 Solution

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.