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.


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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