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Edit. Here is an answer to the question that you asked in a comment. I will add more references when I have more time.

For a group $G$ let $r(G)$ denote the least cardinality of a generating set of $G$.

Theorem. For every $n$ there exists a finitely generated group $\Gamma$ such that for every finitely generated group $\Gamma'$ quasi-isometric to $\Gamma$ we have $r(\Gamma')\ge n$.

Proof. I will need several ingredients.

1. Suppose that $G$ is a group, $k$ is a field, $V$ is a $kG$-module, then $$ \dim_k H^1(G; V)\le r(G) \dim_k V $$

2. Below, by an abuse of terminology, a Kleinian group will mean a finitely generated nonelementary discrete subgroup of the isometry group of the hyperbolic 3-space $\mathbb H^3$ with nonempty domain of discontinuity $\Omega\subset \mathbb C P^1$ and compact quotient $(\mathbb H^3 \cup \Omega)/\Gamma$. I will use the notation $\Lambda=\Lambda(\Gamma)$ for the limit set of $\Gamma$, $\lambda$ is the complement of $\Omega$ in $\mathbb C P^1$.

There exists an 18-dimensional $\mathbb C G$-module $V$ of the group $G=\PSL(2,\mathbb C)$ such that the following holds. Let $\Gamma$ be a Kleinian subgroup of $G=\PSL(2,\mathbb C)$ (i.e. is an orientation-preserving Kleinian group). Then $$ \dim_{\mathbb C} H^1(\Gamma; V)\ne 0. $$ The module $V$ is the sum of two irreducible $\mathbb C G$-modules of dimensions 7 and 11 respectively. See Chapter VII, section 1 in (In the case when $\Gamma$ is torsion-free one can take of course, $V=\mathbb C$.

Kra, Irwin, Automorphic forms and Kleinian groups, Mathematics Lecture Note Series. Reading, Mass.: W. A. Benjamin, Inc., Advanced Book Program. XIV, 464 p. (1972). ZBL0253.30015.

3. Let $\Gamma_1, \Gamma_2$ be two Kleinian groups whose limit sets are homeomorphic to Sierpinski carpets and each component of $\Omega(\Gamma_i)$, $i=1,2$, is a round disk. Then $\Gamma_1, \Gamma_2$ are quasiisometric to each other if and only if they are commensurable, in the sense that there exists an isometry $g$ of $\mathbb H^3$ such that $$ \Gamma_1\cap g \Gamma_2 g^{-1} $$ has finite index in both $\Gamma_1, g \Gamma_2 g^{-1}$. (This, as well as item 5 below, in different degree of generality, is due to Schwartz, Kleiner, Bonk and others.) Note that the groups $\Gamma_1, \Gamma_2$ are necessarily nonelementary Gromov-hyperbolic.

4. There are infinitely many Kleinian groups $\Gamma_i$ as in 3 which are pairwise noncommensurable (in the same sense as in 3). To construct such groups one uses, for instance, compact arithmetic hyperbolic manifolds of the simplest type whose fundamental groups are pairwise noncommensurable. These manifolds contain (nonseparating) totally-geodesic compact surfaces. Cutting these manifolds open along those surfaces results in Kleinian groups $\Gamma_i$.

5. Suppose that $\Gamma'$ is a finitely generated group quasi-isometric to a Kleinian group $\Gamma$ as in 3. Then $\Gamma'$ acts isometrically properly discontinuously on the hyperbolic 3-space $\mathbb H^3$, so that the image $\bar{\Gamma}$ of $\Gamma'$ in the isometry group of $\mathbb H^3$ has the same limit set as $\Gamma$ and $\bar{\Gamma}\cap \Gamma$ has finite index in both $\Gamma$ and $\bar{\Gamma}$. In other words, modulo a finite normal subgroup, $\Gamma'$ is isomorphic to a Kleinian group $\bar\Gamma$ of the same type as in 3.

6. Suppose that $\Gamma_1,...,\Gamma_n$ are nonelementary hyperbolic groups which pairwise non-quasi-isometric to each other. Then every $(L,A)$-quasi-isometry of $$ \Gamma= \Gamma_1\times...\times\Gamma_n $$ is within $C(L,A)$-distance from a quasi-isometry that preserves the product decomposition, i.e. is of the form $$ f(x_1,...,x_n)=(f_1(x_1),...,f_n(x_n)), $$ where each $f_i: \Gamma_i\to \Gamma_i$ is an $(L',A')$-quasi-isometry. In particular, if $\Gamma'$ is a finitely generated group quasi-isometric to $\Gamma$, then $\Gamma'$ is isomorphic to a product $$ \Gamma'_1\times ... \times \Gamma_n' $$ with $\Gamma_i'$ quasi-isometric to $\Gamma_i$, $i=1,...,n$.

