3
$\begingroup$

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

$\endgroup$
3
$\begingroup$

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 $G$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ @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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.