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](https://en.wikipedia.org/wiki/Eulerian_poset), then $\mu(1,H) = \mu(1,G) \mu(H,G)$ for any $H \in \mathcal{L}(G)$ (see [this post](https://mathoverflow.net/q/280917/34538)). 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](https://en.wikipedia.org/wiki/Eulerian_path), not with [Eulerian posets](https://en.wikipedia.org/wiki/Eulerian_poset).



Is an Eulerian subgroup lattice boolean?
**Yes**, and a much more general statement is true.


First, note that if $\mathcal{L}(G)$ is an [Eulerian lattice](https://en.wikipedia.org/wiki/Eulerian_poset) 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](https://infoscience.epfl.ch/record/130417/files/fulltext.pdf)) 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. $\square$


