Let $G$ be a finite group and $H$ a subgroup such that the interval $[H,G]$ is a boolean lattice.

Let $L_1, \dots , L_n$ be the maximal subgroups of $G$ containing $H$.

Let the alternative sum of the interval $[H,G]$ defined as follows:

$$\chi([H,G]):= \sum_{r=0}^n (-1)^{r} \sum_{ \ i_1 < i_2 < \cdots < i_r } [L_{i_1} \wedge \cdots \wedge L_{i_r}: H] $$
*Notation*: $L_{i_1} \wedge \cdots \wedge L_{i_r} = G$ for $r=0$.

**Theorem**: $\chi([H,G]) > 0$.

*Proof*: Observe that

$$\chi([H,G]) = \frac{\vert G \vert - \vert \bigcup_i L_i \vert}{\vert H \vert} $$ but a boolean lattice is distributive so by a result of Oystein Ore (see here) $\exists g \in G$ with $\langle H,g \rangle = G$, which precisely means that $g \not \in L_i \ \forall i$, and so $\chi([H,G])> 0$ $\square$

Let $K_1, \dots , K_n$ be the minimal overgroups of $H$.

Let the dual alternative sum of the interval $[H,G]$ defined as follows:

$$\hat{\chi}([H,G]):= \sum_{r=0}^n (-1)^{r} \sum_{ \ i_1 < i_2 < \cdots < i_r } [G: K_{i_1} \vee \cdots \vee K_{i_r}] $$
*Notation*: $K_{i_1} \vee \cdots \vee K_{i_r} = H$ for $r=0$.

**Question**: Is $\hat{\chi}([H,G]) > 0$ ?

*Remark*: after GAP checking, it is true for $[G:H]<32$ (recall that $[H,G]$ is assumed boolean).