Suppose that two finitely generated groups quasi-isometrically embed into each other. Does it follow that the two groups are quasi-isometric? Recall that a quasi-isometry is a quasi-isometric embedding that is quasi-surjective, see e.g. https://www.math.ucdavis.edu/~kapovich/EPR/pc_lectures3.pdf
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2$\begingroup$ The answer is certainly negative, the question is only what's the cost for a reasonable answer. Here are a few remarks: 1) the answer is obviously negative for, say, connected graphs of bounded degree (e.g., consider a 3-regular tree and the wedge of the latter and a half-line. 2) If $G$ is the first Grigorchuk group, then $G$ is QI to $G\times G$ and contains a QI-embedded line. Hence $G$ and $G\times\mathbf{Z}$ QI-embed into each other. I guess they are not QI but I can't prove it. $\endgroup$– YCorNov 1, 2014 at 22:55
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1$\begingroup$ 3) Let $H$ be the infinite-dimensional space $\ell^1$ (thanks Alain for a correction, I first said Hilbert by mistake), and let $S=SOL$ (some polycyclic group of Hirsch length 3 and exponential growth). Then $H\times S$ and $H$ embed QI into each other, but are not QI (because only one of the two has simply connected asymptotic cones). I expect this argument could be done by replacing $H$ with some finitely generated group $L$: I'd need a f.g. group $L$ with $L$ containing a QI-embedded tree, with $L\times L$ embedding QI into $L$, and $L$ having a simply connected asymptotic cone. $\endgroup$– YCorNov 2, 2014 at 21:17
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$\begingroup$ Since there're recent interest to this question, I'd be curious about the case when we assume the groups to be word-hyperbolic. $\endgroup$– YCorJul 1, 2018 at 17:11
2 Answers
Let $C_n$ be a cyclic groups of order $n$. Then the wreath products $C_2\wr\mathbf{Z}$ and $C_3\wr\mathbf{Z}$ embed QI into each other (for the reverse direction, observe that $C_2\wr\mathbf{Z}$ has a subgroup of index 2 isomorphic to $C_2^2\wr\mathbf{Z}$). But $C_2\wr\mathbf{Z}$ and $C_3\wr\mathbf{Z}$ are not QI, this follows from work of Eskin-Fisher-Whyte ($F\wr\mathbf{Z}$ and $F'\wr\mathbf{Z}$ for $F,F'$ finite groups are QI iff $|F|,|F'|$ have some common power.)
1) LAMPLIGHTER GROUPS
As mentioned by Yves, lamplighter groups over $\mathbb{Z}$ provide counterexamples thanks to Eskin, Fisher, and Whyte's work. Other counterexamples are given by lamplighter groups over one-ended groups thanks to the recent preprint arxiv:2105.04878. More precisely, if $F_1,F_2$ are two finite groups and if $H$ is a finitely presented one-ended group, then $F_1\wr H$ and $F_2 \wr H$ quasi-isometrically embed into each other, but they are not quasi-isometric if $|F_1|,|F_2|$ do not have the same prime divisors. (The converse holds if $H$ is non-amenable. If $H$ is amenable, $|F_1|,|F_2|$ must be powers of a common number and $H$ must admit specific auto-quasi-isometries (which always the case if $H$ is free abelian).) The advantage of these lampligthers is that one can avoid the heavy machinery of coarse differentiation.
2) COXETER GROUPS
Counterexamples can also be found among Coxeter groups. I just give one example, based on the following statement:
Proposition: Let $\Gamma$ be a finite simplicial graph and $u \in V(\Gamma)$ a vertex. Let $\Gamma(u)$ denote the graph obtained by gluing two copies of $\Gamma \backslash \{u\}$ along $\mathrm{link}(u)$. Then the right-angled Coxeter groups $C(\Gamma)$ and $C(\Gamma(u))$ are commensurable.
Sketch of proof. Let $u_1,\ldots, u_k$ denote the vertices of $\Gamma \backslash \mathrm{star}(u)$. Then, by a standard ping-pong argument, one proves that the reflexion subgroup $\langle u_1, \ldots, u_k,u_1^u, \ldots, u_k^u \rangle$ admits $\{u_1, \ldots, u_k,u_1^u, \ldots, u_k^u\}$ as a basis. The commutation graph of this family is precisely $\Gamma(u)$. $\square$
(Some details can be found in arxiv:1910.04230; see Proposition 3.14.)
Now, we apply the proposition to the following graph:
Clearly, $\Gamma(u)$ contains an induced copy of $\Gamma \sqcup \{ \mathrm{pt} \}$, so $C(\Gamma)$ quasi-isometrically embeds into the free product $C(\Gamma) \ast \mathbb{Z}/2\mathbb{Z}$. The reverse quasi-isometric embedding is clear. But $C(\Gamma)$ is one-ended, because $\Gamma$ has no separating complete subgraph, so $C(\Gamma)$ and $C(\Gamma) \ast \mathbb{Z}/2\mathbb{Z}$ cannot be quasi-isometric.
3) RIGHT-ANGLED ARTIN GROUPS
Given a simplicial graph $\Gamma$, the associated right-angled Artin group is defined by the following presentation: $$A_\Gamma = \langle V(\Gamma) \mid [u,v]=1, \ \{ u,v \} \in E(\Gamma) \rangle$$ where $V(\Gamma)$ and $E(\Gamma)$ denote the vertex- and edge-sets of $\Gamma$ respectively.
