What are the rank 3 boolean intervals [H,G], with G simple group? The rank $n$ boolean lattice $B_{n}$ is the subset lattice of $\{1,2, \dots , n\}$.
The lattice  $B_{3}$ is the following:

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.

Edit (29/08/2021): See the recent paper Boolean lattices in finite alternating and symmetric groups by Andrea Lucchini, Mariapia Moscatiello, Pablo Spiga and myself.
 A: 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.
A: Example I found by hand: $\mathrm{Co}_1$ with subgroups of order 2690072985600, 235146240, 138568320, 117573120, 11547360, 2099520, and 1049760
