Subgroup of PSL(2,R) generated by two elements with elliptic commutator is not free.

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Suppose $[A,B]$ is elliptic for some $A,B \in PSL(2;\mathbb{R})$. Then $\langle A,B \rangle < PSL(2;\mathbb{R})$ is not free. Is there any nice way to see this? If $[A,B]$ is elliptic of finite order then it is clear, but what about elliptic of infinite order?

The claim is from: http://comet.lehman.cuny.edu/keenl/LiftingFreeSubgroups.pdf (the bottom paragraph of page 12).