Given all this, here is how one proves the theorem. Given $n$, one takes pairwise noncommensurable Kleinian groups $\gamma_i, i=1,...,2n$, as in 3. Take $$ \Gamma= \Gamma_1\times...\times\Gamma_{2n}. $$ Let $\Gamma'$ be a group quasi-isometric to $\Gamma$. Then, modulo a finite normal subgroup, $\Gamma'$ is isomorphic to
$$ \Gamma'_1\times ... \times \Gamma_{2n}' $$ where each $\Gamma_i'$ is a Kleinian group commensurable to $\Gamma_i$, $i=1,...,2n$. Now, each $\Gamma'_i$ is either (i) a subgroup of $G=\PSL(2,\mathbb C)$ or (ii) it admits a nontrivial homomorphism to $\mathbb Z_2$ (the orientation-homomorphism), depending on where it preserve the orientation on $\mathbb H^3$. Suppose that $n$ of the direct factors in $\Gamma'$ are of type (ii). Then $$ H^1(\Gamma'; \mathbb Z_2)\ge n $$ and, hence, $r(\Gamma')\ge n$. Alternatively, suppose that $n$ of the direct factors in $\Gamma'$ are of type (i). After relabeling, we have that $\Gamma_1',...,\Gamma'_n$ satisfy this. It suffices to prove that their direct product $$ \Gamma'':=\Gamma'_1\times ... \times \Gamma_{n}' $$ satisfies $r(\Gamma')\ge n$. Let $V$ denote the of $\mathbb C G$-module as in 2. Then $V$ is also a ${\mathbb C} \Gamma''$-module and, by the Kunneth formula, we have $$ \dim_{\mathbb C} H^1(\Gamma''; V)\ge \sum_{i=1}^n H^1(\Gamma'_i; V)\ge n. $$ Hence, $r(\Gamma')\ge r(\Gamma'')\ge n$. qed

Let $G_1,...,G_m$ be pairwise non-isomorphic maximal lattices in isometry groups of real-hyperbolic spaces $\mathbb H^{k_i}$ (possibly of different dimensions $k_i\ge 3$) with nontrivial abelianizations. (Note that every lattice is contained in a maximal lattice as a finite index subgroup.) For instance, we can take some pairwise noncommensurable arithmetic subgroups of the simplest type which do not preserve orientation: All such lattices have nontrivial homomorphisms to $\mathbb Z_2$. Take $$ G=G_1\times ... \times G_m. $$ Suppose that we have an inclusion $G\to H$ as a finite index subgroup. Then, by Mostow Rigidity, the group $H$ acts isometrically (possibly, with finite kernel) on the product of hyperbolic spaces $$ X=\mathbb H^{k_1}\times ... \times \mathbb H^{k_m} $$ extending the product action of $G$. In view of maximality of each $G_i$ (and the assumption that these groups are pairwise nonisomorphic), this action of $H$ cannot permute the factors and, thus, again, by maximality, the action has to be equal to the action of $G_i$ on the corresponding factor in the product. Thus, $H$ is an extension of $G$ by a finite subgroup $K$, the kernel of the action of $H$ on $X$. The group $G$ and, hence, $H$, has a surjective homomorphism to a product of $m$ nontrivial abelian groups (abelianizations of the groups $G_i$). Thus, the rank of $H$ (the minimal number of generators) is at least $m$. The same argument works when we allow some $k_i=2$ as long as the corresponding group $G_i$ is a triangle reflection group.

1. Suppose that $G$ is a group, $k$ is a field, $V$ is a $kG$-module, then $$ dim_k H^1(G; V)\le r(G) dim_k V $$

2. Below, by an abuse of terminology, a Kleinian group will mean a finitely generated nonelementary discrete subgroup of the isometry group of the hyperbolic 3-space $\mathbb H^3$ with nonempty domain of discontinuity $\Omega\subset \mathbb C P^1$ and compact quotient $(\mathbb H^3 \cup \Omega)/\Gamma$. I will use the notation $\Lambda=\Lambda(\Gamma)$ for the limit set of $\Gamma$, $\lambda$ is the complement of $\Omega$ in $\mathbb C P^1$.

