Is there any easy example of a finitely generated group with a Cantor set of ends that is not quasi-isometric to a finitely generated free group?


Thanks in advance.



F.g group with infinite ends not Q.I to a free group
The free product $\mathbb{Z}^2 \ast \mathbb{Z}$ has infinitely-many ends, but is not quasi-isometric to a free group: furthermore, it is not hyperbolic since it has $\mathbb{Z}^2$ as a subgroup.


