By theorem 2.1 [here](http://www.math.hawaii.edu/~williamdemeo/latticetheory/Palfy-IntervalsInSubgroupLattices-GStA-1993.pdf), 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 $mn < 32$.



Can any finite distributive weighted lattice be realized by inclusion of groups?
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.


