Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 27 / 3 / a / Solution

Let $T=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$ and $S=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$. For $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ in the modular group, left multiplication by powers of $T$ and by $S$ performs subtraction and exchange on its first-column entries. The Euclidean algorithm reduces $(a,c)$ to $(\pm 1,0)$, since the determinant condition makes them coprime. The resulting matrix is $\pm T^m$, and $-I=S^2$ is already generated. Hence $S,T$ generate $S{L}_2(\mathbb{Z})$.

The standard fundamental domain of the modular group is $D=\{z\in \mathbb{H}:|\mathrm{Re}z|\le 1/2,|z|\ge 1\}$, with the vertical sides identified by $T$ and the circular sides by $S$. To prove every orbit meets it, choose a coprime integer pair $(c,d)$ minimizing $|cz+d|$. A minimum exists because only finitely many pairs have this quantity below any fixed bound: $|c|\mathrm{Im}z\le |cz+d|$ bounds $c$, then $d$. Complete the pair to a determinant-one matrix. Its image has maximal imaginary part in the orbit, since $\mathrm{Im}(\gamma z)=\mathrm{Im}z/|cz+d{|}^2$. Translate its real part into $[-1/2,1/2]$. If its modulus were below one, inversion would increase its imaginary part, a contradiction. This is the reduction to the standard modular region.