 Solution

Suppose first that $\varphi :G\to H$ is a $(\lambda ,\epsilon )$-quasi-isometry. If $k\in \ker \varphi$, the lower quasi-isometry inequality givesso the kernel of a group homomorphism lies in the finite word-metric ball of radius $\lambda \epsilon$ and is finite. Coarse surjectivity gives an $R$ such that every $h\in H$ is within $R$ of $\varphi (G)$. The finite ball $B_H(1,R)$ therefore contains representatives for every coset of $\varphi (G)$, so $H/\varphi (G)$ is finite.

Conversely, suppose $\ker \varphi$ is finite and $phi(G)$ is a finite-index subgroup of $H$. The map factors asThe first arrow is a finite-kernel quotient quasi-isometry, the middle arrow is an isomorphism of finitely generated groups, and the last arrow is a finite-index subgroup quasi-isometry. Their composition is a quasi-isometry. Hence the quasi-isometry criterion for a group homomorphism is