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 $mn < 32$.

  • $\begingroup$ Have you tried $(36,6,1)$? That seems like the first tricky $(n^2,n,1)$ case... $\endgroup$ – Nick Gill Feb 10 '15 at 15:29
  • 1
    $\begingroup$ 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... $\endgroup$ – Nick Gill Feb 10 '15 at 15:38
  • $\begingroup$ @NickGill: I've checked by GAP that your construction realizes $(36,6,1)$. $\endgroup$ – Sebastien Palcoux Feb 10 '15 at 15:57
  • $\begingroup$ 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. $\endgroup$ – 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.


Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.