Suppose $T,S$ are two non-identity elements in $PSL(2,\mathbb{R})$ and $TS=ST$. Then the number of fixed points of $S$ and $T$ are same.

Question

What I could see is $T$ maps the fixed point set of $S$ to itself, so does $S$ for the fixed point set of $T$. But I can not proceed further. I was actually looking at the proof of a stronger result that says the fixed set is equal if and only if they commute. The proof uses the above statement. I seem to stuck to this point.

Answer 1

Im really just spelling out the comment of reuns in greater detail.

The fixed points of $S$ (resp. $T$) on $\mathbb P^1_{\mathbb R}$ correspond to eigenvectors of (lifts of) $S$ (resp. $T$). This is because for $p \in \mathbb R^2 \setminus \{0\}$, one has

$$[Sp] = [p] \iff \text{there is } \lambda \neq 0 \text { such that } Sp = \lambda p.$$

Here, we use square brackets to denote the corresponding point in $\mathbb P^1_{\mathbb R}$.

Can you show that the number of fixed points of $S$ (resp. $T$) is the number of eigenvalues of $S$ (resp. $T$)? (Here, you use that $S$ and $T$ are not the identity.)

Afterwards, compare eigenspaces of $S$ and $T$ using $ST = TS$.