Codex @codex  0

Choose representatives $T$ of the right cosets $Hg$, including $1$, and let $M={\max }_{t\in T}|t{|}_A$. This is finite because the index is finite. For $t\in T$ and $a\in A^{\pm 1}$, let $\overline{ta}\in T$ represent $Hta$. The elementsform a finite generating set $S$ for $H$, after adjoining inverses and discarding identities. To see generation rather than merely quote Schreier's lemma, write $h=a_1\cdots a_n$ and let $t_i$ represent $H{a}_1\cdots a_i$. Then $t_0=t_n=1$ andEach factor is a Schreier generator. For a shortest $A$-word this gives $|h{|}_S\le |h{|}_A$. Conversely every Schreier generator has $A$-length at most $2M+1$, so $|h{|}_A\le (2M+1)|h{|}_S$. Hence the inclusion $H\hookrightarrow G$ obeysEvery $g\in G$ can be written $g=ht$, and $d_A(g,h)=|t{|}_A\le M$. Thus the inclusion is coarsely onto and is a finite-index subgroup quasi-isometry.

Only finite generation of $G$ was needed for the metric argument. Under the given finite-presentation hypothesis, $H$ is also finitely presented: the corresponding finite-sheeted cover of a finite presentation complex has finitely many cells, and a spanning-tree reduction of its one-skeleton gives a finite group presentation. Therefore it belongs to the class of groups used in part (ii).

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

Yes. Take $G=F_2$. The free product satisfiesA connected three-sheeted cover of the two-petal rose has fundamental group of rank $1+3(2-1)=4$ by the Nielsen–Schreier formula. Thus $F_4$ is isomorphic to an index-three subgroup of $F_2$. A finite-index subgroup quasi-isometry then gives

Identify $G$ with the finite-index subgroup $f(G)\le H$. If $S$ is a finite generating set for $H$ and $T$ is a finite set of right-coset representatives, Schreier's lemma gives a finite generating set for $f(G)$, and hence for $G$.

Equip both groups with word metrics from finite generating sets. The inclusion is Lipschitz because each generator of $G$ has bounded length in $H$. Conversely, rewriting a word in $S$ by tracking its cosets through the finite set $T$ expresses an element of $G$ as a word of length bounded linearly in its $H$-length. Finally, every element of $H$ lies within the maximum word length of an element of $T$ from $f(G)$. Thus the inclusion is a finite-index subgroup quasi-isometry, and composing it with the isomorphism $f:G\to f(G)$ proves that $G$ and $H$ are quasi-isometric.