Is it true that two random (w.r.t. Haar measure) rotations in $SO(3)$ generate a free group?



Random rotations in SO(3) and free group
Yes. Here's what should be a proof: the set of pairs of elements satisfying any particular relation is Zariski closed, hence has measure zero (to show that it is not $SO(3) \times SO(3)$ it suffices to know that at least one subgroup generated by two elements is free), and there are countably many relations.


