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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.

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Yes, and a much more general statement is true.

First, note that if $\mathcal{L}(G)$ is an Eulerian lattice then $\mu(1,G)=\pm 1$.

Theorem: $\mu(1,G)=\pm 1$ iff $G$ is cyclic of square-free order iff $\mathcal{L}(G)$ is boolean.

Proof: Théorème 3.1. of the paper "Fonction de Möbius d'un groupe fini et anneau de Burnside" (1984) by Kratzer and Thévenaz (available here) states the following (with $n_0$ the square-free part of $n$): $$\mu(1,G) \in \frac{|G|}{|G:G'|_0} \mathbb{Z}$$

But if $\mu(1,G)=\pm 1$ then $|G|= |G:G'|_0$, and so $G'=1$. It follows that $G$ is abelian with $|G|$ square-free, so $G$ is cyclic of square-free order and $\mathcal{L}(G)$ is boolean. The converse comes from a theorem of Ore stating that $G$ is cyclic iff $\mathcal{L}(G)$ is distributive. $\square$

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