1. Introduction
Grigorchuk exhibited continuously many quasi-isometry classes of residually finite three-generator groups by producing continuously many growth types [Reference Grigorchuk4, Thm 7.2]. Continuously many means having the cardinality of $\mathbb{R}$ . Here, we describe another family of such groups by building upon Bowditch’s method for distinguishing quasi-isometry classes [Reference Bowditch2] and use consequences of the theory of special cube complexes to obtain residual finiteness [Reference Agol1].
Consider the rank-2 free group $F_2=\langle a,b\rangle$ . Let $w_n = [a,b^{2^{2^n}}][a^2,b^{2^{2^n}}]\cdots [a^{100},b^{2^{2^n}}]$ for $n\in{\mathbb{N}}$ .
Each subset $S\subseteq{\mathbb{N}}$ is associated to the following group:
In Section 3, we show that $G(S)$ is residually finite when $S\subseteq{\mathbb{N}}_{\gt 100}$ . We also observe that $G(S)$ and $G(S')$ are not quasi-isometric when $S\Delta S'$ is infinite.
In fact, our proof of residual finiteness for $G(S)$ works in precisely the same way to prove the residual finiteness for the original examples of Bowditch having torsion. But it appears to fail for Bowditch’s torsion-free examples. We refer to Remark 3.3.
We also produced an uncountable family of pairwise non-isomorphic residually finite groups in [Reference Chong and Wise3], and perhaps an appropriate subfamily also yields continuously many quasi-isometry classes.
Our simple approach arranges for certain infinitely presented small-cancellation groups to be residually finitely presented small-cancellation groups. This approach is likely to permit the construction of other interesting families of finitely generated groups.
2. Review of Bowditch’s result
We first recall some small-cancellation background. See [Reference Lyndon and Schupp5, Ch.V].
Definition 2.1. For a presentation, a piece $p$ is a word appearing in more than one way among the relators. Note that for a relator $r=q^n$ , subwords that differ by a ${\mathbb{Z}}_n$ -action are regarded as appearing in the same way. A presentation is $C'\!\left(\dfrac{1}{6}\right)$ if $|p|\lt \dfrac{1}{6}|r|$ for any piece $p$ in a relator $r$ .
A major subword $v$ of a relator $r$ is a subword of a cyclic permutation of $r^{\pm }$ with $|v| \gt \dfrac{|r|}{2}$ . A word $u$ is majority-reduced if $u$ does not contain a major subword of a relator. We will use the following well-known property for $C'\!\left(\dfrac{1}{6}\right)$ groups [Reference Lyndon and Schupp5, Ch.V Thm 4.5].
Proposition 2.2. Let $\langle x_1, x_2, \dots \mid r_1, r_2, \dots \rangle$ be a $C'\!\left(\dfrac{1}{6}\right)$ presentation. A non-empty cyclically reduced majority-reduced word in the generators must represent a nontrivial element in the group.
We now recall definitions leading to the main theorem of [Reference Bowditch2]. Let ${\mathbb{N}}^+ = \{n\in{\mathbb{Z}} \;:\; n\geq 1\}$ . Let ${\mathbb{N}}_{\gt k}=\{n\in{\mathbb{N}} \;:\; n\gt k\}$ for some $k\in{\mathbb{N}}^+$ .
Definition 2.3. Two subsets $L, L'\subseteq{\mathbb{N}}^+$ are related if for some $k\geq 1$ :
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1. for any $m\in L$ with $m\gt k$ , there is $m'\in L'$ with $m'\in \left[\dfrac{m}{k}, km\right]$ ; and
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2. for any $m'\in L'$ with $m'\gt k$ , there is $m\in L$ with $m\in \left[\dfrac{m'}{k}, km'\right]$ .
We write $L\sim L'$ if $L$ and $L'$ are related, and write $L\not \sim L'$ otherwise.
Remark 2.4. This is a simplified but equivalent form of Bowditch’s definition [Reference Bowditch2, Def. before Lem 3] who used $m\gt (k+1)^2$ and $m'\gt (k+1)^2$ . The equivalence is easy by proving relatedness via $(k+1)^2$ on one direction, and the other direction is clear since $(k+1)^2\gt k$ .
Remark 2.5. As pointed out by the referee, it is equivalent to say $L, L'\subseteq{\mathbb{N}}^+$ are related if there is $k\geq 1$ such that the sets $M=L\cap{\mathbb{N}}_{\gt k}$ and $M'=L'\cap{\mathbb{N}}_{\gt k}$ satisfy that $|\log M, \log M'|\leq k$ . Here, $|Z, Z'| = \inf \{|z-z'| \;:\; z\in Z,z'\in Z'\}$ denotes the Hausdorff distance between sets $Z$ and $Z'$ . This observation could clarify the proofs below, especially Lemma 3.4, for some readers.
