Freeness of the group generated by two hyperbolic isometries If $f$ and $g$ are two hyperbolic isometries of the hyperbolic space $\mathcal H^n$, we know that $f$ has 2 fixed points on $\partial \mathcal H^n$ and similiarly for $g$.
Is it true that $f$ and $g$ generate a non-abelian free group if $\mathrm{fix}(f)\cap \mathrm{fix}(g) = \varnothing$ ? If so, could you provide a reference for this fact ?
 A: That is not true.
A simple example is the following one: consider the hyperbolic isometry
\begin{equation*}
\phi:=
\left[
\begin{array}{l l}
4 &0\\
0 &\frac{1}{4}\\
\end{array}
\right]
\in\text{PSL}(2,\mathbb{R})=\text{Isom}^+\left(\mathbb{H}^2\right)
\end{equation*}
and the finite order elliptic element
\begin{equation*}
\rho:=
\left[
\begin{array}{l l}
\frac{1}{2} &-\frac{\sqrt{3}}{2}\\
\frac{\sqrt{3}}{2} &\frac{1}{2}\\
\end{array}
\right]
\in\text{PSL}(2,\mathbb{R})=\text{Isom}^+\left(\mathbb{H}^2\right)
\end{equation*}
with $\rho^6=1$. The product
\begin{equation*}
\psi:=\phi\rho=
\left[
\begin{array}{l l}
4 &0\\
0 &\frac{1}{4}\\
\end{array}
\right]
\cdot
\left[
\begin{array}{l l}
\frac{1}{2} &-\frac{\sqrt{3}}{2}\\
\frac{\sqrt{3}}{2} &\frac{1}{2}\\
\end{array}
\right]
=
\left[
\begin{array}{l l}
2 &-2\sqrt{3}\\
\frac{\sqrt{3}}{8} &\frac{1}{8}\\
\end{array}
\right]
\end{equation*}
is again hyperbolic as $\text{tr}(\phi\rho)=2+\frac{1}{8}>2$. Moreover it is easy to see that its fixed points are disjoint from those of $\phi$.
The subgroup $\Gamma:=\langle\phi,\psi\rangle$ cannot be a non-abelian free group as it contains the torsion element $\rho$.
