= Solution
For $\tau=x+iy\in\mathfrak h$ and $\gamma=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in\Gamma(1)$,
$$
\operatorname{Im}(\gamma\tau)=\frac{y}{|c\tau+d|^2}.
$$
The set $\{c\tau+d:(c,d)\in\mathbb Z^2\setminus\{0\}\}$ is a <lattice> in $\mathbb C$, so it has a shortest nonzero vector. Dividing $(c,d)$ by their greatest common divisor can only shorten it; hence a minimizing pair may be chosen coprime and completed to the bottom row of some $\gamma\in SL_2(\mathbb Z)$. Consequently the orbit contains a point $\tau_0$ of maximal imaginary part.
Apply a power of $T$ so that $-1/2\leq\operatorname{Re}\tau_0\leq1/2$. If $|\tau_0|<1$, then
$$
\operatorname{Im}(-1/\tau_0)=\frac{\operatorname{Im}\tau_0}{|\tau_0|^2}>\operatorname{Im}\tau_0,
$$
contradicting maximality. Thus $|\tau_0|\geq1$, and $\tau_0$ lies in the <standard fundamental domain of the modular group>.
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