This post is the infinite version of this one, and is motivated by an exchange with Carmela Musella and Maria De Falco. We are interested in relative versions of the following Ore's theorem and corollary. For a reference, see Section 1.2 in Schmidt's book Subgroup lattices of groups.
Theorem: A group $G$ is locally cyclic iff its subgroup lattice $L(G)$ is distributive.
Corollary: A group $G$ is cyclic iff $L(G)$ is distributive and satisfies one of the following properties:
- maximal condition,
- for any subgroup $K \neq 1$, the interval $[K,G]$ is finite,
- for any subgroup $K \neq 1$, the index $|G:K|$ is finite,
- for any $K,K' \in L(G)$, $|G:K| = |G:K'|$ $\Leftrightarrow$ $K=K'$.
Let $G$ be a group and $H$ a subgroup.
Definitions: the group $G$ is called $H$-cyclic if there is $g \in G$ such that $\langle H,g \rangle = G$.
It is called locally $H$-cyclic if for any finite subset $S \subset G$, the subgroup $\langle H \cup S \rangle$ is $H$-cyclic.
Remark: the group $G$ is $H$-cyclic iff the hypergroup $G/H$ is singly generated, as $\langle H,g \rangle = \langle Hg \rangle$.
Assume that the interval $[H,G] \subseteq L(G)$ is distributive.
Question 1: Is $G$ locally $H$-cyclic?
Assume moreover that $[H,G]$ satisfies the following property ($x$), with $x \in \{ 1,2,3,4 \}$:
(1) maximal condition,
(2) for any $K \in (H,G]$, the interval $[K,G]$ is finite,
(3) for any $K \in (H,G]$, the index $|G:K|$ is finite,
(4) for any $K,K' \in [H,G]$, $|G:K| = |G:K'|$ $\Leftrightarrow$ $K=K'$.
Question 2($x$): $G$ is $H$-cyclic?
Note that Q1 > Q2(1) > Q2(2) > Q2(3) > Q2(4).
Remark: All these questions have a positive answer if the interval $[H,G]$ is finite (see here).
Observation: with (2) we have the following immediate (but promising) result: $$\forall g \in G \setminus H, \ \exists g' \in G \text{ such that } \langle H,g,g' \rangle = G \ \ \ \ \ (*)$$
If $H = 1$, the property $(*)$ is known as $\frac{3}{2}$-generation (see this poster of Scott Harper, and this post).
Proposition: If $G$ is $\frac{3}{2}$-generated then every proper quotient of $G$ is cyclic (proof).
Conjecture (Breuer-Guralnick-Kantor): A finite group is $\frac{3}{2}$-generated iff every proper quotient is cyclic.
Theorem (Guralnick-Kantor): Every finite simple group is $\frac{3}{2}$-generated.
If $G$ satisfies $(*)$, the hypergroup $G/H$ can be called $\frac{3}{2}$-generated, as $\langle H,g,g' \rangle = \langle Hg, Hg' \rangle$.
