Example: $G$ is quasi-isometric to 1 if and only if $G$ is finite.

Definition: A metric space $X$ is proper if closed balls of finite radius in $X$ are compact.  The action of a group $\Gamma$ on a metric space $X$ is cocompact if $X/\Gamma$ is compact in the quotient topology.

The Švarc-Milnor Lemma: Let $X$ be a proper geodesic metric space.  Let $\Gamma$ act cocompactly and properly discontinuously on $X$.  (Properly discontinuously means that for all compact $K\subseteq X, |\{\gamma\in \Gamma | \gamma K\cap K \neq \emptyset \}| < \infty$.)  Then $\Gamma$ is finitely generated and, for any $x_0\in X$, the map $\Gamma \rightarrow X$ $\gamma \mapsto \gamma.\ x_0$

is a quasi-isometry (where $\Gamma$ is equipped with the word metric).

Proof: We may assume that $\Gamma$ is infinite and $X$ is non-compact.  Let $R$ be large enough that the $\Gamma$-translates of $B=B(X, R)$ cover $X$.  Set $S=\{s\in \Gamma\setminus 1 | s\bar{B}\cap \bar{B} \neq \emptyset \}$

Let $r=\inf\{d(\bar{B},\gamma\bar{B}) | \gamma\in\Gamma , \gamma \not\in S \cup \{1\} \}$.  Let $\lambda=\max_{s\in S} d(x_0, sx_0)$.   We want to prove that:
(a) $S$ generates,
(b) $\forall \gamma\in\Gamma$, $\underbrace{ \lambda^{-1}d(x_0,\gamma x_0)}_{i} \leq \ell_S(\gamma)\leq \underbrace{ \frac{1}{r}d(x_0,\gamma x_0) + 1}_{ii}$
(c) $\forall x\in X$, there exists $\gamma\in\Gamma$ such that $d(x,\gamma x_0) \leq R$

Note: $d_S(1,\gamma)=\ell_S(\gamma)$ and $d_S(\gamma, \delta)=\ell_S(\gamma^{-1}\delta)$.

(c) is obvious.
(b-i) is also obvious.
To complete the proof we need to show (a) and (b-ii).

Assume $\gamma\not\in S\cup\{1\}$.  Let $k$ be such that $R + (k-1)r \leq d(x_0, \gamma x_0) < R + kr$

As $\gamma\not\in S\cup\{1\}$, $k>1$.  Choose $x_1, \cdots, x_{k+1} =\gamma x_0$, such that $d(x_0, x_1) and $d(x_i, x_{i+1}) for each $i>0$.   Choose $1=\gamma_0,\gamma_1,\cdots, \gamma_{k-1},\gamma_k=\gamma$ such that $x_{i+1}\in\gamma_i\bar{B}$ for each $i$.  Let $s_i=\gamma_{i-1}^{-1}\gamma_i$, so $\gamma=s_1\cdots s_k$.  Now $d(\bar{B}, s_i\bar{B}) \leq d(\gamma_{i-1}^{-1} x_i, s_i\gamma_i^{-1}x_{i+1}) = d(x_i, x_{i+1})

So, $s_i\in S\cup\{1\}$.  Therefore $S$ generates $\Gamma$.

Also, $\ell_S(\gamma) \leq k \leq \frac{1}{r}d(x_0, \gamma x_0) + \frac{r-R}{r} < \frac{1}{r}d(x_0,\gamma x_0)+1$

as required.

Corollary: If $K\subseteq \Gamma$ is a finite index subgroup of a finitely generated group then $K$ is quasi-isometric to $\Gamma$.

Two groups $G_1$ and $G_2$ are commensurable if they have isomorphic subgroups of finite index.  Clearly, if $G_1$ and $G_2$ are commensurable then they are quasi-isometric.

Example: $Sol = \mathbb{R}^2 \rtimes_E \mathbb{R}$.
Semidirect product is taken over the matrix This means that $Sol=\{(x,y,t) | x,y,t\in\mathbb{R} \}$, but $(x,y,t)(x',y',t')=(x+e^tx', y+e^{-1}y', t+t')$

Let $A\in SL_2(\mathbb{Z})$ with eigenvalues $\lambda, \lambda^{-1}$ with $\lambda >1$.  Let $\Gamma=\mathbb{Z}^2 \rtimes_A \mathbb{Z}$. $\Gamma_A$ sits inside $Sol$ as a uniform lattice, meaning $Sol/\Gamma_A$ is a compact space.

Exercise 11: What is this quotient?

So, $\Gamma_A$ is a quasi-isomorphic to $Sol$  But, Bridson-Gersten showed that $\Gamma_A$ and $\Gamma_{A'}$ are commensurable if and only if the corresponding eigenvalues $\lambda, \lambda'$ have a common power.

Exercise 12: Let $T_k$ be the infinite regular $2k$ valent tree.  Prove that for all $k, k' \geq 2$, $T_k$ is quasi-isometric to $T_{k'}$.