TITLE: Can any finite distributive weighted lattice be realized by inclusion of groups?
QUESTION [5 upvotes]: By theorem 2.1 here, any finite distributive lattice  $\mathcal{L}$ can be realized as an intermediate subgroups lattice.  
A weighted lattice $(\mathcal{L},\tau)$ is a lattice $\mathcal{L}$ with a weight $\tau: \mathcal{L} \to  \mathbb{N}$ satisfying;
- $b \le b' \Rightarrow \frac{\tau(b')}{\tau(b)}\in \mathbb{N}$
- $\tau(l) = 1$ for $l$ the least element of $\mathcal{L}$.      
Let $(H \subset G)$ be an inclusion of finite groups, then it realizes $(\mathcal{L}(H \subset G),\tau)$, with $\tau(K) = [K:H]$ and $\mathcal{L}(H \subset G)$ the lattice of intermediate subgroups $H \subseteq K \subseteq G $.  
Question: Can any  finite distributive weighted lattice be realized by inclusion of groups?  
Remark: It's obviously true for the lattice with two elements and any weight thanks to the inclusion $(S_{n-1} \subset S_{n})$.
I don't know if it's true for the lattice with three elements, weighted $(mn,n,1)$ or even just $(n^{2},n,1)$, but I have checked by GAP that it's true for $mn < 32$.

REPLY [4 votes]: Let me hastily summarise my comments above: a subgroup inclusion chain of length 3 corresponds precisely to an imprimitive permutation group on a set of size $mn$ with a unique system of imprimitivity (and we require that the blocks in this system have size $m$).  The corresponding lattice will then be $(mn,n,1)$.
Such a group is given by $Sym(m) \wr Sym(n)$, although for particular $m$ and $n$ there are no doubt many others. This answers the specific question given by the OP.
As for generalizations: one naturally wonders how to deal with $(\ell mn, mn, n, 1)$. I think for this you can use $(Sym(\ell) \wr Sym(m)) \wr Sym(n)$, with similar iterations working for longer chains. Indeed one can probably generalize this construction even more so that given any intermediate subgroup lattice, you can stick a chain "on the top of it" by taking wreath products. This at least reduces the general question somewhat.