There exists an 18-dimensional $\mathbb C G$-module $V$ of the group $G=PSL(2,\mathbb C)$ such that the following holds. Let $\Gamma$ be a Kleinian subgroup of $G=PSL(2,\mathbb C)$ (i.e. is an orientation-preserving Kleinian group). Then $$ dim_{\mathbb C} H^1(\Gamma; V)\ne 0. $$ The module $V$ is the sum of two irreducible $\mathbb C G$-modules of dimensions 7 and 11 respectively. See Chapter VII, section 1 in (In the case when $\Gamma$ is torsion-free one can take of course, $V=\mathbb C$.

Given all this, here is how one proves the theorem. Given $n$, one takes pairwise noncommensurable Kleinian groups $\gamma_i, i=1,...,2n$, as in 3. Take $$ \Gamma= \Gamma_1\times...\times\Gamma_{2n}. $$ Let $\Gamma'$ be a group quasi-isometric to $\Gamma$. Then, modulo a finite normal subgroup, $\Gamma'$ is isomorphic to
$$ \Gamma'_1\times ... \times \Gamma_{2n}' $$ where each $\Gamma_i'$ is a Kleinian group commensurable to $\Gamma_i$, $i=1,...,2n$. Now, each $\Gamma'_i$ is either (i) a subgroup of $G=PSL(2,\mathbb C)$ or (ii) it admits a nontrivial homomorphism to $\mathbb Z_2$ (the orientation-homomorphism), depending on where it preserve the orientation on $\mathbb H^3$. Suppose that $n$ of the direct factors in $\Gamma'$ are of type (ii). Then $$ H^1(\Gamma'; \mathbb Z_2)\ge n $$ and, hence, $r(\Gamma')\ge n$. Alternatively, suppose that $n$ of the direct factors in $\Gamma'$ are of type (i). After relabeling, we have that $\Gamma_1',...,\Gamma'_n$ satisfy this. It suffices to prove that their direct product $$ \Gamma'':=\Gamma'_1\times ... \times \Gamma_{n}' $$ satisfies $r(\Gamma')\ge n$. Let $V$ denote the of $\mathbb C G$-module as in 2. Then $V$ is also a ${\mathbb C} \Gamma''$-module and, by the Kunneth formula, we have $$ dim_{\mathbb C} H^1(\Gamma''; V)\ge \sum_{i=1}^n H^1(\Gamma'_i; V)\ge n. $$ Hence, $r(\Gamma')\ge r(\Gamma'')\ge n$. qed

$\DeclareMathOperator\PSL{PSL}$Let $G_1,...,G_m$ be pairwise non-isomorphic maximal lattices in isometry groups of real-hyperbolic spaces $\mathbb H^{k_i}$ (possibly of different dimensions $k_i\ge 3$) with nontrivial abelianizations. (Note that every lattice is contained in a maximal lattice as a finite index subgroup.) For instance, we can take some pairwise noncommensurable arithmetic subgroups of the simplest type which do not preserve orientation: All such lattices have nontrivial homomorphisms to $\mathbb Z_2$. Take $$ G=G_1\times ... \times G_m. $$ Suppose that we have an inclusion $G\to H$ as a finite index subgroup. Then, by Mostow Rigidity, the group $H$ acts isometrically (possibly, with finite kernel) on the product of hyperbolic spaces $$ X=\mathbb H^{k_1}\times ... \times \mathbb H^{k_m} $$ extending the product action of $G$. In view of maximality of each $G_i$ (and the assumption that these groups are pairwise nonisomorphic), this action of $H$ cannot permute the factors and, thus, again, by maximality, the action has to be equal to the action of $G_i$ on the corresponding factor in the product. Thus, $H$ is an extension of $G$ by a finite subgroup $K$, the kernel of the action of $H$ on $X$. The group $G$ and, hence, $H$, has a surjective homomorphism to a product of $m$ nontrivial abelian groups (abelianizations of the groups $G_i$). Thus, the rank of $H$ (the minimal number of generators) is at least $m$. The same argument works when we allow some $k_i=2$ as long as the corresponding group $G_i$ is a triangle reflection group.

1. Suppose that $G$ is a group, $k$ is a field, $V$ is a $kG$-module, then $$ \dim_k H^1(G; V)\le r(G) \dim_k V $$

2. Below, by an abuse of terminology, a Kleinian group will mean a finitely generated nonelementary discrete subgroup of the isometry group of the hyperbolic 3-space $\mathbb H^3$ with nonempty domain of discontinuity $\Omega\subset \mathbb C P^1$ and compact quotient $(\mathbb H^3 \cup \Omega)/\Gamma$. I will use the notation $\Lambda=\Lambda(\Gamma)$ for the limit set of $\Gamma$, $\lambda$ is the complement of $\Omega$ in $\mathbb C P^1$.

