The rank $n$ boolean lattice $B_n$, is the subset lattice of $\{1,2, \dotsm n \}$.
Let $[H,G]$ be a boolean interval of finite groups. Its Euler totient is defined by $$\varphi(H,G):=\sum_{K \in [H,G]} (-1)^{\ell(K,G)} |K:H|.$$
with $\ell(K,G)$ the rank of the boolean lattice $[K,G]$. Its dual Euler totient is defined by $$\hat{\varphi}(H,G):=\sum_{K \in [H,G]} (-1)^{\ell(H,K)} |G:K|.$$
The Euler totient $\varphi(H,G)$ is exactly the number of cosets $Hg$ generating $G$ individually.
It is nonzero by a theorem of Oystein Ore (see here).
Question: Is the dual Euler totient $\hat{\varphi}(H,G)$ also nonzero?
Remark: If the boolean interval $[H,G]$ is group-complemented (i.e. $\forall K \in [H,G]$, $KK^{\complement} = K^{\complement}K$), then we observe that $\hat{\varphi}(H,G) = \varphi(H,G)$, so that it is nonzero in this case.
Remark: At index $|G:H|<32$, there are, up to equivalence, 612 boolean intervals. They are all group-complemented except $[D_8,L_2(7)]$ and $[S_3,L_2(7)]$ (both of rank $2$), but their dual Euler totients are also nonzero, so that the question admits a positive answer at index $|G:H|<32$.
Remark: This Euler totient $\varphi$ (and dual $\hat{\varphi}$) extends to any interval of finite groups (using the Möbius function). And then, $\varphi(\{e\},\mathbb{Z}/n) = \varphi(n)$, the usual Euler's totient. So the naming is relevant.
Remark: We just find this paper of Marius Tarnauceanu, containing some analogous (but not same) investigations (see section 4).
Suggestion of Russ Woodroofe (see this answer): Observe that $\hat{\varphi}(H,G)$ is exactly the Mobius invariant of the bounded coset poset $\hat{P} = \hat{C}(H,G)$ of $H$ in $G$. So it is suffisient to show that the poset $\hat{P}$ is Cohen-Macaulay (notion by R. Stanley) and the nontrivial reduced Betti number of the order complex $\Delta(P)$ is nonzero. For so, it is sufficient to prove the existence of, first, a dual EL-labeling on $\hat{P}$, and next, a maximal decreasing chain on it (for more details see this paper). Of course this strategy can work only if all these sufficient conditions are also necessary for boolean intervals.
