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Let $G$ be a torsion-free group. For an element $g \in G \setminus \{1_G\}$ we define the radical of $g$ in $G$ as $$ Rad_G(g) = \left\{r \in G \mid r^a = g^b \mbox{ for some } a,b \in \mathbb{Z}\setminus\{0\}\right\} \cup \{1_G\}. $$

For which torsion-free groups can we show that $Rad_G(g)$ is an infinite cyclic subgroup of $G$ for every nontrivial element? So far I have been able to establish it for the following classes:

  • residually finitely generated torsion-free nilpotent groups,
  • torsion-free hyperbolic groups,
  • relatively hyperbolic groups (in the sense of Bowdich), where the associated subgroups already have the property (this includes for example toral relatively-hyperbolic groups).

Are there any easy examples I am missing?

EDIT: there is also an equivalent definition.

For an element of infinite order we define $$ plog_G(g) = \max\{k \in \mathbb{N} \mid r^k = g \mbox{ for some } r\in G\}. $$ If $plog_G(g)$ is finite, then it makes sense to define $$ \sqrt[G]{g} = \{r \in G \mid r^{plog_G(g)} = g\}. $$ For a torsion-free group $G$ and an element $g \in G \setminus\{1\}$ the following are equivalent:

  • $Rad_G(g)$ is an infinite cyclic subgroup of $G$,
  • $|\sqrt[G]{g}| = 1$ and $plog_G(g^n) = n\cdot plog_G(g)$ for every $n \in \mathbb{N}$.
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  • $\begingroup$ "Clearly a subgroup": no, it's not a subgroup in general. For instance in the fundamental group of the Klein bottle $\langle x,y|x^2=y^2\rangle$, both $x$ and $y$ belong to $Rad_G(x)$, but $xy$ does not. Stlll the question makes sense (you're asking when is the subset $Rad_G(g)$ a cyclic subgroup for every $g\neq 1$. $\endgroup$
    – YCor
    Commented Jun 9, 2018 at 0:28
  • $\begingroup$ Of course you have finite direct products of such groups. Other examples are torsion-free finitely generated subgroups in $\mathrm{SL}_2(A)$, where $A$ is is the ring of integral elements in $\mathbf{Q}$. For instance, this includes torsion-free finite-index subgroups in $SL_2(\mathbf{Z}[\sqrt{2}])$, which are far from relatively hyperbolic and are not virtually residually torsion-free nilpotent. $\endgroup$
    – YCor
    Commented Jun 9, 2018 at 1:55
  • $\begingroup$ Actually, the property of having "cyclic radicals" is preserved w.r.t. to graph products and amalgamation along a common retract, but that's not easy to see. I am using this property to study certain behaviour and I was wondering which groups do have it, on top of the obvious ones. $\endgroup$
    – Michal Ferov
    Commented Jun 9, 2018 at 2:53
  • $\begingroup$ As is usual in MO, you missed the torsion-free Tarski monsters (and many other groups where the centralizer of every non-identity element is cyclic). You also missed the R. Thompson group $F$. $\endgroup$
    – user6976
    Commented Jun 9, 2018 at 4:23
  • $\begingroup$ By the way, your property is equivalent to a combination of two properties, which are worth being treated separately: (a) that $Rad_G(g)$ is locally cyclic for every $g\neq 1$. This means that any two elements with a common power have a common root. (b) every locally cyclic subgroup is cyclic. For instance, any torsion -free subgroup of $SL_2$ satisfies (a). And (a) is a 1st-order invariant, since it's equivalent to the condition that any two elements with a common power commute. Hence it's equivalent to a combination of infinitely many 1st-order formulas, and passes to inductive limits. $\endgroup$
    – YCor
    Commented Jun 9, 2018 at 7:52

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