There exists an 18-dimensional $\mathbb C G$-module $V$ of the group $G=\PSL(2,\mathbb C)$ such that the following holds. Let $\Gamma$ be a Kleinian subgroup of $G=\PSL(2,\mathbb C)$ (i.e. is an orientation-preserving Kleinian group). Then $$ \dim_{\mathbb C} H^1(\Gamma; V)\ne 0. $$ The module $V$ is the sum of two irreducible $\mathbb C G$-modules of dimensions 7 and 11 respectively. See Chapter VII, section 1 in (In the case when $\Gamma$ is torsion-free one can take of course, $V=\mathbb C$.

Given all this, here is how one proves the theorem. Given $n$, one takes pairwise noncommensurable Kleinian groups $\gamma_i, i=1,...,2n$, as in 3. Take $$ \Gamma= \Gamma_1\times...\times\Gamma_{2n}. $$ Let $\Gamma'$ be a group quasi-isometric to $\Gamma$. Then, modulo a finite normal subgroup, $\Gamma'$ is isomorphic to
$$ \Gamma'_1\times ... \times \Gamma_{2n}' $$ where each $\Gamma_i'$ is a Kleinian group commensurable to $\Gamma_i$, $i=1,...,2n$. Now, each $\Gamma'_i$ is either (i) a subgroup of $G=\PSL(2,\mathbb C)$ or (ii) it admits a nontrivial homomorphism to $\mathbb Z_2$ (the orientation-homomorphism), depending on where it preserve the orientation on $\mathbb H^3$. Suppose that $n$ of the direct factors in $\Gamma'$ are of type (ii). Then $$ H^1(\Gamma'; \mathbb Z_2)\ge n $$ and, hence, $r(\Gamma')\ge n$. Alternatively, suppose that $n$ of the direct factors in $\Gamma'$ are of type (i). After relabeling, we have that $\Gamma_1',...,\Gamma'_n$ satisfy this. It suffices to prove that their direct product $$ \Gamma'':=\Gamma'_1\times ... \times \Gamma_{n}' $$ satisfies $r(\Gamma')\ge n$. Let $V$ denote the of $\mathbb C G$-module as in 2. Then $V$ is also a ${\mathbb C} \Gamma''$-module and, by the Kunneth formula, we have $$ \dim_{\mathbb C} H^1(\Gamma''; V)\ge \sum_{i=1}^n H^1(\Gamma'_i; V)\ge n. $$ Hence, $r(\Gamma')\ge r(\Gamma'')\ge n$. qed

Edit. Here is an answer to the question that you asked in a comment. I will add more references when I have more time.

For a group $G$ let $r(G)$ denote the least cardinality of a generating set of $G$.

Theorem. For every $n$ there exists a finitely generated group $\Gamma$ such that for every finitely generated group $\Gamma'$ quasi-isometric to $\Gamma$ we have $r(\Gamma')\ge n$.

Proof. I will need several ingredients.

1. Suppose that $G$ is a group, $k$ is a field, $V$ is a $kG$-module, then $$ dim_k H^1(G; V)\le r(G) dim_k V $$

2. Below, by an abuse of terminology, a Kleinian group will mean a finitely generated nonelementary discrete subgroup of the isometry group of the hyperbolic 3-space $\mathbb H^3$ with nonempty domain of discontinuity $\Omega\subset \mathbb C P^1$ and compact quotient $(\mathbb H^3 \cup \Omega)/\Gamma$. I will use the notation $\Lambda=\Lambda(\Gamma)$ for the limit set of $\Gamma$, $\lambda$ is the complement of $\Omega$ in $\mathbb C P^1$.

There exists an 18-dimensional $\mathbb C G$-module $V$ of the group $G=PSL(2,\mathbb C)$ such that the following holds. Let $\Gamma$ be a Kleinian subgroup of $G=PSL(2,\mathbb C)$ (i.e. is an orientation-preserving Kleinian group). Then $$ dim_{\mathbb C} H^1(\Gamma; V)\ne 0. $$ The module $V$ is the sum of two irreducible $\mathbb C G$-modules of dimensions 7 and 11 respectively. See Chapter VII, section 1 in (In the case when $\Gamma$ is torsion-free one can take of course, $V=\mathbb C$.

Kra, Irwin, Automorphic forms and Kleinian groups, Mathematics Lecture Note Series. Reading, Mass.: W. A. Benjamin, Inc., Advanced Book Program. XIV, 464 p. (1972). ZBL0253.30015.

