I want to prove that every non-elementary subgroup (call it $\Gamma$ ) of $PSL (2, \mathbb{R})$ has infinitely many hyperbolic elements in wich one fixed point is distinct. (Should I put the definition of non-elementary in this question?) .
I proved before that always exists an hyperbolic element in $\Gamma.$ Let $h$ be that element and suppose that $r_1, r_2$ are its fixed points in $\mathbb{R}\cup \{ \infty \}.$
I deduced that if I find an element $T \in PSL (2, \mathbb{R}) $ such that $T$ sends a third point $r_3$ to either $r_1$ or $r_2$ then $T^{-1}hT$ fixes $r_3.$
The my problem is here: It is clear that $T^{-1}hT$ is also a hyperbolic element because trace is invariant under conjugation. But how can I guarantee that $T^{-1}hT$ is also an element of $\Gamma?$ If that's false, is there another way for proving the claim?
Thanks in advance. I'm beginning at hyperbolic geometry.