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It is well known that, the Hurwitz groups are quotients of the $(2,3,7)$ triangle group $\Gamma=\langle a,b\colon a^2=b^3=(ab)^7=1\rangle $ in $PSL(2,\mathbb{R})$. If $G$ is a Hurwitz group, then there will be an epimorphism from $\Gamma$ to $G$ with the kernel, a surface group.

Let $f_1,f_2\colon \Gamma\rightarrow G$ be two epimorphisms with surface kernels $\varLambda_1$, and $\varLambda_2$.

If $\varLambda_1=\varLambda_2$, then we get one Riemann surface $\mathbb{H}/\varLambda_1$ from these two epimorphisms having automorphism group $G$.

But, if $\varLambda_1\neq \varLambda_2$, are the Riemann surface $\mathbb{H}/\varLambda_1$, and $\mathbb{H}/\varLambda_2$ necessarily non-isomorphic, with automorphism group $G$?

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Hint: since $\Gamma_1$ and $\Gamma_2$ are hyperbolic Fuchsian groups, the compact Riemann surfaces they uniformize are isomorphic (i.e., biholomorphic, or equivalently, isometric) iff $\Gamma_1$ and $\Gamma_2$ are conjugate in $\operatorname{PSL}_2(\mathbb{R})$. – Pete L. Clark Mar 24 '11 at 17:47

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