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

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 $mn < 32$.

• Have you tried $(36,6,1)$? That seems like the first tricky $(n^2,n,1)$ case... – Nick Gill Feb 10 '15 at 15:29
• I could be wrong, but if you let $G=S_m \wr S_n$ and consider the natural action on $mn$ points, then the point-stabilizer has index $mn$ and there is a unique subgroup in between of index $n$. Sorry, no time to check this through but if it works it deals with $(mn,n,1)$.... And iterated wreath products will probably deal with chain lattices... – Nick Gill Feb 10 '15 at 15:38
• @NickGill: I've checked by GAP that your construction realizes $(36,6,1)$. – Sebastien Palcoux Feb 10 '15 at 15:57
• Well that sounds promising. My thinking is that the three element lattice is equivalent to having an imprimitive permutation group on a set of size $mn$ with a unique system of imprimitivity. There are no doubt many of these, but the wreath product is certainly one such. And I can't see a reason why iterating wouldn't work, so that probably sorts chains out. I'll try and write a proper answer later today when I have time. – Nick Gill Feb 11 '15 at 8:51

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.