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

Question

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?

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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.

Answer 1

The answer is no, there exist non-simple torsion-free Tarski monsters. Theorem 31.4 of Ol’shanskii's book "Geometry of defining relations in groups" ZBL0676.20014 MR1191619 is:

Theorem. There is a non-abelian group all of whose proper subgroups are infinite cyclic and the intersection of any two of them is non-trivial.

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.