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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
  • $\begingroup$ Can you spell the question out completely? You are looking for $8$ groups $H_I$ one for each $I⊆\{1,2,3\}$, such that $H_J≤H_I$ whenever $J⊆I$, and with $G:=H_{\{1,2,3\}}$ simple? Or are there extra conditions like maybe $H_I∩H_J=H_{I∩J}$ and/or $\langle H_I∪H_J\rangle =H_{I∪J}$? Or that every subgroup between $H:=H_\varnothing$ and $G$ is one of the $H_I$? $\endgroup$ – Gro-Tsen Aug 16 at 20:42
  • $\begingroup$ @Gro-Tsen: we are looking for the classification of all the intervals $[H,G]$ in the subgroup lattice $\mathcal{L}(G)$ for all finite simple group $G$ such that $[H,G]$ is lattice-equivalent to the subset lattice of $\{1,2,3\}$ (i.e. the Boolean lattice of rank $3$). More generally, we are interested in such a classification for all rank $\ge 3$. Is it clearer? $\endgroup$ – Sebastien Palcoux Aug 16 at 21:22
  • $\begingroup$ So I think this means “yes” to all the extra conditions I listed. The way the question is written, it didn't seem clear (without the additional clarification you gave) that you are demanding lattice-equivalence with the lattice structure induced from the subgroup lattice and not just order-equivalence (or perhaps lattice-equivalence with a lattice structure that would come from the order but not the lattice of all subgroups). $\endgroup$ – Gro-Tsen Aug 17 at 11:13
  • $\begingroup$ @Gro-Tsen: yes! $\endgroup$ – Sebastien Palcoux Aug 17 at 11:14
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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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  • $\begingroup$ There seems to be a mistake in your answer: the stabilizer $G_1$ of $\pi_1=\{\{1,2\},\{3,4\},\{5,6\},\{7,8\}\}$ in $A_8$ is NOT a maximal subgroup, according to the following GAP computation: gap> G:=AlternatingGroup(8);; P1:=[[1,2],[3,4],[5,6],[7,8]];; G1:=PartitionStabilizerPermGroup(G,P1);; IntermediateSubgroups(G,G1).inclusions; [ [ 0, 1 ], [ 0, 2 ], [ 1, 3 ], [ 2, 3 ] ] Note that we still get an interesting example. I posted an answer containing some other computation following your answer. $\endgroup$ – Sebastien Palcoux Aug 14 at 14:50
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    $\begingroup$ Hi Sebastien, the case you found is an anomaly, see Table 1 in "A classification of the maximal subgroups of the finite alternating and symmetric groups", by Martin Liebeck, Cheryl Praeger and Jan Saxl. There are no other such examples. $\endgroup$ – John Shareshian Aug 14 at 15:29
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Some computations following the answer of John Shareshian, pointing out one interesting anomaly:

For $G=A_8$ and the partitions $\pi_1=\{\{1,2\},\{3,4\},\{5,6\},\{7,8\}\}$, $\pi_2=\{\{1,2,3,4\},\{5,6,7,8\}\}$.

gap> G:=AlternatingGroup(8);
Alt( [ 1 .. 8 ] )
gap> P1:=[[1,2],[3,4],[5,6],[7,8]];
[ [ 1, 2 ], [ 3, 4 ], [ 5, 6 ], [ 7, 8 ] ]
gap> P2:=[[1,2,3,4],[5,6,7,8]];
[ [ 1, 2, 3, 4 ], [ 5, 6, 7, 8 ] ]
gap> G1:=Stabilizer(G,P1,OnSetsDisjointSets);
Group([ (5,6)(7,8), (3,4)(7,8), (1,2)(7,8), (3,4)(5,6), (3,6,4,5)(7,8), (3,5,8)(4,6,7), (1,3,5,8)(2,4,6,7) ])
gap> G2:=Stabilizer(G,P2,OnSetsDisjointSets);
Group([ (5,8,7), (5,8)(6,7), (5,7)(6,8), (3,4)(7,8), (2,3,4), (1,4)(2,3), (1,3)(2,4), (1,5,4,7,3,6)(2,8) ])

