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!

Application: Generalization of a theorem of Øystein Ore in group theory: the infinite case


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 $G$.

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  • $\begingroup$ @Sebastien Palcoux: Olshanskiy ''s book "Geometry of defining relations" or use Google. $\endgroup$ – user6976 Oct 15 '19 at 9:14
  • $\begingroup$ 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 $G$." $\endgroup$ – Sebastien Palcoux Oct 15 '19 at 10:43
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    $\begingroup$ By the way, Mark implicitly means "torsion-free". Indeed, there exist infinite 2-generated groups in which every proper subgroup is finite cyclic, and which have infinite exponent (they are actually easier to produce than their better known finite exponent cousins — and the torsion-free ones are also easier). $\endgroup$ – YCor Oct 15 '19 at 10:50
  • $\begingroup$ @SebastienPalcoux: Your Google is damaged. google.com/… $\endgroup$ – user6976 Oct 15 '19 at 11:01

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