I'm looking to demonstrate that there exists a free group on two generators $ G\subseteq SO(3) $ (this is for homework, so hints are preferred to complete answers I've turned it in now, so full answers welcome!). I'd like to do something similar to the first answer in this MathOverflow post as follows:
Let $A$ denote the set of all reduced words and $w\in A$. $A$ is equinumerous with $(\mathbb{Z}/2\mathbb{Z})[x]$, hence countable; and each $w$ defines a $\tilde{w}\in C^1(SO(3)^2\to SO(3))$ via substitution. I think, for each nontrivial $w$, the set $Z_w=\tilde{w}^{\leftarrow}(\{I_3\})$ ought to be nowhere dense. Then I can conclude that, as $SO(3)^2$ is a complete metric space, it is Baire, so there must exist some $(x,y)\in SO(3)^2\setminus\cup_{w\in A\setminus\{e\}}{Z_w}$. Thus $\langle x,y\rangle\subseteq SO(3)$ is free, as desired.
Now, to prove each $Z_w$ is nowhere dense: $\tilde{w}$ is continuous, so $Z_w$ is closed, so it suffices to show each $Z_w$ is interiorless. I feel like an argument should run along the lines of the proof that any Hamel basis of an infinite-dimensional Banach space is uncountable: $Z_w$ is "like" a manifold of non-vanishing codimension. One way to obtain such a result is the Implicit Function Theorem: we have one $SO(3)$'s worth of equations, so there should be one $SO(3)$ "perpendicular" to the solutions.
But I don't know how to show that $\tilde{w}$ has nonsingular derivative. The nicest parameterization of $SO(3)$ seems to be the unit quaternions in polar form, but even then, I can't figure out a nice way to write the (total) derivative (considering $(\mathbb{H}_{0,\text{unit}}\times\mathbb{R})^2$ as a real 6-manifold). The key has to lie in the Baker-Campbell-Hausdorff formula, so that I can write a product of quaternion exponentials as a quaternion exponential of a sum, because everything above holds for $SO(2)$ parameterized by $\mathbb{C}$, except that all the "higher-order" terms of the BCH formula vanish.
So: how do polar quaternions multiply, according to the Baker-Campbell-Hausdorff formula?