free groups of rotations The question of which pairs of rotations of the sphere are independent goes back to Hausdorff, who produced such a pair a century ago. "Independent" means "are free generators of a free group". The following assertion might not be true, but its statement is certainly elegant! Maybe there is a simple counterexample?
Question:  If A and B are noncommuting rotations of a sphere of infinite order, then A and B are independent.
One can also consider finite order (as did Hausdorff who used order 2 and 3): 
If A and B are noncommuting rotations of a sphere, then A and B A Inverse(B) are independent.
Can such strong and general assertions possibly be true?
Stan Wagon
 A: 
Question: If A and B are noncommuting rotations of a sphere of infinite order, then A and B are independent.

No, this is too strong. Let $w \in F_2$ be a word which does not vanish in the abelianization. Then it's known that the induced map
$$\text{SO}(3) \times \text{SO}(3) \ni (A, B) \mapsto w(A, B) \in \text{SO}(3)$$
is surjective. So we can find $A, B$ such that, for example, $A [A, B] = -1$. With a little more care (to check that the derivative of the word map is surjective, so that the word map is a submersion) we should be able to show that there are enough of these that we can arrange for $A$ and $B$ to both have infinite order. This gives a counterexample to both of your questions. 
What is true is that if $A$ and $B$ are chosen randomly then with probability $1$ they generate a free group. In fact the subspace of pairs $(A, B)$ which don't generate a free group is a countable union of Zariski-closed subspaces (and in particular has measure zero, but this is a stronger statement). 
