Slick argument to find free group in $SO(4)$

Question

It is well known that ${\rm SO}(3)$ contains the free group ${\rm F}_2$, so clearly ${\rm SO}(n)$ for $n>3$ does, too. However, the standard proof with ${\rm SO}(3)$ is somewhat annoying (the one sketched here). I've been able to prove in a nice way that if ${\rm SO}(4)$ contains ${\rm F}_2$, then so does ${\rm SO}(3)$, using ${\rm Spin}(4) \cong {\rm Spin}(3) \times {\rm Spin}(3)$ and a lemma about free groups.

It seems there may be a slicker argument that ${\rm SO}(4)$ contains ${\rm F}_2$ compared to ${\rm SO}(3)$, as we have more room to work, in a sense. Is there a nice way to obtain this?

Answer 1

There is a standard argument which works for all algebraic groups - a relation is an algebraic equation, so a generic pair of elements in any algebraic group will generate a free group.

See

Epstein, D. B. A., Almost all subgroups of a Lie group are free, J. Algebra 19, 261-262 (1971). ZBL0222.22012.

Answer 2

Here is a concrete example of a free group in $SO(4)$. Let $p=\cos\rho+i\sin\rho$ and $q=\cos\rho+j\sin\rho$ two quaternions, where $\rho$ is a transcendental number. The quaternions $p$ and $q$ generate a free group in the multiplicative group $H$ of quaternions. Next we should take advantage of the fact that the quaternions are represented by $4\times4$ real matrices, with $p$ and $q$ represented by orthogonal matrices. Details here: Dekker T. J., Decompositions of sets and spaces. III., Indag. Math 19 (1957): 104-107.