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.

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    $\begingroup$ What about an infinite cyclic group? $\endgroup$ Jun 24, 2021 at 9:32
  • $\begingroup$ @MikaeldelaSalle I meant to exclude that possibility! I'll edit the question to rule it out. $\endgroup$
    – ADL
    Jun 24, 2021 at 9:33
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    $\begingroup$ 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. $\endgroup$ Jun 24, 2021 at 10:06
  • $\begingroup$ 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. $\endgroup$ Jun 24, 2021 at 10:39
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    $\begingroup$ @AchimKrause: If $G/Z(G)$ is finite, then the commutator subgroup $[G,G]$ is finite, by a theorem of Schur. $\endgroup$ Jun 24, 2021 at 14:34

1 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.


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