$\mathrm{SL}(2,\mathbb{Z})$ finitely generated by using the Schwarz-Milnor lemma

Question

Recently, I have been studying the modular group $G=\mathrm{SL}(2,\mathbb{Z})$, and I am trying to prove $G$ is finitely generated by using the Schwarz-Milnor lemma in geometric group theory.I am reading about geometric group theory from the book by Clara Löh. Below is the proof I am considering:

$G$ act on the upper half plane by isometry. $G$ acting via Möbius transformations on $X=\mathbb{H}^2$,and $\mathbb{H}^2$ with hyperbolic metric is geodesic metric space. $G$ act on $\mathbb{H}^2$ by Möbius transformation, then the fundamental domain corresponding action $B=\{z : -\frac{1}{2}<z \leq \frac{1}{2}, |z|>1\}$.

Here $X$ is a geodesic metric space hence it is (1,$\epsilon$) quasi-geodesic metric space every $\epsilon > 0$, we can choose $\epsilon=\frac{1}{2}$ and fix it.

Furthere more $\cup_{g \in G}gB= \mathbb{H}^2$, and $B'=\{x \in X : d_{\mathrm{hyp}}(x,y)<1, \exists y \in B\}$, then the set $S=\{g\in G : gB' \cap B'\neq \phi\}$ is fnite set By applying the Schwarz-Milnor lemma (see the author Clara Löh, Chapter 5, Proposition 5.41), we can directly conclude that $G=\mathrm{SL}(2, \mathbb{Z})$ is finitely generated.

Is this the right way of thinking, or am I missing something? Thank you very much!

Answer 1

The diameter of $B$ has to be finite which in your case isn't as you took $B = \{ z : - \frac{1}{2} < z \le \frac{1}{2} , |z| > 1 \}$ and without the finiteness assumption, you can get $\mathbb{R}$ acting on $\mathbb{R}^2$ as a counter-example as it satisfies the second and third assumptions of the Svarc–Milnor lemma ( Prop 5.4.1 in Clara Loh's Geometric Group Theory).