Lemma 2.6. The relation $\sim$ in Definition 2.3 is an equivalence relation on subsets of ${\mathbb{N}}^+$ .
Proof. The relation $\sim$ is reflexive via $k=1$ . The relation $\sim$ is symmetric by definition. Hence, it suffices to show $\sim$ is transitive.
Let $S, S', S''\subseteq{\mathbb{N}}^+$ . Suppose $S\sim S'$ via $k$ and $S'\sim S''$ via $k'$ . We claim that $S\sim S''$ via $kk'$ . Let $m\in S$ with $m\gt kk'$ . There is $m'\in S'$ with $m'\in\!\left[\dfrac{m}{k},km\right]$ by $S\sim S'$ via $k$ , hence $m'\gt k'$ . Then there is $m''\in S''$ with $m''\in \!\left[\dfrac{m'}{k'},k'm'\right]$ by $S'\sim S''$ via $k'$ . Thus, $m''\in \left [\dfrac{m}{kk'}, kk'm\right ]$ . Similarly, there is $m\in \left [\dfrac{m''}{kk'}, kk'm''\right ]$ for any $m''\in S''$ with $m''\gt kk'$ .
Example 2.7. All finite sets are related. All uniform nets are related. $\{2^n\}_{n\in{\mathbb{N}}}\sim \{3^n\}_{n\in{\mathbb{N}}}$ .
For sets $S,S'$ , their symmetric difference is $S\Delta S' = (S - S')\cup (S' - S)$ .
Example 2.8. If $S,S'\subseteq{\mathbb{N}}^+$ with infinite $S\Delta S'$ , then $\{2^{2^n}\}_{n\in S} \not \sim \{2^{2^m}\}_{m\in S'}$ [Reference Bowditch2, Lem 4].
With the notion of $\sim$ , the following is a simplified version of the main theorem in [Reference Bowditch2].
Theorem 2.9. Let $G$ and $G'$ be the finitely generated $C'\!\left(\dfrac{1}{6}\right)$ groups presented below. If $G$ is quasi-isometric to $G'$ , then $\{|w_i|\}_{i\in I} \sim \{|w'_{\!\!j}|\}_{j\in J}$ :
3. Proving the family of groups have desired properties
3.1. Small cancellation
Proposition 3.1. For any infinite subset $S\subseteq{\mathbb{N}}_{\gt 100}$ , the associated group $G(S)$ is $C'\!\left(\dfrac{1}{6}\right)$ . Furthermore, $G_k(S) = \left \langle a,b\ \middle|\ b^{2^{2^{k}}}, w_n \;:\; n\in S, n\lt k\right \rangle$ is $C'\!\left(\dfrac{1}{6}\right)$ for each $k\in{\mathbb{N}}$ .
Proof. For the first statement, it suffices to show that $w_n$ and $w_m$ have small overlap for $n\gt m\gt 100$ . The longest piece between $w_n$ and $ w_m$ is $b^{-2^{2^{m}}}a^{100}b^{2^{2^{m}}}$ . Thus, $C'\!\left(\dfrac{1}{6}\right)$ holds since:
For the second statement, we additionally show that $w_n$ and $b^{2^{2^k}}$ satisfy the $C'\!\left(\dfrac{1}{6}\right)$ condition for $100\lt n\lt k$ . Their longest piece is $b^{2^{2^n}}$ , which is shorter than $\dfrac{1}{6}$ of the lengths of $w_n$ and $b^{2^{2^k}}$ .
3.2. Residual finiteness
Observe that $G_k(S)=G(S)/\langle \!\langle b^{2^{2^k}}\rangle \!\rangle$ since $w_m\in \langle \!\langle b^{2^{2^k}}\rangle \!\rangle$ for $m\geq k$ . Indeed, $w_n = \left[a,b^{2^{2^n}}\right]\left[a^2,b^{2^{2^n}}\right]\cdots$ $\left[a^{100},b^{2^{2^n}}\right]$ is trivialised when $b^{2^{2^n}}$ becomes trivial.
Proposition 3.2. For any infinite subset $S\subseteq{\mathbb{N}}_{\gt 100}$ , the associated group $G(S)$ is residually finite.
Proof. Since $G_k(S)$ is a finitely presented $C'\!\left(\dfrac{1}{6}\right)$ group, the hyperbolic group $G_k(S)$ is cocompactly cubulated by [Reference Wise6]. Thus, $G_k(S)$ is residually finite by [Reference Agol1].
