# Are groups with every proper, non-trivial subgroup infinite cyclic simple?

In the 1970s Ol'shanskii constructed a non-cyclic finitely generated group $$G$$ with the following properties:

1. Every proper, non-trivial subgroup of $$G$$ is infinite cyclic.
2. If $$X^m=Y^n$$ for $$X, Y\in G$$ with $$m,n\neq0$$, then $$\langle X, Y\rangle$$ is cyclic i.e., any two maximal subgroups of $$G$$ have trivial intersection.

Ol'shanskii gave an easy proof that such a group is simple, which roughly goes: Suppose $$N$$ is a proper, non-trivial normal subgroup of $$G$$. If $$N$$ is maximal then $$G/N$$ is cyclic of prime order, so $$G$$ is virtually-$$N$$, so $$G$$ is a torsion-free virtually-$$\mathbb{Z}$$ group, so must itself be cyclic. If $$N$$ is not maximal then $$N$$ is contained in a maximal subgroup $$M$$ such that $$M^g\cap M\neq1$$ for all $$g\in G$$, so as $$M^g$$ is also maximal and as maximal subgroups intersect trivially (by (2)) we have that $$M^g=M$$ for all $$g\in G$$, i.e. $$M$$ is normal in $$G$$, which is impossible by the previous case.

Property (2) was used here. I was wondering if this can be dropped. So:

Question. Suppose $$G$$ is a non-cyclic finitely generated group with every proper, non-trivial subgroup of $$G$$ infinite cyclic. Is $$G$$ simple?

If $$G$$ instead satisfies that it is infinite and every proper, non-trivial subgroup has order $$p$$ for a fixed prime $$p$$ then $$G$$ is a "Tarski monster" group and is indeed simple: If $$N$$ is a proper, non-trivial normal subgroup of $$G$$ and $$g\not\in N$$ then $$N\cap\langle g\rangle=1$$, as both subgroups have prime order, so $$N\langle g\rangle=N\rtimes\langle g\rangle$$ has order $$p^2$$, a contradiction. However, this proof uses primality so does not extend to the setting here.

• What about an infinite cyclic group? Jun 24, 2021 at 9:32
• @MikaeldelaSalle I meant to exclude that possibility! I'll edit the question to rule it out.
• Small observation: Any normal subgroup is necessarily central. Assume $\langle x \rangle$ is normal, and there is $y$ with $yxy^{-1} = x^{-1}$. Then $\langle x,y\rangle$ cannot be cyclic, so it must be all of $G$. The subgroup $\langle x,y^2\rangle$ is abelian, thus a proper subgroup, and must be cyclic. So $G$ is a normal extension $\mathbb{Z}\to G \to C_2$ with sign action on $\mathbb{Z}$, and thus isomorphic to the infinite dihedral group, which contains $2$-torsion, contradiction. Jun 24, 2021 at 10:06
• Also note that if we do have nontrivial center $\langle x \rangle$, then $G/\langle x \rangle$ is a pretty weird group: It has the property that every proper subgroup is finite cyclic. It feels like one should be able to finish from here, but I haven't figured out how yet. Jun 24, 2021 at 10:39
• @AchimKrause: If $G/Z(G)$ is finite, then the commutator subgroup $[G,G]$ is finite, by a theorem of Schur. Jun 24, 2021 at 14:34
Such a group necessarily has infinite cyclic centre: take any non-commuting $$x$$ and $$y$$, then the centralizer of $$\langle x \rangle \cap \langle y \rangle \cong \mathbb{Z}$$ contains $$\langle x, y \rangle$$, which is the whole group. Thanks to Ashot Minasyan for pointing out Ol’shanskii's theorem.