What are the rank 3 boolean intervals [H,G], with G simple group?

Question

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.

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.

We have used "SimpleGroupsIterator" and this code of A. Hulpke (he has found $A_8$ and $PSp(6,2)$ himself). — Sebastien Palcoux, Commented Jul 31, 2016 at 16:33
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 $H_I$? — Gro-Tsen, Commented Aug 16, 2019 at 20:42
@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? — Sebastien Palcoux, Commented Aug 16, 2019 at 21:22
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). — Gro-Tsen, Commented Aug 17, 2019 at 11:13

John Shareshian · Accepted Answer · 2018-05-22 03:37:41Z

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.

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. — Sebastien Palcoux, Commented Aug 14, 2019 at 14:50
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. — John Shareshian, Commented Aug 14, 2019 at 15:29

ahulpke · Accepted Answer · 2021-08-28 22:55:38Z

Since it is a bit too long for a comment: In the development version of GAP, my laptop computed the result for $A_{32}$ in about 20 minutes and 1.5GB:

rec( inclusions := [ [ 0, 1 ], [ 0, 2 ], [ 0, 3 ], [ 0, 6 ], [ 1, 4 ],
      [ 1, 5 ], [ 1, 7 ], [ 2, 5 ], [ 2, 8 ], [ 2, 9 ], [ 3, 4 ], [ 3, 9 ],
      [ 3, 12 ], [ 4, 11 ], [ 4, 13 ], [ 5, 10 ], [ 5, 11 ], [ 6, 7 ],
      [ 6, 8 ], [ 6, 12 ], [ 7, 10 ], [ 7, 13 ], [ 8, 10 ], [ 8, 14 ],
      [ 9, 11 ], [ 9, 14 ], [ 10, 15 ], [ 11, 15 ], [ 12, 13 ], [ 12, 14 ],
      [ 13, 15 ], [ 14, 15 ] ],
  subgroups := [ <permutation group of size 3221225472 with 8 generators>,
      <permutation group of size 9663676416 with 8 generators>,
      <permutation group of size 86973087744 with 34 generators>,
      <permutation group of size 260919263232 with 12 generators>,
      <permutation group of size 338228674560 with 11 generators>,
      <permutation group of size 7044820107264 with 38 generators>,
      <permutation group of size 21134460321792 with 10 generators>,
      <permutation group of size 63403380965376 with 10 generators>,
      <permutation group of size 106542032486400 with 7 generators>,
      <permutation group of size 2219118333788160 with 5 generators>,
      <permutation group of size 685597979049984000 with 3 generators>,
      <permutation group of size 10571633173463040000 with 7 generators>,
      <permutation group of size 31714899520389120000 with 13 generators>,
      <permutation group of size 437763136697395052544000000 with 6 generators\
> ] )

Thank you! See the classification in the following paper: doi.org/10.1017/fms.2020.49 — Sebastien Palcoux, Commented Aug 29, 2021 at 4:43

Daniel Sebald · Accepted Answer · 2023-02-04 06:09:14Z

1

Example I found by hand: $\mathrm{Co}_1$ with subgroups of order 2690072985600, 235146240, 138568320, 117573120, 11547360, 2099520, and 1049760

answered Feb 4, 2023 at 6:09

Daniel Sebald

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Sebastien Palcoux · Accepted Answer · 2019-08-16 19:48:40Z

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?

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 $H$...... — John Shareshian, Commented Aug 17, 2019 at 3:34
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. — John Shareshian, Commented Aug 17, 2019 at 3:37

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What are the rank 3 boolean intervals [H,G], with G simple group?

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What are the rank 3 boolean intervals [H,G], with G simple group?

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