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The rank $n$ boolean lattice $B_{n}$ is the subset lattice of $\{1,2, \dots , n\}$.
The lattice $B_{3}$ is the following:

enter image description here

Question: What are the rank $3$ boolean intervals of the form $[H,G]$, with $G$ a simple group?

Remark: For $\vert G \vert \leq 4000000$, we have found (by GAP):

  • $A_8$ (of order $20160$) with a subgroup of index $315$,
  • $PSU(3,5)$ (of order $126000$) with a subgroup of index $6000$,
  • $PSp(6,2)$ (of order $1451520$) with a subgroup of index $2835$,
  • $PSU(4,3)$ (of order $3265920$) with a subgroup of index $25515$.

Can we have a classification in general?

There is a large class of examples given by the BN-pairs, as pointed out in Example 4.21 of this paper:

Let $G$ be a finite group with a BN-pair, $H$ be the corresponding Borel subgroup and $(W, S)$ be the associated Coxeter system. Let $n :=|S|$ be the rank of the BN-pair. Then the interval $[H, G]$ is Boolean of rank $n$. Any finite simple group $G$ of Lie type (over a finite field of characteristic $p$) admits a BN-pair (except Tits group). If moreover, $G$ is a Chevalley group, then $n$ is the number of vertices in its Dynkin diagram.

The above interval with $G = A_8 $ or $ PSp(6,2)$ comes from a BN-pair (where $G$ is the Chevalley group $A_3(2)$ or $C_3(2)$), whereas the one with $G = PSU(3,5) $ or $ PSU(4,3)$ does not.

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  • $\begingroup$ We have used "SimpleGroupsIterator" and this code of A. Hulpke (he has found $A_8$ and $PSp(6,2)$ himself). $\endgroup$ – Sebastien Palcoux Jul 31 '16 at 16:33
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Here is another infinite class of examples: Say $q>3$ is a prime power and $a$ is a square-free odd integer. Then $L_2(q)$ embeds in $L_2(q^a)$, and the lattice of subgroups above the image of such an embedding is a Boolean algebra whose height is the number of prime divisors of $a$. There might well be lots of infinite classes of examples of a similar nature. It is known that every overgroup of a subfield subgroup in a Lie type simple group is another subfield subgroup. To get Boolean algebras, one should choose the field extension so that each subfield subgroup in question is self-normalizing (and the square-free property holds).

Although I have not thought this through, here is a different possible infinite class of examples: Fix $n$ and let $\pi_1,\pi_2,\ldots,\pi_k$ be nontrivial equipartitions of $\{1,\ldots,n\}$ such that each $\pi_i$ ($i<k$) refines $\pi_{i+1}$. Let $G_i$ be the stabilizer of $\pi_i$ in $A_n$ (so each $G_i$ is an imprimitive maximal subgroup). Let $H=\bigcap_{i=1}^k G_i$. It might well be the case that $[H,A_n]$ is a Boolean algebra of height $k$. This is true when $k=2$, as shown in a paper of Aschbacher and myself.

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  • $\begingroup$ Does your above definition of $H$ allows us to deduce the index $|A_n:H|$? $\endgroup$ – Sebastien Palcoux Dec 23 '18 at 11:09

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