This question is about the proof of theorem 6.13.12 in Cellular automata and groups, Ceccherini.

Let $G,H$ be finitely generated. We say that $\phi:G\to H$ is a quasi-isometric embedding if

i) For all $A\subseteq G$ finite there exists $B\subseteq H$ finite such that $g_1^{-1}g_2\in A$ implies $\phi(g_1)^{-1}\phi(g_2)\in B$

ii) For all $B\subseteq H$ finite there exists $A\subseteq G$ finite such that $\phi(g_1)^{-1}\phi(g_2)\in B$ implies $g_1^{-1}g_2\in A$.

Then, $\phi$ is quasi-isometric embedding iff there exist $\alpha,\beta$ such that $$\frac{1}{\alpha}d_S(g_1,g_2)-\beta\leq d_T(\phi(g_1),\phi(g_2))\leq\alpha d_S(g_1,g_2) +\beta\tag{1}$$ where $S\subseteq G$, $T\subseteq H$ are finite generating sets and $d_S,d_T$ are the induced metrics.

Proving i) implies the right inequality in (1) is easy: take $A$ to be $S$ and then, for $g_1,g_2\in G$, write $g_1^{-1}g_2=s_1\dots s_n$ as a length-minimum word in $S$, and bound $$d_T(\phi(g_1),\phi(g_2)) \leq \sum_{k=1}^{n-1}d_T(\phi(g_1s_1\dots s_k),\phi(g_1s_1\dots s_{k+1}))\leq |B|n=|B|d(g_1,g_2)$$ where $|B| = \max_{h\in B}d_T(h,e_H)$.

The book says that for ii) implies the right inequality in (1) the proof is similar and is omitted. But I can't complete the details, because an analogous proof would involve taking $B$ as the generating set $T$, but this don't work since we don't know if $T\subseteq \phi(G)$. How can I prove this?

Also, if we add to i) and ii) that $\phi$ is quasi-onto, then (1) follows easily. This made me think that I am trying to prove something false, but I couldn't find a counterexample.


If I have understood correctly then I think the claim is false as stated. For a counterexample, let $$H = \langle x,y,z \mid [x,y]=z,[x,z]=[y,z]=1 \rangle$$ be the Heisenberg group, let $G$ be its subgroup $\langle z \rangle$, and let $\phi(z)=z$.

Then (i) is true with $B=A$. For (ii), note that $\phi(g_1)^{-1}\phi(g_2) \in B \Rightarrow g_1^{-1}g_2 \in G \cap B$, so (ii) is true with $A = G \cap B$.

We can take $S = \{z\}$ and $T = \{x,y,z\}$ for generating sets of $G$ and $H$ (note that whether (1) is true does not depend on the choice of $S$ and $T$, although changing generating sets could change $\alpha$ and $\beta$). Then $d_S(1,z^{n^2}) = n^2$, but $[x^n,y^n]=z^{n^2}$ in $H$, so $d_T(1,z^2) \le 4n$, and the left equality of (1) fails.

  • $\begingroup$ you are completely right. one million thanks $\endgroup$ – Veridian Dynamics Aug 13 '18 at 12:19

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