Fuchsian groups and surfaces It's a fact that if two Fuchsian group are conjugate, the corresponding surfaces are isometric. Is the converse true ? 
Take 2 isometric Riemann surfaces $S$ and $S'$(which are covered by the upper half-plane) or equivalently 2 hyperbolic surfaces. You can endow their universal covering with the complex(or hyperbolic) structure, so they can be identified to the hyperbolic plane. Then $S = \mathbb{H} / \Gamma$ and $S'= \mathbb{H} / \Gamma'$ where $\Gamma$ and $\Gamma'$ are two fuchsian groups. Are they conjugate ?
 A: Let $\varphi\colon S\to S'$ be an isometry.  Since $\mathbb{H}$ is simply connected,  $\varphi$ lifts to a map $\widetilde{\varphi}\colon \mathbb{H}\to \mathbb{H}$ making the following diagram commute:
$$
\begin{array}{ccc}
\mathbb{H} & \xrightarrow{\widetilde{\varphi}} & \mathbb{H} \\
\downarrow & & \downarrow \\
S & \xrightarrow{\varphi} & S'
\end{array}
$$
Then $\widetilde{\varphi}$ is a local isometry.  Since $\mathbb{H}$ is simply connected and geodesically complete, it follows that $\widetilde{\varphi}$ is an isometry.  We claim that $\widetilde{\varphi}^{-1}\;\Gamma'\,\widetilde{\varphi} = \Gamma$.
Let $\gamma'\in\Gamma'$, and let $p\colon\mathbb{H}\to S$ and $p'\colon\mathbb{H}\to S'$ be the covering maps.  We know that $\gamma'$ is a covering transformation for $p'$, i.e. $p'\gamma'=p'$.  Since $p'\widetilde{\varphi} = \varphi p$, we have
$$
p\widetilde{\varphi}^{-1}\gamma'\widetilde{\varphi} \,=\, \varphi^{-1}p' \gamma'\widetilde{\varphi} \,=\, \varphi^{-1} p' \widetilde{\varphi} = p.
$$
Thus $\widetilde{\varphi}^{-1}\gamma'\widetilde{\varphi}$ is a covering transformation for $p$, so $\widetilde{\varphi}^{-1}\gamma'\widetilde{\varphi} \in \Gamma$.  This proves that $\widetilde{\varphi}^{-1}\Gamma'\widetilde{\varphi} \leq \Gamma$, and a similar argument shows that $\Gamma \leq \widetilde{\varphi}^{-1}\Gamma'\widetilde{\varphi}$.
