Let $[H,G]$ be a rank $n$ boolean interval of finite groups (i.e. $[H,G] \simeq B_n$ as lattice).
Let the set $E = \{ g \in G \ | \ \langle H,g \rangle = G \}$
Remark: If $g \in E$ then $Hg \subset E$; so $|E|$ is a multiple of $|H|$.
Question: $|E| / |H| \ge 2^{n-1}$? $ \ $ [checked by GAP for $|G:H| < 32$]
(the left-hand side can be interpreted as an alternating sum of indices, see here)
Remark: If this lower bound is correct then it is optimal because realized by $[1 \times S_2^{n-1} , S_2 \times S_3^{n-1} ]$.
The proof of the following theorem could help to answer the question.
Theorem: $E \neq \emptyset $.
proof by induction on the rank $n$:
If $n = 1$ then $H$ is a maximal subgroup of $G$, so $E = G - H \neq \emptyset$.
Now, we assume it is true for every rank $<n$ and we will prove it for rank $n$.
Let $M$ be a maximal subgroup of $G$ containing $H$, and $M^{\complement}$ the lattice-complement of $M$ in $[H,G]$
(i.e. $M \vee M^{\complement} = G$ and $M \wedge M^{\complement} = H$; existence and unicity come from the boolean structure).
By induction there are $a, b \in G$ such that $\langle H,a \rangle = M$ and $\langle H,b \rangle = M^{\complement}$. Let $g=a b$ then $a=g b^{-1}$ and $b=a^{-1}g$, so $\langle H,a,g \rangle = \langle H,g,b \rangle = \langle H,a,b \rangle = M \vee M^{\complement} = G$.
Now, $\langle H,g \rangle = \langle H,g \rangle \vee H = \langle H,g \rangle \vee (M \wedge M^{\complement})$ but by distributivity $\langle H,g \rangle \vee (M \wedge M^{\complement}) = (\langle H,g \rangle \vee M \rangle) \wedge (\langle H,g \rangle \vee M^{\complement} \rangle)$.
So $ \langle H,g \rangle = \langle H,a,g \rangle \wedge \langle H,g,b \rangle = G$. $\square$
Remark: this result is also true in the distributive case (due to O. Ore, see here), but it can be reduced to the boolean case (see here) which is easier to prove as above.
