Suppose first that $\phi:G\to H$ is a $(\lambda,\varepsilon)$-quasi-isometry. If $k\in\ker\phi$, the lower quasi-isometry inequality gives
$$
\lambda^{-1}|k|_G-\varepsilon\leq d_H(1,\phi(k))=0,
$$
so the <kernel of a group homomorphism> lies in the finite word-metric ball of radius $\lambda\varepsilon$ and is finite. Coarse surjectivity gives an $R$ such that every $h\in H$ is within $R$ of $\phi(G)$. The finite ball $B_H(1,R)$ therefore contains representatives for every coset of $\phi(G)$, so $H/\phi(G)$ is finite.

Conversely, suppose $\ker\phi$ is finite and $phi(G)$ is a <finite-index subgroup> of $H$. The map factors as
$$
G\longrightarrow G/\ker\phi\xrightarrow{\ \cong\ }\phi(G)\hookrightarrow H.
$$
The 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
$$
\boxed{\phi\text{ is a quasi-isometry}\iff |\ker\phi|<\infty\text{ and }[H:\phi(G)]<\infty.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 143 / 2 / c / Solution

For <finitely generated groups> $G,H$, a <group homomorphism> $\phi:G\to H$ is a <quasi-isometry> exactly when its <kernel> is finite and its image is a <finite-index subgroup> of $H$. The finite kernel controls collapse of distances, and finite index is exactly the coarse-surjectivity condition.

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