Note that $G_1$ is NOT a maximal subgroup of $G$ (whereas $G_2$ is):

gap> IntermediateSubgroups(G,G1);
rec( inclusions := [ [ 0, 1 ], [ 0, 2 ], [ 1, 3 ], [ 2, 3 ] ], subgroups := [ Group([ (1,8)(2,7), (1,8,3)(2,7,4), (1,5,2,6)(3,8,4,7), (1,4,2,3)(5,7,6,8), (1,2)(7,8), (3,4)(7,8), (5,6)(7,8), (2,3)(6,7) ]), Group([ (1,8)(2,7), (1,8,3)(2,7,4), (1,5,2,6)(3,8,4,7), (1,4,2,3)(5,7,6,8), (1,2)(7,8), (3,4)(7,8), (5,6)(7,8), (2,3)(6,8) ]) ] )
gap> IntermediateSubgroups(G,G2);
rec( inclusions := [ [ 0, 1 ] ], subgroups := [  ] )

Finally, $[H,G]$ is Boolean of rank $3$ (and not $2$ as expected) (it is exactly the example given in the question):

gap> H:=Intersection(G1,G2);
Group([ (5,8)(6,7), (5,7)(6,8), (3,4)(5,7,6,8), (1,2)(5,7,6,8), (1,3)(2,4)(5,8)(6,7), (1,5,4,8)(2,6,3,7) ])
 gap> IntermediateSubgroups(G,H);
rec( inclusions := [ [ 0, 1 ], [ 0, 2 ], [ 0, 3 ], [ 1, 4 ], [ 1, 5 ], [ 2, 4 ], [ 2, 6 ], [ 3, 5 ], [ 3, 6 ], [ 4, 7 ], [ 5, 7 ], [ 6, 7 ] ],
  subgroups := [ Group([ (1,5)(2,6)(3,7)(4,8), (3,4)(7,8), (2,3,4)(6,7,8), (1,4)(2,3), (5,8)(6,7), (5,7)(6,8), (1,3)(2,4) ]), Group([ (1,5)(2,6)(3,7)(4,8), (3,4)(7,8), (2,4,3)(6,7,8), (1,4)(2,3), (5,8)(6,7), (5,7)(6,8), (1,3)(2,4) ]), Group([ (3,5,8)(4,6,7), (3,4)(5,7,6,8), (1,3)(2,4)(5,7)(6,8) ]), Group([ (1,3)(2,4), (2,3,4), (5,7)(6,8), (1,2,3,4)(5,8,7,6), (1,2,3,4)(5,6), (1,5)(2,6)(3,7)(4,8) ]), Group([ (5,7)(6,8), (2,3,5)(4,7,6), (1,2)(3,4)(5,6)(7,8) ]), Group([ (5,7)(6,8), (1,3,5)(4,7,6), (1,2)(3,4)(5,6)(7,8) ]) ] )

Note that this anomaly does not appear for $G=S_8$:

gap> G:=SymmetricGroup(8);
Sym( [ 1 .. 8 ] )
gap> G1:=Stabilizer(G,P1,OnSetsDisjointSets);
Group([ (7,8), (5,6), (3,4), (1,2), (3,5,4,6), (3,5,8)(4,6,7), (1,3,5,8)(2,4,6,7) ])
gap> G2:=Stabilizer(G,P2,OnSetsDisjointSets);
Group([ (7,8), (5,8,7), (5,8)(6,7), (5,7)(6,8), (3,4), (2,3,4), (1,4)(2,3), (1,3)(2,4), (1,5,4,7,3,6)(2,8) ])
gap> H:=Intersection(G1,G2);
Group([ (5,6), (5,7,6,8), (3,4)(5,7,6,8), (1,2)(5,7,6,8), (1,3)(2,4)(5,7,6,8), (1,5,3,7,2,6,4,8) ])
gap> IntermediateSubgroups(G,H);
rec( inclusions := [ [ 0, 1 ], [ 0, 2 ], [ 1, 3 ], [ 2, 3 ] ],
  subgroups := [ Group([ (1,2), (3,4), (5,6), (7,8), (1,3,5,7)(2,4,6,8), (1,3)(2,4) ]), Group([ (1,2,3,4), (1,2), (5,6,7,8), (5,6), (1,5)(2,6)(3,7)(4,8) ]) ] )

