Commensurability classes in groups

Question

Suppose $G$ is a group. Let’s call $a, b \in G$ commensurable, iff $\exists m, n \in \mathbb{Z} \setminus \{0\}$, such that $a^n = b^m$. One can see, that commensurability is an equivalence relationship: $$a = a$$ $$(a^n = b^m) \to (b^m = a^n)$$ $$((a^n = b^m) \cap (b^p = c^q)) \to (a^{np} = c^{mq})$$

Thus we can divide the elements of a group into commensurability classes.

All finite-order elements of a group form a commensurability class

If $a$ and $b$ are finite-order elements, then we can take $n$ and $m$ as their respective orders.

If $a^k = e$ and $a^n = b^m$, $b^{mk} = e$.

My question is:

What is the possible structure of commensurability classes of infinite-order elements?

What do I know about them:

Suppose, $a$ and $b$ are commensurable commuting infinite-order elements of $G$. Then $\exists c \in G$, such that $a, b \in \langle c \rangle$

Suppose $a^n = b^m$. Then $c = a^q b^p$, where $p$ and $q$ are integers, such that $pn + qm = gcd(n, m)$.

Now an important question to which I do not know an answer is:

Does there exist a group that contains a pair of commensurable non-commuting infinite-order elements?

The negative answer to this question implies that all commensurability classes of infinite order elements are of the form $Q \setminus \{e\}$, where $Q$ is a maximal locally cyclic subgroup of $G$. However, I do not know, how to prove that the answer is negative (or does such group actually exist?)

Answer 1

Consider the group $S(\mathbb N)$ of all bijections of the set of natural numbers. Now consider the elements $(2,4,1,3)\sigma$ and $(1,2,3,4)\sigma$, where $\sigma$ is any infinite-order bijection of $\{5,6,...\}$.

Answer 2

There are plenty of examples. For instance, for any $n,m\ge 2$, in the group with presentation $\langle x,y|x^m=y^n\rangle$, the elements $x,y$ have infinite order, do not commute, and are commensurate (I prefer "commensurate" to "commensurable" since there's no choice). (For $n=m=2$ this is the $\pi_1$ of the Klein bottle). Idem, $x,y$ are commensurate in the Baumslag-Solitar group $\langle t,x|tx^nt^{-1}=x^m\rangle$ for $y=txt^{-1}$.

The first question "What is the possible structure of" seems too imprecise for an answer. Anyway there are many natural things one can do with this. For instance, that a class of some infinite order element is invariant under conjugation is an interesting property; this is called a commensurated infinite cyclic subgroup.