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$M_{13}$ 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 $M_{13}$?

share|improve this question
    
Another possible characterization in a slightly different direction: $M_{13}$ is the only 6-transitive groupoid whose non-trivial elements have support at least $\frac13(n-1)$. –  Nick Gill Jul 16 at 20:32
3  
What precisely are you actually asking? -- To me your question seems pretty vague. –  Stefan Kohl Jul 16 at 20:33
    
@StefanKohl, roughly speaking, I would like to know if $M_{13}$ can be characterized as a subset of a permutation group in some basic way. Alternatively, I would like some context and motivation explaining (from a group theoretic point of view) the construction of $M_{13}$. Hmmm, perhaps that's still too vague... –  Nick Gill Jul 16 at 20:40
3  
Since 3 people have upvoted Stefan Kohl's comment let me put the question another way, more bluntly: $M_{13}$ 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 $M_{13}$? –  Nick Gill Jul 16 at 21:35
    
The name of Scott C. was correct, now it is wrong. I do not edit as you seem in the process of doing so. –  quid Jul 16 at 22:18

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