# Given a lattice L with n elements, are there finite groups H < G such that L $\cong$ the lattice of subgroups between H and G?

If there is no restriction on $n$, this is a famous open problem. I'm wondering if any recent work has been done for small $n>6$. I believe the question is answered (positively) for $n=6$ by Watatani (1996) MR1409040 and Aschbacher (2008) MR2393428. I also believe we can answer it for $n=7$, with one possible exception. The exceptional case is shown below.

So my two questions are these:

1) Does anyone know of recent work on this special case of the problem (specifically for $n=7$ or $n=8$)?

2) Has anyone found a finite group $G$ with a subgroup $H$ such that the interval

$[H, G] = \{K : H \leq K \leq G \}$

is the lattice shown above?

If not, I'd be happy to hear if anyone has ideas about how we might prove that such a group exists (without necessarily producing the group). Or, if you can put restrictions on such a group, that might be useful. For example, I believe I can show that there is no finite solvable group with this property.

(By the way, I have searched many of the groups in GAP's small groups library, and this was helpful for finding some of the other difficult 7-element cases.)

Correction: We don't really have an answer to this question for $n=7$. Rather, we can show that every lattice with at most 7 elements (except possibly the one above) is the congruence lattice of a finite algebra. Pálfy and Pudlák MR593011 showed the two questions are equivalent when $n$ is not restricted, but their proof doesn't go through for fixed $n$. Nonetheless, we can prove that a (minimal) finite algebra with the lattice above as congruence lattice must be a transitive G-set (modulo constant operations), so the existence of such a finite algebra is equivalent to the existence of finite groups H < G with [H,G] isomorphic to the lattice above. (This is the reason I phrased my question in terms of subgroup lattices.)

Update: (2012/02/22) If the interval $[H,G]$ is the lattice above, with $H$ core-free, I believe I can prove that a minimal normal subgroup of $G$ must be non-abelian, which isn't much, but it's a start! In particular, I think it rules out the affine case of the Aschbacher-O'Nan-Scott Theorem.

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One simple observation that might help: you might as well assume that H has trivial core in G. For any normal subgroup N of G, HN is a modular element in the lattice [H,G]. Let M be the subgroup that is the only element of its connected component in (H,G). Since the set of lattice theoretic complements to M in [H,G] is not an antichain, M is not modular in [H,G]. Therefore, given the assumption above, M has trivial core in G and G is a primitive permutation group on M\G. Now you can attempt to use the Aschbacher-O'Nan-Scott Theorem. –  John Shareshian Jan 15 '12 at 22:22
@John: Thank you very much for this suggestion. I will think about how to apply this observation about M, and the Aschbacher-O'Nan-Scott Theorem, to get this lattice. Thanks again. –  William DeMeo Jan 16 '12 at 1:29
I checked all nonsolvable primitive permutation groups of degree at most 30 with ONanScottType=1 (affine) in GAP's library of primitive groups, and (barring bugs in my code) none has the lattice above as upper interval in its subgroup lattice. I wonder if there's an "obvious" theoretical reason we could at least rule out the affine groups. –  William DeMeo Feb 5 '12 at 9:55
@me Yes, I think we can rule out (subgroups of) affine groups (see update in question post). –  William DeMeo Feb 22 '12 at 23:24