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.

  • $\begingroup$ I think you should have -1 on the diagonal. $\endgroup$ Sep 9, 2020 at 22:18
  • $\begingroup$ Sorry I meant that $\Gamma< \operatorname{PSL}(2,\mathbb{R})$, and I edited the question to reflect this. So I don't think such things matter. $\endgroup$
    – Chris Z
    Sep 11, 2020 at 15:04

1 Answer 1


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 $\text{Stab}(\infty;\Gamma')$.


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