I am reading the hyperbolic geometry by Martelli's book 《An Introduction to Geometry Topology》, the question is from p136 proposition 5.1.4's proof $ (3) \Rightarrow ( 4 ) $, it ask me to adapting the proof of lemma 4.2.1: enter image description here
So if $\varphi_1 $ and $\varphi_2$ is all isometry in $\mathsf{Isom}(\mathbb{H^n}) $ or some model, i think it can't be $\varphi_1 \varphi_2=\varphi_2 \varphi_1\Rightarrow \mathsf{Fix} \varphi_1 = \mathsf{Fix} \varphi_1$, I think it shoud be $\varphi_1 \varphi_2=\varphi_2 \varphi_1\Rightarrow \mathsf{Fix} \varphi_1 \cap \mathsf{Fix} \varphi_1 \neq \varnothing$
But I really dont know how to prove that, I have proved in $\mathsf{Isom}(\mathbb{H^2}) $, because it have only one fixed point ,but in higher dimension, the elliptic type have at least one, even infinity(I am not sure).
I have try in the direction of the order of elliptic is finite, but I dont know what can it bring me, I ongly can write $\mathsf{Fix} \varphi_1$ into $\{\varphi_2 x_1,....,\varphi_2^n x_1,\varphi_2 x_2,....,\varphi_2^n x_2,......\} $, which the order of $\varphi_2$ is n.