Let $\Gamma \subset \mathrm{SL}(2,\mathbb{R})$ be a lattice. If $N\_1, N\_2$ are a pair of independent parabolic subgroups contained in $\Gamma$, why must $\Gamma$ contain a hyperbolic element? By parabolic subgroup, I mean "subgroup containing only parabolic elements, other than the identity." This is used in the proof of Theorem 10.1, here:


[https://people.math.harvard.edu/~ctm/papers/home/text/papers/abel/abel.pdf](https://people.math.harvard.edu/%7Ectm/papers/home/text/papers/abel/abel.pdf)


If it helps, $\Gamma$ here is the stabilizer of a 1-form defining a Teichmuller curve.



Parabolic elements and hyperbolic elements in SL(2,R)
This very special case follows from "general principles" (namely a version of the ping-pong lemma) but it is also possible to give a direct proof, as follows.




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Suppose that $a$ and $b$ are the distinct points at infinity fixed by the two parabolic subgroups $A$ and $B$ (note change of notation). Since $\mathrm{SL}(2, \mathbb{R})$ is three-transitive, we can conjugate $\Gamma$ and so assume that $a = \infty$ and $b = 0$. We deduce that elements of $A$ now have the form
$$
\begin{pmatrix}
1 & r\\
0 & 1
\end{pmatrix}
$$
while elements of $B$ have the form
$$
\begin{pmatrix}
1 & 0\\
s & 1
\end{pmatrix}
$$
Taking an inverse if needed, we obtain such elements of $A$ and $B$ where $r$ and $s$ are non-zero and have the same sign. We now multiply to find
$$
\begin{pmatrix}
1 & r\\
0 & 1
\end{pmatrix}
\begin{pmatrix}
1 & 0\\
s & 1
\end{pmatrix}
=
\begin{pmatrix}
1 + rs & r\\
s & 1
\end{pmatrix}
$$
This has trace $2 + rs > 2$ so is hyperbolic, as desired.




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Hmm. Now that I write this, I realise that there is a third proof using the classification of fixed points of isometries, and the intermediate value theorem. It helps to draw a picture and think dynamically.


