For context, I am following Sullivan's proof to the No Wandering Theorem, we have the following setup:
Let $S_1 \xrightarrow{f_1} S_2 \xrightarrow{f_2} S_3 \xrightarrow{f_3} S_4 \xrightarrow{f_4} ...$ be a chain of analytic surjective maps between hyperbolic Riemann surfaces.
Let $\Gamma_\infty= \bigcup_i \Gamma_i $ where $\Gamma_i$ is the fundamental group of the hyperbolic Riemann surface $S_i$. We have $\Gamma_1 \subset \Gamma_2 \subset ... \subset \Gamma_\infty$.
The claim is that assuming one of the $\Gamma_i$ is non-abelian, we must have that $\Gamma_\infty$ is discrete. I'm not sure why this is, but here's what I figured so far:
We have to use the result "If a subgroup of PSL($2, \mathbb{C})$ contains no elliptic elements, it is either elementary or discrete."
Since $\Gamma_\infty$ is a subgroup of PSL($2,\mathbb{C})$, we can apply this result.
It is here where I don't know how to show that $\Gamma_\infty$ is not elementary and contains no elliptics elements.
Any help would be greatly appreciated. Thank you.