Each $g\in G(S)-\{1\}$ is represented by a cyclically reduced word $v$ with minimal length. Then $v$ is majority-reduced since otherwise, $v$ contains a major subword of a relator, which can reduce the length of $v$ . Moreover, $v$ does not contain a majority subword of $b^{2^{2^{|v|}}}$ since $|v|\lt \dfrac{1}{2}\cdot 2^{2^{|v|}}=\dfrac{1}{2}\left |b^{2^{2^{|v|}}}\right |$ . Hence, $v\neq 1_{G_{|v|}}$ by Proposition 2.2 since $v$ is majority-reduced in $G_{|v|}$ . Thus, $G(S)$ is residually residually finite and hence residually finite.
Remark 3.3. Bowditch’s original examples were $B(S) = \langle a, b \mid \big (a^{2^{2^n}}b^{2^{2^n}}\big )^7 \;:\; n\in S\subseteq{\mathbb{N}}\rangle$ . As in Proposition 3.2, $B(S)$ is residually finite since it is residually finitely presented $C'\!\left(\dfrac{1}{6}\right)$ using the quotients to $B/\langle \!\langle a^{2^{2^n}}, b^{2^{2^n}}\rangle \!\rangle$ for $n\geq 3$ . However, the analogous argument fails for Bowditch’s torsion-free examples $B'(S) = \langle a, b \mid a\big (a^{2^{2^n}}b^{2^{2^n}}\big )^{12} \;:\; n\in S\subseteq{\mathbb{N}}\rangle$ .
3.3. Pairwise non-quasi-isometric
We first prove a lemma about the relation $\sim$ .
Lemma 3.4. $S\sim nS\sim (S+n)$ for $n\in{\mathbb{N}}^+$ and $S\subseteq{\mathbb{N}}^+$ .
Proof. First, $S\sim nS$ via $n$ . Indeed, for any $s\in S$ , $ns\in \left [\dfrac{s}{n},ns\right ]$ ; for any $ns\in nS$ , $s\in \left [\dfrac{ns}{n},n\cdot ns\right ]$ .
Moreover, $S\sim (S+n)$ via $n+1$ . For any $s\in S$ , $s+n\leq (n+1)s$ , so $s+n\in \left [\dfrac{s}{n+1},(n+1)s\right ]$ . On the other hand, for $s+n\in S+n$ , $(n+1)s\geq s+n$ implies $s\geq \dfrac{s+n}{n+1}$ . Hence, $s\in \left [\dfrac{s+n}{n+1}, (n+1)(s+n)\right ]$ .
Proposition 3.5. Let $S,S'\subseteq{\mathbb{N}}^+$ have infinite $S\Delta S'$ , then $\{|w_n|\}_{n\in S}\not \sim \{|w_m|\}_{m\in S'}$ .
Proof. $\{|w_n| \;:\; n\in S\} = \{10100+200\cdot 2^{2^n} \;:\; n\in S\} = 10100+200\cdot \{2^{2^n} \;:\; n\in S\}$ . By Lemma 3.4, $\{|w_n| \;:\; n\in S\}\sim \{2^{2^n} \;:\; n\in S\}$ . Similarly, $\{|w_m| \;:\; m\in S'\}\sim \{2^{2^m} \;:\; m\in S'\}$ . By Example 2.8, $\{2^{2^n} \;:\; n\in S\}\not \sim \{2^{2^m} \;:\; m\in S'\}$ , so $\{|w_n|\}_{n\in S}\not \sim \{|w_m|\}_{m\in S'}$ by Lemma 2.6.
Corollary 3.6. If $S,S'\subseteq{\mathbb{N}}_{\gt 100}$ have infinite $S\Delta S'$ , then $G(S)$ and $G(S')$ are not quasi-isometric.
Proof. $\{|w_n|\}_{n\in S}\not \sim \{|w_m|\}_{m\in S'}$ , hence $G(S)$ and $G(S')$ are not quasi-isometric by Theorem 2.9.
For $A,B\subseteq N$ , declare $A\sim _{_\Delta }\!B$ if $\left |A\Delta B\right |\lt \infty$ . As noted by Bowditch, each $\sim _{_\Delta }\!$ equivalence class is countable. Hence, there are continuously many $\sim _{_\Delta }\!$ equivalence classes. Our construction thus produces continuously many pairwise non-quasi-isometric groups $G(S)$ , which are $C'\!\left(\dfrac{1}{6}\right)$ and residually finite.
Acknowledgements
We are grateful to the referee for helpful comments. Research is supported by NSERC.