 Solution

Let $S=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ and $T=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$. The modular group acts on the complex upper half-plane by fractional linear transformations, andFor fixed $z=x+iy$, the values $|cz+d|$ with primitive integer pairs $(c,d)$ have a positive minimum: the conditions $|cz+d|\le R$ bound both $|c|\le R/y$ and $|d|\le R+|cx|$, leaving finitely many pairs. By Bezout identity, a minimizing pair is the bottom row of a matrix in $S{L}_2(\mathbb{Z})$. Applying that matrix gives a point of maximal imaginary part in its orbit.

Translate it by a power of $T$ into $-1/2<x\le 1/2$. It must have modulus at least one, since otherwise $S$ would increase its imaginary part. If it is on the unit circle with negative real part, apply $S$, which there changes $x$ to $-x$. This gives an element of the specified half-open standard fundamental domain of the modular group. It proves existence by reduction to the standard modular region.

For uniqueness, if $z$ lies in the closed standard region then $y\ge \sqrt{3}/2$ andfor every primitive pair. If $c=0$, this follows from $d=\pm 1$; if $|c|\ge 2$, it follows from $|c|y\ge \sqrt{3}$. If $|c|=1$, then $d=0$ uses $|z|\ge 1$, while $|d|\ge 1$ givesIf both $z$ and $z^{\prime }=\gamma z$ lie in the region, applying this inequality to $\gamma$ and ${\gamma }^{-1}$ forces their imaginary parts to agree, hence $|cz+d|=1$.

The equality cases describe every possible boundary identification. For $c=0$ the map is an integer translation; the half-open vertical strip retains one representative. For $|c|=1,d=0$, equality requires $|z|=1$, and the map is an inversion followed by an integer translation. Inversion exchanges the two halves of the circle, so retaining only the right half removes the duplication; the only translation that returns the exchanged endpoint is the one identifying the two corners. The remaining equality cases have $|c|=|d|=1$, $|z|=1$ and $x=\pm 1/2$, and give only those same corner identifications. Our region retains the right corner alone. Therefore $z^{\prime }=z$, proving

The equality analysis also gives the elliptic stabilizers of the modular group. The central matrices $\pm I$ fix every point. The point $i$ is fixed by $S$, with $S^2=-I$, and the retained corner ${\rho }_{+}=e^{i\pi /3}$ is fixed by $R=TS$, with $R^3=-I$. There are no other noncentral stabilizers, and the possible bottom rows in the equality cases give exactlyThe corresponding stabilizer orders in $PS{L}_2(\mathbb{Z})$ are two, three and one; the requested group is $S{L}_2(\mathbb{Z})$, so the center must be included.