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

For a finite generating set $S$ of $G$, the Cayley graph ${\mathrm{Cay}}_S(G)$ has vertex set $G$ and an edge from $g$ to $gs$ for every $g\in G$ and $s\in S\cup S^{-1}$; one may retain orientations and labels, or forget them.

As unlabelled undirected graphs, the Cayley graph of the infinite cyclic group $\mathbb{Z}$ with generator $1$ and the Cayley graph of the infinite dihedral groupwith generators $a,b$ are both the two-way infinite line. The first group is an abelian group and the second is not, so they are not isomorphic. Thus the requested example is the infinite line as a Cayley graph of $\mathbb{Z}$ and $D_{\infty }$.

A map $f:X\to Y$ is a quasi-isometry if there are $\lambda \ge 1$ and $\epsilon ,R\ge 0$ such thatfor all $x,x^{\prime }\in X$, and every point of $Y$ lies within distance $R$ of $f(X)$. The Milnor–Švarc lemma says that a group acting properly discontinuously, cocompactly and isometrically on a proper geodesic metric space is finitely generated, and each orbit map from a word metric is a quasi-isometry.

Now let $S,T$ be finite generating sets of $G$. PutReplacing each letter in an $S$-word by a $T$-word and conversely givesThus the identity map is a bilipschitz equivalence, hence a quasi-isometry. All finite generating sets of a finitely generated group give quasi-isometric Cayley graphs.

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