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Suppose a cocompact Fuchsian group contained a nonidentity parabolic isometry of the hyperbolic plane $\varphi$. After conjugating in the upper half-plane model, write $\varphi (z)=z+c$ with $c\ne 0$. The supplied estimate givesso 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 $\varphi$ 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.

Letbe the orientation-preserving $(2,3,7)$ hyperbolic triangle group. DefineEvery defining relator of $G$ maps to the identity, and $A,B,C$ lie in the image, so this is a surjective group homomorphism. Since$\Delta (2,3,7)$ is a non-elementary Fuchsian group.

Solved by gpt-5.6-sol high.

The equality $x=a^2$ and the relator $[b,x]=1$ show that $x$ commutes with both $a$ and $b$. Since $c=(ab{)}^{-1}$, it also commutes with $c$, and then with $y=b^3$. Hence $x\in Z(G)$. Similarly, $y=b^3$ commutes with $b$ and, by $[a,y]=1$, with $a$; it therefore commutes with $c$ and $x$. Thus

Quotienting by $⟨x,y⟩$ givesA non-elementary Fuchsian group has trivial center: two hyperbolic elements with different pairs of boundary fixed points have only the identity in their common centralizer in ${\mathrm{PSL}}_2(\mathbb{R})$. Therefore the image in the quotient of every element of $Z(G)$ is trivial. It follows that

Solved by gpt-5.6-sol high.