3. Let $\Gamma_1, \Gamma_2$ be two Kleinian groups whose limit sets are homeomorphic to Sierpinski carpets and each component of $\Omega(\Gamma_i)$, $i=1,2$, is a round disk. Then $\Gamma_1, \Gamma_2$ are quasiisometric to each other if and only if they are commensurable, in the sense that there exists an isometry $g$ of $\mathbb H^3$ such that $$ \Gamma_1\cap g \Gamma_2 g^{-1} $$ has finite index in both $\Gamma_1, g \Gamma_2 g^{-1}$. (This, as well as item 5 below, in different degree of generality, is due to Schwartz, Kleiner, Bonk and others.) Note that the groups $\Gamma_1, \Gamma_2$ are necessarily nonelementary Gromov-hyperbolic.

4. There are infinitely many Kleinian groups $\Gamma_i$ as in 3 which are pairwise noncommensurable (in the same sense as in 3). To construct such groups one uses, for instance, compact arithmetic hyperbolic manifolds of the simplest type whose fundamental groups are pairwise noncommensurable. These manifolds contain (nonseparating) totally-geodesic compact surfaces. Cutting these manifolds open along those surfaces results in Kleinian groups $\Gamma_i$.

5. Suppose that $\Gamma'$ is a finitely generated group quasi-isometric to a Kleinian group $\Gamma$ as in 3. Then $\Gamma'$ acts isometrically properly discontinuously on the hyperbolic 3-space $\mathbb H^3$, so that the image $\bar{\Gamma}$ of $\Gamma'$ in the isometry group of $\mathbb H^3$ has the same limit set as $\Gamma$ and $\bar{\Gamma}\cap \Gamma$ has finite index in both $\Gamma$ and $\bar{\Gamma}$. In other words, modulo a finite normal subgroup, $\Gamma'$ is isomorphic to a Kleinian group $\bar\Gamma$ of the same type as in 3.

6. Suppose that $\Gamma_1,...,\Gamma_n$ are nonelementary hyperbolic groups which pairwise non-quasi-isometric to each other. Then every $(L,A)$-quasi-isometry of $$ \Gamma= \Gamma_1\times...\times\Gamma_n $$ is within $C(L,A)$-distance from a quasi-isometry that preserves the product decomposition, i.e. is of the form $$ f(x_1,...,x_n)=(f_1(x_1),...,f_n(x_n)), $$ where each $f_i: \Gamma_i\to \Gamma_i$ is an $(L',A')$-quasi-isometry. In particular, if $\Gamma'$ is a finitely generated group quasi-isometric to $\Gamma$, then $\Gamma'$ is isomorphic to a product $$ \Gamma'_1\times ... \times \Gamma_n' $$ with $\Gamma_i'$ quasi-isometric to $\Gamma_i$, $i=1,...,n$.

Given all this, here is how one proves the theorem. Given $n$, one takes pairwise noncommensurable Kleinian groups $\gamma_i, i=1,...,2n$, as in 3. Take $$ \Gamma= \Gamma_1\times...\times\Gamma_{2n}. $$ Let $\Gamma'$ be a group quasi-isometric to $\Gamma$. Then, modulo a finite normal subgroup, $\Gamma'$ is isomorphic to
$$ \Gamma'_1\times ... \times \Gamma_{2n}' $$ where each $\Gamma_i'$ is a Kleinian group commensurable to $\Gamma_i$, $i=1,...,2n$. Now, each $\Gamma'_i$ is either (i) a subgroup of $G=PSL(2,\mathbb C)$ or (ii) it admits a nontrivial homomorphism to $\mathbb Z_2$ (the orientation-homomorphism), depending on where it preserve the orientation on $\mathbb H^3$. Suppose that $n$ of the direct factors in $\Gamma'$ are of type (ii). Then $$ H^1(\Gamma'; \mathbb Z_2)\ge n $$ and, hence, $r(\Gamma')\ge n$. Alternatively, suppose that $n$ of the direct factors in $\Gamma'$ are of type (i). After relabeling, we have that $\Gamma_1',...,\Gamma'_n$ satisfy this. It suffices to prove that their direct product $$ \Gamma'':=\Gamma'_1\times ... \times \Gamma_{n}' $$ satisfies $r(\Gamma')\ge n$. Let $V$ denote the of $\mathbb C G$-module as in 2. Then $V$ is also a ${\mathbb C} \Gamma''$-module and, by the Kunneth formula, we have $$ dim_{\mathbb C} H^1(\Gamma''; V)\ge \sum_{i=1}^n H^1(\Gamma'_i; V)\ge n. $$ Hence, $r(\Gamma')\ge r(\Gamma'')\ge n$. qed