# Infinite group generated by a single coset

Let $$G$$ be an infinite countable group having a core-free subgroup $$H$$ such that the interval $$[H,G]$$ in the subgroup lattice $$\mathcal{L}(G)$$ is ACC of infinite length, and for every $$K \in (H,G]$$, $$G$$ is generated by a single $$K$$-coset (i.e. there is $$g \in G$$ with $$\langle Kg \rangle = G$$).

Question: Is $$G$$ generated by a single $$H$$-coset?

It is an exercise to reformulate as: let $$[H,G]$$ be an ACC interval of groups and $$(K_i)_{i \in I}$$ its coatoms, i.e. the maximal elements in $$[H,G)$$. Assume that $$I$$ is an infinite countable set, and for every finite subset $$J \subset I$$ we have $$\bigcap_{j \in J} (G \setminus K_j) \neq \emptyset.$$ Question: Is it true that $$\bigcap_{i \in I} (G \setminus K_i) \neq \emptyset$$?

Examples: for $$G = \mathbb{Z}$$ and $$H = \{0\}$$, the ACC is satisfied, the coatoms are $$(p\mathbb{Z})_{p \in \mathbb{P}}$$ and $$\bigcap_{p \in \mathbb{P}} (\mathbb{Z} \setminus p\mathbb{Z}) = \{-1,1 \} \neq \emptyset$$. For $$G = \mathbb{Z} \rtimes C_2$$ and $$H = C_2$$, it works as well.
Any other example (with $$H$$ core-free) is welcome!

The answer is "no". Let $$G$$ be a Tarski torsion-free monster, $$H$$ be the trivial subgroup. Then $$(H,G]$$ consists of $$G$$ and all cyclic subgroups of $$G$$, it has ACC and infinite length. If $$K$$ is a cyclic subgroup and $$a$$ is not in the cyclic centralizer $$C(K)$$, then the coset $$aK$$ contains two non-commuting elements, so it generates $$G$$ but no coset of $$H$$ generates $$G$$.
• I got the expected result in Ol'shanskii's book: Theorem 28.3 (on page 306): "There is a simple torsion-free group $G$ (...) in which every proper subgroup is infinite cyclic (...) any two maximal subgroups in $G$ have trivial intersection". Then on page 307 he wrote: "The lattice of subgroups of $G$ has a very simple structure: it consists of a countable number of lattices of subgroups of an infinite cyclic group pasted together at the trivial subgroup and the whole group $G$." – Sebastien Palcoux Oct 15 '19 at 10:43