I am reading a proof of the following fact:
Take a hyperbolic surface $S$ with finite area and three (possibly degenerate) geodesic boundary components. If $S$ is diffeomorphic to a pair of pants, then it is isometric to a pair of pants.
The following excerpt is taken from Hummel's book Gromov’s Compactness Theorem for Pseudo-holomorphic Curves (Lemma IV.3.5, pp. 63-64).
The proof boils down to showing that, given three isometries $\sigma_1,\sigma_2,\sigma_3\in \text{PSL}(2,\mathbb R)$ of the hyperbolic plane $\mathbb H$ with the relation $\sigma_1\sigma_2=\sigma_3$, the group they generate is determined by the absolute values of their traces, up to conjugation by some isometry and reflection.
Essentially, these isometries correspond to the boundary components under the correspondence $\pi_1(S)\leftrightarrow\Sigma$, where $\Sigma$ is the deck transformation group of the universal cover $\tilde S\subseteq\mathbb H$ of $S$, while the absolute values of the traces are $\ge 2$ and correspond to the lengths of the boundary components (with some formula).
What I don't understand is why, given $a>1$, $\sigma_2$ and $\sigma_3$ are determined by $t_2:=|\text{tr}(\sigma_2)|$ and $t_3:=|\text{tr}(\sigma_3)|$. It seems to me that the equations $|b+c|=t_2$, $|ab+a^{-1}c|=t_3$ can have two different solutions (up to sign). For instance take $a=2$; for $t_2=t_3=2$ there are two solutions: $$\sigma_2=\begin{bmatrix} 2/3 & -1/6 \\ 2/3 & 4/3 \end{bmatrix},\quad \sigma_3=\begin{bmatrix} 4/3 & -1/3 \\ 1/3 & 2/3 \end{bmatrix}$$ $$\sigma_2=\begin{bmatrix} -2 & 9/2 \\ -2 & 4 \end{bmatrix},\quad \sigma_3=\begin{bmatrix} -4 & 9 \\ -1 & 2 \end{bmatrix}$$ I guess they are conjugated in some obvious way that I don't see, or maybe one of the two groups does not act properly discontinuously...