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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.

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    $\begingroup$ If you restrict to torsion-free finitely generated groups: ($\infty$ many ends)$\Leftrightarrow$ (non-trivial free product), while (quasi-isometric to a free group) $\Leftrightarrow$ (free non-abelian). $\endgroup$
    – YCor
    Commented Jan 16, 2016 at 0:55

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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.

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There is a considerable literature on the subject, see the paper of Papazoglu (who is usually Papasoglu) and Whyte.

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