Let $G$ be a finite group. Consider all the subgroups $H$ such that its subgroups lattice $\mathcal{L}(H)$ is distributive (i.e. the group $H$ is cyclic, by Ore's theorem), and among them, let $\{ H_{i_1} \ \vert \ i_1 \in I_1 \}$ be the set of subgroups having a maximal order. Now for a fixed $i_1$, consider all the intermediate subgroups $K$ of $(H_{i_1} \subseteq G)$, i.e. $H_{i_1} \subseteq K \subseteq G$, such that the intermediate subgroups lattice $\mathcal{L}(H_{i_1} \subseteq K)$ is distributive, and among them, let $\{ H_{i_1,i_2} \ \vert \ i_2 \in I_2 \}$ be the set of subgroups having a maximal order. We iterate the process, and we obtain several chains of (strictly) increasing subgroups: $$\{e\} \subset H_{i_1} \subset H_{i_1,i_2} \subset H_{i_1,i_2,i_3} \subset \dots \subset G$$
Question: Are all these chains of same length?
Example: $G = S_4$ (see the subgroups lattice below)
There are $16$ cyclic subgroups: $3+6$ isomorphic to $\mathbb{Z}/2$, $4$ isomorphic to $\mathbb{Z}/3$ and $3$ isomorphic to $\mathbb{Z}/4$. So among them, the maximal order is $4$. Next for a given subgroup isomorphic to $\mathbb{Z}/4$, there are two intermediate subgroups (i.e. $\mathbb{Z}/4 \subseteq K \subseteq S_4$) with $\mathcal{L}(\mathbb{Z}/4 \subseteq K)$ distributive: $K=Dih_4$ or $S_4$, and the one of maximal order is $S_4$.
Conclusion, we get three chains of same length $2$, and isomorphic to: $\{e\} \subset \mathbb{Z}/4 \subset S_4$
