modular group isomorphic to free product of groups I am trying to prove that the modular group PSL(2,Z) is isomorphic to the free product of groups Z_2*Z_3. Any ideas on how to get around this?
Any hints much appreciated.
 A: If you understand how $PSL(2,\mathbb{Z})$ acts on the upper half plane of $\mathbb{C}$ by fractional linear transformations, i.e. on the hyperbolic plane $\mathbb{H}^2$, then there is a geometric proof. 
Start with the ideal hyperbolic triangle with vertices $0,1,\infty$, and reflect it repeatedly across sides to get a tiling of $\mathbb{H}^2$ by ideal triangles. You can see this tiling in the picture supplied by the first answer of this question. The action of $PSL(2,\mathbb{Z})$ on $\mathbb{H}^2$ by fractional linear transformations preserves this tiling.
Each ideal hyperbolic triangle in this tiling has a unique barycenter, which is the unique point fixed by the order 3 rotation group of that triangle. For instance the barycenter of the $0,1,\infty$ triangle is $\frac{1}{2} + \frac{\sqrt{3}}{2} i$, which is the unique fixed point of the order 3 fractional linear transformation that takes $0 \mapsto 1 \mapsto \infty \mapsto 1$
For any two of the triangles which share a side, connect their barycenters by a hyperbolic geodesic segment cutting through that side. For instance the $0,1,\infty$ and the $-1,0,\infty$ triangles share the side $0,\infty$ and one connects their barycenters by the segment with endpoints $-\frac{1}{2} + \frac{\sqrt{3}}{2} i$, $\frac{1}{2} + \frac{\sqrt{3}}{2} i$ that cuts through the side $0,\infty$ at the point $0+1i$. Subdivide each of these segments into two subsegments where it crosses the side.
The union of these segments is a tree, and this tree is invariant under the action of $PSL(2,\mathbb{Z})$. This tree has a valence 3 vertex at the barycenter of each triangle, and the subgroup of $PSL(2,\mathbb{Z})$ that stabilizes that vertex is isomorphic to $\mathbb{Z}/3$. This tree also has a valence 2 vertex where it crosses each side of each triangle, with stabilizer group isomorphic to $\mathbb{Z}/2$. Each edge of the tree connects some triangle barycenter with a point on a side of that triangle, and the stabilizer of that edge is the trivial subgroup.
Now what you are left to prove is that if a group acts on a tree with two vertex orbits and one edge orbit, and if the vertex stabilizers are isomorphic to $\mathbb{Z}/3$ and $\mathbb{Z}/2$ respectively, and if the edge stabilizers are trivial, then that group is isomorphic to $\mathbb{Z}/3 * \mathbb{Z}/2$. This can be done by hand, or it can be done by applying Bass-Serre theory.
