13}$ on a finite state machine" and this is what makes $M_{13}$ remarkable. In some sense Scott's answer explains why $M_{13}$ is a big deal, but it's not quite in the direction I'm looking for...

**Back to transitivity**: What I would like is a *characterization* of $M_{13}$ via some group-theoretic properties. For instance as a subset of a permutation group, via the notion of transitivity: So, for instance, we can characterize $M_{12}$ and $M_{24}$ as the only subgroups of $S_n$ that don't contain $A_n$ and that are 5-transitive. Here's an analogous statement that (if true) would be the sort of thing I'm looking for.

A characterization of $M_{13}$?... $M_{13}$ is the only subset of $S_n$ that is not almost all of $S_n$ and for which at least $\frac{1}{n}$ of the 6-tuples of distinct elements are universal donors, and at least $\frac{1}{n}$ of the 6-tuples of distinct elements are universal recipients.

The notions of *universal donors and recipients* are described in Conway, Elkies, Martin above and pertain to transitivity for groupoids. The notion of *almost all* is hopefully obvious. I'm deliberately leaving the statement rough because it's just a vague musing.

(In particular I'm not even sure that at least $\frac{1}{n}$ of the 6-tuples of distinct elements *really are* universal recipients for $M_{13}$ - but it's certainly some large fraction. Moreover the statement can't be right because supersets of $M_{13}$ are also included... But I hope it gives a rough idea.)

There might be other ways of approaching this, and that's what I'm looking for with this question. To me the key thing should be that the characterization of $M_{13}$ should not just be an immediate consequence of the fact that it contains a copy of $M_{12}$ which is itself a remarkable group. I don't *think* the statement I suggest falls foul of this...

**Added later**: People are suggesting that this question is too vague, so let me sum up as follows:

$M_{13}$ is a union of cosets of a permutation group. There are many unions of cosets of permutation groups but we don't usually bother studying them. Why do people bother with $M_{13}$?