There is a beautiful way to see that the congruence subgroup $\Gamma(2)$ is free on two generators: the action of $\Gamma(2)$ on $\mathbb{H}$ is free and properly discontinuous, and there is a modular function $\lambda$ with respect to $\Gamma(2)$ coming from Legendre normal form such that $\mathbb{H}/\Gamma(2) \xrightarrow{\lambda} \mathbb{C} - \{ 0, 1 \}$ is an isomorphism. (Details.) It follows that $\mathbb{H}$ is the universal cover of $\mathbb{C} - \{0, 1 \}$, hence that $\Gamma(2)$ is isomorphic to the fundamental group of $\mathbb{C} - \{0, 1 \}$.

However, the action of $\Gamma(1) \simeq PSL_2(\mathbb{Z})$ on $\mathbb{H}$ is not ~~properly discontinuous~~ free; there are problems, which maybe I should call "ramification," at the points $i, e^{ \frac{\pi i}{3} }, e^{ \frac{2\pi i}{3} }$. This is supposed to be responsible for the fact that $\Gamma(1)$ is not free, but is instead the free product of a cyclic group of order $2$ and a cyclic group of order $3$, where the former somehow comes from the behavior at $i$ and the latter somehow comes from the behavior at the sixth roots of unity. That's what I've been told, anyway, but I don't know how the argument actually goes. What general context does it fit into? (Is "monodromy" a keyword here?)

isproperly discontinuous; what it is not is free. $\endgroup$ – Pete L. Clark Mar 18 '10 at 20:53