First, a general statement. A graph $\Gamma$ is join if there exists a partition $V(\Gamma)= A \sqcup B$ such that $A,B$ are both non-empty and such that any vertex of $A$ is adjacent to any vertex of $B$.
Proposition: Let $\Gamma$ be a finite simplicial graph which is not a join and which is not reduced to a single vertex. Then $A_\Gamma$ contains a quasi-isometrically embedded subgroup isomorphic to the free product $A_\Gamma \ast \mathbb{Z}$.
Sketch of proof. Let $\Gamma^e$ denote the extension graph of $\Gamma$, ie., the graph whose vertices are the conjugates $gug^{-1}$, where $g \in A_\Gamma$ and $u \in V(\Gamma)$, and whose edges link two elements when they commute. In Embedability between right-angled Artin groups, Kim and Koberda prove that, if $\Lambda$ is a finite induced subgraph of $\Gamma^e$, then $A_\Lambda$ embeds into $A_\Gamma$. It turns out that $\Gamma^e$ is unbounded since $\Gamma$ is not a join, so that $\Gamma^e$ contains an induced subgraph isomorphic to $\Gamma \sqcup \{ \text{vertex} \}$, hence $A_\Gamma \ast \mathbb{Z} \leq A_\Gamma$.
I don't know if it follows from their argument that the embedding is quasi-isometric, but, in the alternative proof I gave of their theorem in my thesis (see Section 8.5), it is not difficult to show that the embedding I construct is quasi-isometric. $\square$
Corollary: If $\Gamma$ is a finite connected simplicial graph which is not a join, then $A_\Gamma$ and $A_\Gamma \ast \mathbb{Z}$ quasi-isometrically embeds into each other, but they are not quasi-isometric.
The fact that $A_\Gamma$ and $A_\Gamma \ast \mathbb{Z}$ are not quasi-isometric follows from the observation that $A_\Gamma$ is one-ended (as $\Gamma$ is connected and contains at least two vertices).
Now, a concrete example. Consider the right-angled Artin group $$A= \langle a,b,c,d \mid [a,b]=[b,c]=[c,d]=1 \rangle.$$ Of course, $A$ quasi-isometrically embeds into $A \ast \mathbb{Z}$. Conversely, an embedding as mentioned above shows that the subgroup $$\langle a,b,c,d^2,ada^{-1} \rangle= \langle a,b,c,d^2 \rangle \ast \langle ada^{-1} \rangle \simeq A \ast \mathbb{Z}$$ is quasi-isometrically embedded into $A$, which can also be proved directly. However, $A$ and $A \ast \mathbb{Z}$ are not quasi-isometric since $A$ is one-ended.
4) A HYPERBOLIC EXAMPLE
According to arxiv:1812.07799, there exist a torsion-free one-ended hyperbolic group $G$ containing two isomorphic quasiconvex subgroups $H_1,H_2 \leq G$ such that $H_1$ (resp. $H_2$) has finite (resp. infinite) index in $G$. As a consequence of a theorem claimed by Gromov (and proved by Arzhantseva), $G$ contains a subgroup isomorphic to the free product $H_2 \ast \mathbb{Z}$. As a consequence, $G$ and $G\ast \mathbb{Z}$ quasi-isometrically embed into each other but they are not quasi-isometric.
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$\begingroup$ To show that $A_\Gamma$ and $A_\Gamma\ast\mathbf{Z}$ are not QI, it's enough to know that $A_\Gamma$ is 1-ended, which, unless I remember incorrectly, precisely holds when the graph is not a join (although I'm not sure to be aware of a completely immediate proof). (Note that the graph is a join iff the complement graph -same vertices, other edges- is non-connected). $\endgroup$– YCorJul 1, 2018 at 17:08
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$\begingroup$ In fact, $A_\Gamma$ is one-ended iff $\Gamma$ is connected and not reduced to a single vertex. But you're right, it is not necessary to use Stallings' theorem, it would be more elementary to use the fact the RAAG is one-ended. $\endgroup$ Jul 1, 2018 at 17:56
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$\begingroup$ Oh you're right, I got confused, connectedness of $\Gamma$ (rather than the complement graph) is the right condition. $\endgroup$– YCorJul 1, 2018 at 21:30
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1$\begingroup$ (I removed Baumslag-Solitar groups from the list of examples. A recent preprint (arxiv:2204.03983) shows that, given two solvable Baumslag-Solitar groups, if one quasi-isometrically embeds in the other, then they must be quasi-isometric. So the situation is more subtle than I first thought.) $\endgroup$ Apr 12, 2022 at 6:25
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$\begingroup$ I also did the same confusion, thinking that they embed into each other. If, say, we consider the fibre product description $b(x)+b(y)=0$ inside $H^2\times T_{1+3}$ for $BS(1,3)$ and restrict to $T_{1+2}$, what we get is not a cocompact space for $BS(1,2)$, but form some non-unimodular group (the scaling of the real part of $b$ is not the right one). $\endgroup$– YCorApr 12, 2022 at 7:13