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Background: See Noah and Emily's posts on subfactors and planar algebras on the Secret Blogging Seminar.

There are plenty of examples of $3$-super-transitive (3-ST) subfactors; Haagerup, $S_4 < S_5$, and others. There's exactly one known example of a $5$-ST subfactor, the Haagerup-Asaeda subfactor, and one $7$-ST subfactor, the extended Haagerup subfactor.

Below index $4$ there are the $A_n$ and $D_n$ families, which are arbitrarily super-transitive. Ignore those; I'm just interested above index $4$.

Is there anything that's even more super-transitive?

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(Aside: I'm not asking this in the expectation that anyone will have an immediate answer --- if you have one, you should answer here and then hurry off and submit to a journal!) I'm mostly asking here to record a bet with Emily Peters: if such exists, I owe her a bottle of champagne. If you can prove that none exist, she owes me a bottle of champagne. I win by default on my 60th birthday. –  Scott Morrison Oct 7 '09 at 20:39
    
Are the group-subgroup subfactors known to be less than $7$-supertransitive? Else by exploring the maximal inclusions of groups $(H \subset G)$ (and so the primitive permutation groups, with GAP), perhaps we can find such a subfactor more than $7$-supertransitive. –  Sébastien Palcoux Mar 2 at 18:25
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A subgroup subfactor can be at most 3-supertransitive. The 4-box space is at least 15 dimensional. This can be seen by looking at the partition planar algebra (MR2972458). –  Dave Penneys Mar 2 at 22:47
    
I see, thanks Dave. In analogy with the Mathieu groups, we could expected the finiteness of finite depth, index $>4$ and (at least) $4$-supertransitive subfactors. –  Sébastien Palcoux Mar 3 at 1:32
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@DavePenneys: thanks to the paper you cite (here p13-14) if I'm not mistaken, for $k=1,2$ or $3$, $(R^G \subset R^H)$ is $k$-supertransitive iff the action of $G$ on $X=G/H$ is $k$-transitive, and so iff $G/H_G$ is a $k$-transitive group (of course it's false for $k > 3$). So the $3$-$ST$ group-subgroup subfactors are given by the $3$-$T$ groups (classification?). Also, if $G/H_G$ is a $2$-$T$ group then $(R^G \subset R^H)$ is $2$-$ST$, so maximal, so is $(H \subset G)$ and then $G/H_G$ is a primitive group. –  Sébastien Palcoux Mar 3 at 18:15
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