The following statement seems true, but I don't know a proof or a reference for it (and I would like one).



> 
> Let $\Gamma< \operatorname{PSL}(2,\mathbb R)$ be a nonuniform lattice with one cusp. We may conjugate $\Gamma$ so that an element
> $
> \begin{pmatrix}
> 1 & s\\
> 0 & 1
> \end{pmatrix}
> $ generates the cusp of $\Gamma$. Let $\Gamma'$ be such that $\Gamma<\Gamma'$ be an index $2$ subgroup. Then, the element
> $
> \begin{pmatrix}
> 1 & s/2\\
> 0 & 1
> \end{pmatrix}
> $ generates the cusp of $\Gamma'$.
> 
> 
> 


I imagine a similar statement would be true if both $\Gamma$ and $\Gamma'$ had two cusps.



Cusps of hyperbolic surfaces under finite covers
Assume just that $\Gamma$ has index $k$ in $\Gamma'$. Let $C \subset \mathbb R \cup \{\infty\}$ be the set of parabolic points for the action of $\Gamma$. Then $C$ is also the set of parabolic points for the action of $\Gamma'$, because if $\gamma \in \Gamma'$ is parabolic with fixed point $x$ then for some integer $i \ge 1$, $\gamma^i \in \Gamma$ is parabolic with the same fixed point $x$.


In general every $\Gamma'$ orbit of $C$ decomposes as a union of $\Gamma$ orbits, so
$$\#\text{cusps}(\Gamma') \le \#\text{cusps}(\Gamma)
$$
with equality if and only if the every $\Gamma'$ orbit of $C$ is just a single $\Gamma$-orbit. So assuming that
$$\#\text{cusps}(\Gamma') = \#\text{cusps}(\Gamma)
$$
it follows for each $x \in C$ the group $\text{Stab}(x;\Gamma)$ has index $k$ in $\text{Stab}(x;\Gamma')$.


For the case $x=\infty \in C$ it follows that if $\begin{pmatrix}
1 & s\\
0 & 1
\end{pmatrix}$ generates $\text{Stab}(\infty;\Gamma)$ then $\begin{pmatrix}
1 & s/k\\
0 & 1
\end{pmatrix}$ generates $\text{Stab}(\infty;\Gamma')$.


