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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.

rowcol
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  • What do you mean by "non-elementary" subgroup? – Andrea Mori Nov 18 '19 at 17:12
  • @AndreaMori A subgroup $\Gamma $ of $PSL (2, \mathbb{R})$ is called elementary if there exists a finite $\Gamma$-orbit in the closure of the upper-half-plane. Non-elementary is the negation of that definition – rowcol Nov 18 '19 at 17:16
  • The subgroup generated by ${1\ 1}\choose{0\ 1}$ has no finite orbit but contains no hyperbolic element. – Andrea Mori Nov 18 '19 at 17:19
  • This is Theorem 5.1.3 of The geometry of discrete groups from Alan Beardon. – mathcounterexamples.net Nov 18 '19 at 17:20
  • A correct way to phrase your question is: Every nonelementary subgroup of $PSL(2,R)$ contains two hyperbolic elements whose fixed-point sets are distinct. Actually, more is true: You can find hyperbolic elements with disjoint fixed-point sets. I second the suggestion to read Beardon's book. He covers all the basics, such as your question, very well. Lastly, "closure" of the upper half-plane is meaningless since you did not specify where the closure is taken. You can say, closure in the Riemann sphere. In fact, you can simply say that $\Gamma$ has no finite orbits in the Riemann sphere. – Moishe Kohan Nov 18 '19 at 18:38
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    @AndreaMori: OP is sloppy with the definition, but the closure is meant to be taken in the Riemann sphere. Then your example has a fixed point, namely, $\infty$. – Moishe Kohan Nov 18 '19 at 18:41

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