The rank n boolean lattice Bn, is the subset lattice of {1,2,⋯n}.
Let [H,G] be a boolean interval of finite groups. Its Euler totient is defined by φ(H,G):=∑K∈[H,G](−1)ℓ(K,G)|K:H|.
with ℓ(K,G) the rank of the boolean lattice [K,G]. Its dual Euler totient is defined by ˆφ(H,G):=∑K∈[H,G](−1)ℓ(H,K)|G:K|.
The Euler totient φ(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 ˆφ(H,G) also nonzero?
Remark: If the boolean interval [H,G] is group-complemented (i.e. ∀K∈[H,G], KK∁=K∁K), then we observe that ˆφ(H,G)=φ(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 [D8,L2(7)] and [S3,L2(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 φ (and dual ˆφ) extends to any interval of finite groups (using the Möbius function). And then, φ({e},Z/n)=φ(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 ˆφ(H,G) is exactly the Mobius invariant of the bounded coset poset ˆP=ˆC(H,G) of H in G. So it is suffisient to show that the poset ˆP is Cohen-Macaulay (notion by R. Stanley) and the nontrivial reduced Betti number of the order complex Δ(P) is nonzero. For so, it is sufficient to prove the existence of, first, a dual EL-labeling on ˆP, 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.