Cheap proof that $\mathrm{SO}(3)$ contains a rank $2$ free subgroup It's a well-known fact, used in the proof of the Banach-Tarski paradox for example, that $\mathrm{SO}(3)$ contains a subgroup isomorphic to the free group $\mathbb{F}_2$ on two generators.  Is there a cheap way to prove this fact along the following lines?
Any non-trivial reduced word $w$ in $\mathbb{F}_2$ defines a map
$$f_w : \mathrm{SO}(3) \times \mathrm{SO}(3) \rightarrow \mathrm{SO}(3)$$
by sending a pair $(a, b)$ to the value of $w$ when the two generators of the free group are sent to $a$ and $b$.  We want to show that the union of the $f_w^{-1}(I)$ over all $w$ is not the whole of the domain.  One could try to do this by showing that:


*

*For each $w$ the set $f_w^{-1}(I)$ is "small" in $\mathrm{SO}(3) \times \mathrm{SO}(3)$ (e.g. measure zero, meagre, real algebraic).

*The union of countable many "small" sets is "small".
Is there a way to make this scheme work?
It seems intuitive that any non-abelian Lie group should contain a copy of $\mathbb{F}_2$, just for size reasons.  Is this the case, and can a similar argument be used in this general setting?
 A: I start with a few observations. First of all, it suffices to prove that the group $K=SU(2)$ contains a free nonabelian subgroup since $SO(3)$ is isomorphic to the quotient $SU(2)/\{\pm I\}$. Now, the group $G= SL(2, {\mathbb C})$ is a connected 3-dimensional complex manifold; it contains $SU(2)$ as a subgroup. Moreover, if $\phi: SL(2, {\mathbb C})\to {\mathbb C}$ is a holomorphic function which is constant on an open nonempty subset of $SU(2)$ then it is constant on the entire $SL(2, {\mathbb C})$. (This is proven by observing that in some local holomorphic coordinates, $SU(2)\subset SL(2, {\mathbb C})$ appears as ${\mathbb R}^3\subset {\mathbb C}^3$.)   
The same holds for functions on $G\times G$: If a holomorphic function on $G\times G$ is constant on an open nonempty subset of $K\times K$ then it is constant on $G\times G$. 
Now, consider the set of homomorphisms $Hom(F_2, SU(2))$. It is naturally identified with the product $SU(2)\times SU(2)$ (by recording the images of the generators under each homomorphism $\rho$); the same for the set of homomorphisms $Hom(F_2, SL(2, {\mathbb C}))$. Each element $w\in F_2$ defines a holomorphic function $f_w: G\times G\to G$ by the formula:
$$
f_w(\rho):= \rho(w)\in SL(2, {\mathbb C}).  
$$
(The fact that for the two generators $a, b$ of $F_2$ the functions 
$f_a, f_b$ are holomorphic is immediate; the general case follows, by matrix multiplication, from the fact that the products and sums of holomorphic functions are holomorphic.) 
For each $w\in F_2$ define two closed subsets: $J_w:= f_w^{-1}(I)\subset G\times G$ and $I_w:=J_w\cap K\times K\subset K\times K$. 
By the observations made in the 1st paragraph we have the following dichotomy: 
For every $w\in F_2$ either:


*

*$I_w$ is nowhere dense in $K\times K$, or

*$J_w= G \times G$. 
Now, if we knew that for every $w\ne 1$, $J_w$ is a proper subset of $G\times G$, it would follow from the Baire category theorem (since $K\times K$ is metrizable and $F_2$ is countable) that 
$$
Hom(F_2, K)\ne \bigcup_{w\in F_2 -1} I_w .
$$ 
In other words, we would know that there exists a homomorphism 
$$
\rho: F_2\to SU(2)
$$
with trivial kernel, as required. 
The issue then becomes: How do we know that for every $w\in F_2 - 1$ there exists a homomorphism $\rho: F_2\to SL(2, {\mathbb C})$ such that $\rho(w)\ne I$? This would immediately follow if we knew that there exists a monomorphism $\rho: F_2\to SL(2, {\mathbb R})$. 
How to prove existence of such monomorphism is a separate story. Among ways to prove this is to look closely at the modular group $SL(2, {\mathbb Z})$, or to construct a complete hyperbolic structure on a surface with free fundamental group, or to "play ping-pong'', etc.  
