Suppose that two finitely generated groups quasi-isometrically embed into each other. Does it follow that the two groups are quasi-isometric? Recall that a quasi-isometry is a quasi-isometric embedding that is quasi-surjective, see e.g. <https://www.math.ucdavis.edu/~kapovich/EPR/pc_lectures3.pdf>



Cantor-Bernstein for quasi-isometric embeddings?
Let $C\_n$ be a cyclic groups of order $n$. Then the wreath products $C\_2\wr\mathbf{Z}$ and $C\_3\wr\mathbf{Z}$ embed QI into each other (for the reverse direction, observe that $C\_2\wr\mathbf{Z}$ has a subgroup of index 2 isomorphic to $C\_2^2\wr\mathbf{Z}$). But $C\_2\wr\mathbf{Z}$ and $C\_3\wr\mathbf{Z}$ are not QI, this follows from work of Eskin-Fisher-Whyte ($F\wr\mathbf{Z}$ and $F'\wr\mathbf{Z}$ for $F,F'$ finite groups are QI iff $|F|,|F'|$ have some common power.)


