Let $G$ be a finite group and $\mu$ the Möbius function of the subgroup lattice $\mathcal{L}(G)$.

The reduced Euler characteristic of the order complex of the coset poset $\{ Kg \ | \ K<G, \ g \in G \} $ is $$\chi(G) := -\sum_{H \in \mathcal{L}(G)} \mu(H,G)|G:H|.$$ Gaschütz showed that $\chi(G)$ is nonzero for $G$ solvable and the question whether it is nonzero for any finite group is an open problem motivated by K.S. Brown (see DOI: 10.1016/j.aim.2015.10.018).

Consider a dual version of Brown's problem: the question whether $\hat{\chi}(G)$ is nonzero, with $$\hat{\chi}(G) := -\sum_{H \in \mathcal{L}(G)} \mu(1,H)|H|.$$ We have checked by GAP that $\chi(G)$ and $\hat{\chi}(G)$ are nonzero for $|G| \le 100$.

Let $G$ be a finite group such that $\mathcal{L}(G)$ is an Eulerian lattice, then $\mu(1,H) = \mu(1,G) \mu(H,G)$ for any $H \in \mathcal{L}(G)$ (see this post). Then $\hat{\chi}(G) = - \mu(1,G) \varphi(G)$ with $$\varphi(G) = \sum_{H \in \mathcal{L}(G)} \mu(H,G)|H|.$$ But by Crosscut Theorem and inclucion-exclusion principle $\varphi(G) = | \{g \in G \ | \ \langle g \rangle = G \} |.$ So if $\hat{\chi}(G)$ is nonzero as suggested by the dual Brown's problem, then so is $\varphi(G)$, which means that $G$ is cyclic and $\mathcal{L}(G)$ distributive; but it is assumed Eulerian, so it is boolean. Conclusion, the existence of a non-boolean Eulerian subgroup lattice would give a negative answer to the dual Brown's problem.

**Question**: Is an Eulerian subgroup lattice boolean?

*Warning*: By googling "Eulerian subgroup lattice" you find:

J.P. Bohanon and L. Reid, *Families of finite groups with Eulerian subgroup lattices*, in progress.

It seems that this work in progess deals with Eulerian graphs, not with Eulerian posets.