Now, for $G=A_{16}$, and the partitions $\pi_1=\{\{1,2\},\{3,4\},\{5,6\},\{7,8\},\{9,10\},\{11,12\},\{13,14\},\{15,16\}\}$, $\pi_2=\{\{1,2,3,4\},\{5,6,7,8\},\{9,10,11,12\},\{13,14,15,16\}\}$, $\pi_3=\{\{1,2,3,4,5,6,7,8\},\{9,10,11,12,13,14,15,16\}\}$

gap> G:=AlternatingGroup(16);
Alt( [ 1 .. 16 ] )
gap> P1:=[[1,2],[3,4],[5,6],[7,8],[9,10],[11,12],[13,14],[15,16]];
[ [ 1, 2 ], [ 3, 4 ], [ 5, 6 ], [ 7, 8 ], [ 9, 10 ], [ 11, 12 ], [ 13, 14 ], [ 15, 16 ] ]
gap> P2:=[[1,2,3,4],[5,6,7,8],[9,10,11,12],[13,14,15,16]];
[ [ 1, 2, 3, 4 ], [ 5, 6, 7, 8 ], [ 9, 10, 11, 12 ], [ 13, 14, 15, 16 ] ]
gap> P3:=[[1,2,3,4,5,6,7,8],[9,10,11,12,13,14,15,16]];
[ [ 1, 2, 3, 4, 5, 6, 7, 8 ], [ 9, 10, 11, 12, 13, 14, 15, 16 ] ]
gap> G1:=Stabilizer(G,P1,OnSetsDisjointSets);
<permutation group of size 5160960 with 17 generators>
gap> G2:=Stabilizer(G,P2,OnSetsDisjointSets);
<permutation group of size 3981312 with 19 generators>
gap> G3:=Stabilizer(G,P3,OnSetsDisjointSets);
<permutation group of size 1625702400 with 18 generators>
gap> H:=Intersection(Intersection(G1,G2),G3);
<permutation group with 12 generators>
gap> IntermediateSubgroups(G,H);
rec( inclusions := [ [ 0, 1 ], [ 0, 2 ], [ 0, 3 ], [ 1, 4 ], [ 1, 5 ], [ 2, 5 ], [ 2, 6 ], [ 3, 4 ], [ 3, 6 ],[ 4, 7 ], [ 5, 7 ], [ 6, 7 ] ], subgroups := [ Group([ (1,15,4,14)(2,16,3,13), (1,10,8,14)(2,9,7,13)(3,12,6,16)(4,11,5,15), (7,8)(13,14) ]), Group([ (1,9,8,16,6,13)(2,10,7,15,5,14)(3,12)(4,11), (3,7)(4,8), (7,8)(11,12) ]),
      <permutation group of size 1327104 with 18 generators>, <permutation group of size 3981312 with 13 generators>,
      <permutation group of size 5160960 with 9 generators>, <permutation group of size 1625702400 with 6 generators>
     ] )

There is no anomaly, we get a Boolean lattice of rank $3$. Idem for $G=S_{16}$:

gap> G:=SymmetricGroup(16);
Sym( [ 1 .. 16 ] )
gap> G1:=Stabilizer(G,P1,OnSetsDisjointSets);
<permutation group of size 10321920 with 15 generators>
gap> G2:=Stabilizer(G,P2,OnSetsDisjointSets);
<permutation group of size 7962624 with 17 generators>
gap> G3:=Stabilizer(G,P3,OnSetsDisjointSets);
<permutation group of size 3251404800 with 17 generators>
gap> H:=Intersection(Intersection(G1,G2),G3);
<permutation group with 12 generators>
gap> IntermediateSubgroups(G,H);
rec( inclusions := [ [ 0, 1 ], [ 0, 2 ], [ 0, 3 ], [ 1, 4 ], [ 1, 5 ], [ 2, 5 ], [ 2, 6 ], [ 3, 4 ], [ 3, 6 ], [ 4, 7 ], [ 5, 7 ], [ 6, 7 ] ],
  subgroups := [ Group([ (1,13,3,15)(2,14,4,16), (1,13,11,5)(2,14,12,6)(3,15,9,7)(4,16,10,8), (13,14) ]), Group([ (1,13,5,11,3,15)(2,14,6,12,4,16)(7,9)(8,10), (3,7)(4,8), (15,16) ]),
      <permutation group of size 2654208 with 19 generators>, <permutation group of size 7962624 with 10 generators>, <permutation group of size 10321920 with 10 generators>,
      Group([ (1,2,3,4,5,6,7,8), (1,2), (9,10,11,12,13,14,15,16), (9,10), (1,9)(2,10)(3,11)(4,12)(5,13)(6,14)(7,15)(8,16) ]) ] ) 

