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 $mn < 32$.