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.](http://qchu.wordpress.com/2010/03/12/fractional-linear-transformations-and-elliptic-curves/)) 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?)



How does this geometric description of the structure of PSL(2, Z) actually work?
The key word here is "Bass-Serre Theory" -- using the action on the hyperbolic plane, you can easily cook up a nice action of $PSL\_2(\mathbb{Z})$ on a tree. This is all described nicely in Serre's book "Trees".


EDIT: Let me give a few more details. It turns out that a group $G$ splits as a free produce of two subgroups $G\_1$ and $G\_2$ if and only if $G$ acts on a tree $T$ (nicely, meaning that it doesn't flip any edges) with quotient a single edge $e$ (not a loop) such that the following holds. Let $e'$ be a lift of $e$ to $T$ and let $x$ and $y$ be the vertices of $e'$. Then the stabilizers of $x$ and $y$ are $G\_1$ and $G\_2$ and the stabilizer of $e'$ is trivial.


If you stare at the fundamental domain for the action of $PSL\_2(\mathbb{Z})$ on the upper half plane, then you will see an appropriate tree staring back at you. There is a picture of this in Serre's book.


EDIT 2: This point of view also explains why finite-index subgroups $\Gamma$ of $PSL\_2(\mathbb{Z})$ tend to be free. If you restrict the action on the tree $T$ to $\Gamma$, then unless $\Gamma$ contains some conjugate of the order 2 or order 3 elements stabilizing the vertices, then $\Gamma$ will act freely. This means that the quotient $T/\Gamma$ will have fundamental group $\Gamma$. Since $T/\Gamma$ is a graph, this implies that $\Gamma$ is free.