I tried to go further with $G=A_{32}$ (or $S_{32}$):

gap> G:=AlternatingGroup(32); #or SymmetricGroup(32);
Alt( [ 1 .. 32 ] )
gap> P1:=[[1,2],[3,4],[5,6],[7,8],[9,10],[11,12],[13,14],[15,16],[17,18],[19,20],[21,22],[23,24],[25,26],[27,28],[29,30],[31,32]];
[ [ 1, 2 ], [ 3, 4 ], [ 5, 6 ], [ 7, 8 ], [ 9, 10 ], [ 11, 12 ], [ 13, 14 ], [ 15, 16 ], [ 17, 18 ], [ 19, 20 ], [ 21, 22 ], [ 23, 24 ], [ 25, 26 ], [ 27, 28 ], [ 29, 30 ], [ 31, 32 ] ]
gap> P2:=[[1,2,3,4],[5,6,7,8],[9,10,11,12],[13,14,15,16],[17,18,19,20],[21,22,23,24],[25,26,27,28],[29,30,31,32]];
[ [ 1, 2, 3, 4 ], [ 5, 6, 7, 8 ], [ 9, 10, 11, 12 ], [ 13, 14, 15, 16 ], [ 17, 18, 19, 20 ], [ 21, 22, 23, 24 ], [ 25, 26, 27, 28 ], [ 29, 30, 31, 32 ] ]
gap> P3:=[[1,2,3,4,5,6,7,8],[9,10,11,12,13,14,15,16],[17,18,19,20,21,22,23,24],[25,26,27,28,29,30,31,32]];
[ [ 1, 2, 3, 4, 5, 6, 7, 8 ], [ 9, 10, 11, 12, 13, 14, 15, 16 ], [ 17, 18, 19, 20, 21, 22, 23, 24 ], [ 25, 26, 27, 28, 29, 30, 31, 32 ] ]
gap> P4:=[[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16],[17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32]];
[ [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16 ], [ 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32 ] ]
gap> G1:=Stabilizer(G,P1,OnSetsDisjointSets);
<permutation group of size 685597979049984000 with 37 generators>
gap> G2:=Stabilizer(G,P2,OnSetsDisjointSets);
<permutation group of size 2219118333788160 with 37 generators>
gap> G3:=Stabilizer(G,P3,OnSetsDisjointSets);
<permutation group of size 31714899520389120000 with 35 generators>
gap> G4:=Stabilizer(G,P4,OnSetsDisjointSets);
<permutation group of size 437763136697395052544000000 with 34 generators>
gap> H:=Intersection(Intersection(Intersection(G1,G2),G3),G4);
<permutation group with 22 generators>

but on my laptop, GAP shuts down on the following computation after 2min30s...

gap> IntermediateSubgroups(G,H);

If someone can make this computation, I would be very interested in the result.

Remark: For $G=A_{2^n}$ (resp. $S_{2^n}$) and the equipartitions and $H$ as above, we observe that $|H| = 2^{2^n-2}$ (resp. $2^{2^n-1}$) for $n=2,3,4,5$. Is it true in general? In addition, for $n=2,3$, we observe that $H$ is the unique subgroup of such order. Is it true in general?

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    $\begingroup$ With respect to the remark: Let's work in $S_n$. It seems that $H$ is a Sylow $2$-subgroup. To prove this, first use the fact that a Sylow $2$-subgroup $H \leq S_{2^n}$ is an $n$-fold iterated wreath product of ${\mathbf Z}_2$ with itself to show by induction that this subgroup stabilizes an equipartition of each possible type and that these equipartitions have the desired refinement property. Then observe that a $2^n$-cycle stabilizes exactly one equipartition of each type. Then show by induction that no element of odd prime order stabilizes all equipartition stabilized by $H$...... $\endgroup$ – John Shareshian Aug 17 at 3:34
  • $\begingroup$ It is quite reasonable to hope that, with more work than was required in the comment above, one can then show that $[H,G]$ is a Boolean algebra of rank $n-1$, thus rendering further computation unnecessary. $\endgroup$ – John Shareshian Aug 17 at 3:37

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