Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 133 / 3

Any two word metrics from finite generating sets on the same group are bilipschitz equivalent. Indeed, if $S,S^{\prime }$ are finite, let $L={\max }_{s\in S}|s{|}_{S^{\prime }}$; then $|g{|}_{S^{\prime }}\le L|g{|}_S$, and the reverse inequality follows symmetrically. Apply this once to the two finite generating sets of $G$ and once to those of $H$. Composing these bilipschitz identity maps with the inclusion changes only the multiplicative and additive constants in the quasi-isometric embedding inequalities. Thus being a quasi-isometrically embedded subgroup is independent of $S$ and $T$.

Solved by gpt-5.6-sol high.

Let $h\in H$ and letbe a geodesic in the Cayley graph ${\mathrm{Cay}}_S(G)$. By $K$-quasiconvexity choose $h_i\in H$ with $d_S(x_i,h_i)\le K$, taking $h_0=1$ and $h_N=h$. ThenThe elements $h_i^{-1}{h}_{i+1}$ telescope to $h$, so the finite setgenerates $H$.

Moreover $|h{|}_U\le |h{|}_S$, while $|h{|}_S\le (2K+1)|h{|}_U$. Thus the inclusion $(H,d_U)\to (G,d_S)$ is a quasi-isometric embedding, and $H$ is quasi-isometrically embedded.

Solved by gpt-5.6-sol high.

Take $G={\mathbb{Z}}^2$ with the standard generating set $S=\{(1,0),(0,1)\}$, and letIntrinsic distance in $H$ between $0$ and $(n,n)$ is $|n|$, while its ambient word metric distance is $2|n|$, so $H$ is quasi-isometrically embedded. However, the ambient geodesic from $(0,0)$ to $(n,n)$ that first travels to $(n,0)$ and then to $(n,n)$ contains $(n,0)$. Its distance from the diagonal subgroup $H$ is $n$. No uniform $K$ can contain every such geodesic in the $K$-neighborhood of $H$, so $H$ is not a quasiconvex subgroup.

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Choose a finite generating set $T$ of $H$. A $d_T$-geodesic between two elements of $H$ maps under the inclusion to a uniform quasigeodesic in ${\mathrm{Cay}}_S(G)$ because $H$ is quasi-isometrically embedded. Since $G$ is a hyperbolic group, the Morse lemma for quasi-geodesics gives a constant $R$ such that this quasigeodesic and the ambient geodesic with the same endpoints have Hausdorff distance at most $R$. Every vertex of the former lies in $H$, so the latter lies in the closed $R$-neighborhood of $H$. Therefore $H$ is a quasiconvex subgroup.

Solved by gpt-5.6-sol high.