We know that the modular group $\Gamma=\mathrm{PSL}_2(\Ints)$ is equal to the subgroup generated by the two elements $S, T\in\Gamma$ with

This article shows that in fact $\Gamma$ is the free product of the subgroups $\langle S\rangle$ and $\langle ST\rangle$ , denoted $\langle S\rangle \star \langle ST\rangle$ . To see this, we use the theorem relating free products and group actions (q.v.). Let $\Half$ represent the upper half-plane, $\Half=\{z\in\Complex\ \mid\ \Im(z)>0\}$ , and define (see figure)

Note first that $S:S_2\cup S_3\to S_1$ , since $S$ reverses the sign of the real part of its argument. Thus in particular all nontrivial elements in $\langle S\rangle$ map $S_2\to S_1$ .

Second, recall from the article on the modular group that $ST$ rotates the half-plane around the point $\rho$ , and that under that rotation, $ST(S_1\cup S_3)\subsetneq S_2$ and $(ST)^2(S_1\cup S_3)\subsetneq S_2$ . Thus in particular all nontrivial elements of $\langle ST\rangle$ map $S_1\to S_2$ .

Finally, by the above, we see that $S:S_3\to S_1$ and $ST, (ST)^2:S_3\to S_2$ .

Since $\Gamma=\langle S,ST\rangle$ , we can apply the theorem relating free products and group actions, choosing $s$ to be any point in $S_3$ , to conclude that $$ \Gamma =\langle S\rangle \star \langle ST\rangle\cong \Ints/2\Ints \star \Ints/3\Int $$