= Solution
Set $C=AB=\left(\begin{smallmatrix}0&-1\\1&2\lambda\end{smallmatrix}\right)$. It fixes $z=-\lambda+i\sqrt{1-\lambda^2}$ because $z^2+2\lambda z+1=0$. Write $\lambda=\cos\theta$, $0<\theta<\pi/2$; the <eigenvalues> of $C$ are $e^{\pm i\theta}$.
The stabilizer of an interior point of $\mathbb H$ in $PSL_2(\mathbb R)$ is conjugate to the compact rotation group. Its intersection with a <discrete subgroup> is finite: otherwise compactness gives accumulating elements, and their quotients accumulate at the identity. Thus $C$ has finite order in the projective group. Its eigenvalues then satisfy $e^{2im\theta}=1$ for some positive integer $m$, forcing $\theta/\pi\in\mathbb Q$. Hence $\boxed{\lambda=\cos(\pi y)\text{ for some }y\in\mathbb Q}$. This is a necessary condition only.
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