Baumslag-Solitar Group I am working on an exercise which says Hyperbolic groups do not contain any Baumslag-Solitar group $\operatorname{BS}(m,n)$ as subgroup.
(Hint is $\operatorname{BS}(m,n)$ is not virtually cyclic).
My attempt is I am trying to show that $\operatorname{BS}(m,n)$ contains $\mathbb{Z \oplus Z}$ using the fact that $\operatorname{BS}(m,n)$ is not virtually cyclic where $\operatorname{BS}(m,n)$ denotes the Baumslag-Solitar group.
Any hints are appreciated.
Thanks in advance
 A: I'll convert my comments into a sketch proof. I should warn though that this exercise is not easy (the original proof used "biautomaticity", while I'll use a fancy, but standard, theorem). I suspect that there are results in your course notes/book which make the problem easier.
Also, the "hint" given is meaningless. For example, non-abelian free groups are not virtually cyclic, but every hyperbolic group contains a non-abelian free group or is itself virtually cyclic... Possibly the hint meant "centralisers of elements are virtually cyclic", which is in fact the key theorem we'll use.
Firstly, start with the following (standard) result for hyperbolic groups. I've copied the statement from Corollary III.$\Gamma$.3.10 (p462) of Bridson and Haelfliger's book Metric spaces of non-positive curvature (although the proof there is more general then we need, as it is for "semihyperbolic" groups). It is equivalent to the fact that centralisers of elements are virtually cyclic (why?).
Theorem A. Let $H$ be hyperbolic. If $a\in H$ is an element of infinite order then $\langle a\rangle$ has finite index in its centraliser.
Therefore, all we need to do is find for every Baumslag-Solitar group $\operatorname{BS}(m, n)$ an element $g\in \operatorname{BS}(m, n)$ of infinite order such that $\langle g\rangle$ does not have finite index in $C_{\operatorname{BS}(m, n)}(g)$. Lets break this down into a series of facts which need to be proven.

Fact 1. Every non-trivial element of $\operatorname{BS}(m, n)$ has infinite order.

This follows from the fact that $\operatorname{BS}(m, n)$ is an HNN-extension of a torsion-free group (namely $\mathbb{Z}$).

Fact 2. If $\operatorname{BS}(m, n)=\langle a, t\mid t^{-1}a^mt=a^n\rangle$ with $|m|,|n|>1$ or $|m|=|n|=1$ then $C_{\operatorname{BS}(m, n)}(a^n)$ contains a copy of $\mathbb{Z}^2$.

If $|m|,|n|>1$ then $\langle a^n, t^{-1}ata\rangle$ is abelian, and so by Fact 1 is either $\mathbb{Z}$ or $\mathbb{Z}^2$. Use the fact that $\operatorname{BS}(m, n)$ is an HNN-extensions to prove that this subgroup is non-cyclic. The case $|m|=|n|=1$ is similar.

Fact 3. If $\operatorname{BS}(1, n)=\langle a, t\mid t^{-1}at=a^n\rangle$ (so $m=1$) then $C_{\operatorname{BS}(1, n)}(a)$ contains a non-cyclic locally cyclic subgroup.

The hint here is that $\langle t^{p}at^{-p}\mid p\in\mathbb{N}\rangle$ is not finitely generated, but is locally cyclic (every finitely generated subgroup is cyclic). The key observation needed to prove this is that $(t^pat^{-p})^n=t^{p-1}at^{-(p-1)}$).
Facts 2 and 3 mean that $\langle a\rangle$ does not have finite index in $C_{\operatorname{BS}(m, n)}(a)$, while $a$ has infinite order by Fact 1. The result then follows as if $\operatorname{BS}(m, n)$ embeds into a hyperbolic group $H$ then $C_{\operatorname{BS}(m, n)}(a)\leq C_{H}(a)$, contradicting Theorem A.
