Groups that don't contain quasi-hyperbolic plane

Is there any known example of a one-ended finitely presented group with exponential growth that does not contain a quasi-isometric copy of the hyperbolic plane?

This question is motivated by the following question of Papasoglu mentioned in the paper 'Quasi-hyperbolic planes in hyperbolic group' by Bonk–Kleiner which asks whether every one-ended finitely presented group $$G$$ contains a quasi plane- the image of a uniform embedding $$P\rightarrow G$$ where $$P$$ is a complete Riemannian plane with bounded geometry.

Thank you for reading this.

• I don't know. It's true for polycyclic groups and Baumslag-Solitar groups $\mathrm{BS}(m,n)$ for $0<m<n$. However I'm not sure for $\mathbf{F}_2\times\mathbf{Z}$.
– YCor
Feb 9 at 16:03
• Thank you so much. Can you please give me a reference or any insights about why it is true for any polycyclic group with exponential growth? Feb 9 at 18:18
• For polycyclic groups I just checked it, I don't think it's written. A polycyclic group of exp. growth is QI to some real triangulable Lie group. And the latter contains a QI-embedded 2-dimensional affine group $\mathbf{R}\rtimes\mathbf{R}$.
– YCor
Feb 9 at 19:31
• And I tend to believe that $F_2\times\mathbf{Z}$ has no QI-embedded $\mathbf{H}^2$. Intuitively the QI-embedded planes therein should be Euclidean.
– YCor
Feb 9 at 19:32
• Maybe this is trivial but can I ask why that QI-embedded 2-dimensional affine group obstruct QI-embedding of the hyperbolic plane? Feb 9 at 21:07

It is a result of Buyalo and Schroeder [BS, Corollary 1.2] that for every $$n\ge 2$$ there is no QI embedding of the hyperbolic space $$\mathbb{H}^n_\mathbf{R}$$ into any product of $$n-1$$ trees with a Euclidean space.
In particular, the 1-ended group $$F_2\times\mathbf{Z}$$ (as well as $$F_2\times\mathbf{Z}^d$$ for arbitrary $$d\ge 1$$) contains no QI-copy of the hyperbolic plane, thus answering your question.