An Eulerian subgroup lattice is Boolean (see here), so it is natural to wonder whether it is also true for an interval of finite groups. The smallest non-Boolean Eulerian lattice is the following:

It is the face lattice of the square polytope (see here), let's call this lattice $P_4$.

By some tools used in this paper and the computation of Gordon Royle cited here, we can prove that if $P_4=[H,G]$ as a lattice, then $|G:H| \ge 135$ (and so $|G| \ge 270$, because $H \neq 1$). Moreover, we checked (GAP) that $|G| \ge 512$, and for $G$ simple, $|G| \ge |{\rm PSL}(2,191)|= 3483840$.

The existence of a lattice which is not the lattice of an interval of finite groups is an open problem.

In the following paper (p72): Overgroup lattices in finite groups of Lie type containing a parabolic
Michael Aschbacher conjectures (after John Shareshian) that a lattice $L$ such that the poset $\overline{L}:=L \setminus \{\hat{0},\hat{1} \}$ is disconnected with connected components the posets $\overline{B}_{n_1}, \dots , \overline{B}_{n_r}$, with $B_{n_i}$ a Boolean lattice of rank $n_i \ge3$ (and $r \ge 2$), is not the lattice of an interval of finite groups. But if all the $n_i$ are equal and odd, then $L$ is a non-Boolean Eulerian lattice (here the smallest example).

All these evidences lead to wonder:

**Question**: Is there a non-Boolean Eulerian interval of finite groups?

From an eventual positive answer for the relative version of K.S. Brown's problem and dual (i.e. $\sum_{K \in [H,G]} \mu(K,G)|G:K|$ and $\sum_{K \in [H,G]} \mu(H,K)|K:H|$ are nonzero) and the property here, we can deduce the following extension of (dual) Ore's theorem on Boolean intervals:

*Conjecture*: Let $[H,G]$ be an Eulerian interval of finite groups. Then:

- $\exists V$ irr. $\mathbb{C}$-rep. of $G$ such that $G_{(V^H)} = H$ (see here).
- $\exists g \in G$ such that $\langle Hg \rangle = G$ (see here).

This conjecture could help the investigation.