Suppose a cocompact <Fuchsian group> contained a nonidentity <parabolic isometry of the hyperbolic plane> $\phi$. After conjugating in the <upper half-plane model>, write $\phi(z)=z+c$ with $c\ne0$. The supplied estimate gives
$$
d(iy,\phi(iy))=d(iy,c+iy)\leq\frac{|c|}{y}\longrightarrow0
\qquad(y\to\infty),
$$
so the infimum of the displacement function is zero.

A Fuchsian action is properly discontinuous, and the action is cocompact by hypothesis. Parts b and c therefore imply that $\phi$ fixes a point of the <hyperbolic plane>. A nonidentity parabolic isometry has no fixed point inside the plane, only one on its ideal boundary. This contradiction excludes nontrivial parabolic elements.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 133 / 4 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-133/4/d/solution