= Solution
Let $T=\left(\begin{smallmatrix}1&1\\0&1\end{smallmatrix}\right)$ and $S=\left(\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}\right)$. For $\gamma=\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)$ 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 $(\pm1,0)$, 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 $SL_2(\mathbb Z)$.
The <standard fundamental domain of the modular group> is $D=\{z\in\mathbb H:|\operatorname{Re}z|\le1/2,\ |z|\ge1\}$, 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: $|c|\operatorname{Im}z\le|cz+d|$ bounds $c$, then $d$. Complete the pair to a determinant-one matrix. Its image has maximal imaginary part in the orbit, since $\operatorname{Im}(\gamma z)=\operatorname{Im}z/|cz+d|^2$. 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>.
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