Let $G$ be a finite group then the fundamental theorem of cyclic groups can be formulated as follows:

*Theorem*: $G$ is cyclic iff it admits no two different subgroups with the same order.

*proof*: see here p44 together with Lagrange theorem. $\square$

But there is an other characterization of the finite cyclic groups, using the lattice theory:

*Theorem*: $G$ is cyclic iff its subgroups lattice $\mathcal{L}(G)$ is distributive.

*proof*: see here theorem 4 p267. $\square$

So we get immediately the following statement for a finite group $G$:

If $G$ admits no two different subgroups with the same order, then its subgroups lattice is distributive.

We ask about a generalization of this statement for an inclusion $(H \subset G)$ of finite groups:

**Question**: Is it true that if $(H \subset G)$ admits no two different intermediate subgroups $H \subset K \subset G$ with the same order, then its intermediate subgroups lattice $\mathcal{L}(H \subset G)$ is distributive?

*Remark*: It is checked (by GAP) for $[G:H] \le 31$.

The converse is false because the inclusion $(S_2 \times S_2 \subset S_3 \times S_3)$ is a counter-example (see here).