There is a free group $F_2$ in $SO(3)$ I'd like an exposition of the proof that there is a subgroup of $SO(3)$ isomorphic to $F_2$, the free group on two elements. This is a key step in the proof of the Banach-Tarski paradox. The usual generators chosen for this are rotations about the $x$ and $z$ axes by $\theta=\arccos\left(\frac13\right)$. Wikipedia attempts to describe a proof, but it is too terse for my liking; perhaps someone here can fill in the gaps.

Let $a$ be a rotation of $\theta=\arccos\left(\frac{1}{3}\right)$ about the first, $x$ axis, and $b$ be a rotation of $\theta$ about the $z$ axis (there are many other suitable pairs of irrational multiples of $\pi$, that could be used here as well).
The group of rotations generated by $a$ and $b$ will be called $H$. 
  Let $\omega$ be an element of $H$ which starts with a rotation on the $z$ axis, of the form $\omega=\ldots a^{k_1}b^{k_2} \ldots b^{n}$ [I think this is trying to say that the shortest word expressing $\omega$ is of the form $xb$ for some word $x$].
It can be shown by induction that $\omega$ maps the point $(1,0,0)$ to $\displaystyle\left(\frac i {3^N}, \frac{j\sqrt 2}{3^N}, \frac k {3^N}\right)$, for some $i,j,k \in \mathbb Z,N \in \mathbb N$. Analysing $i,j$ and $k$ modulo 3, one can show that $j\neq 0$. The same argument repeated (by symmetry of the problem) is valid for the opposite rotation about the $z$ axis, as well as rotations about the $x$ axis. This shows that for any non trivial word $\rho \in H$, then $\rho \neq e$. Therefore, the group $H$ is a free group, isomorphic to $F_2$.

Edit: To clarify what I am looking for, I want a concise but rigorous proof of the above statement. The WP proof says "one can show" too much.
 A: For the "by induction" statement: Let $T$ be the set of matrices with entries $a_{ij}$ such that $a_{ij}\in\Bbb Z$ if $i+j$ is even and $a_{ij}\in\Bbb Z\sqrt 2$ if $i+j$ is odd, that is, of the form $$\begin{pmatrix}\Bbb Z&\Bbb Z\sqrt2&\Bbb Z\\\Bbb Z\sqrt2&\Bbb Z&\Bbb Z\sqrt2\\\Bbb Z&\Bbb Z\sqrt2&\Bbb Z\end{pmatrix}.$$ Then $T$ forms a subring of $M(3,\Bbb R)$. Clearly $I,0\in T$ and $T$ is closed under addition. To show that $T$ is closed under multiplication, suppose $A,B\in T$, with elements $(a_{ij}),(b_{ij})$. Then the product is $c_{ij}=\sum_{k=1}^3a_{ik}b_{kj}$. If $i+j$ is even, then every element in the sum is either $a_{ik}b_{kj}\in\Bbb Z\cdot \Bbb Z$ if $k$ is even or $a_{ik}b_{kj}\in\sqrt 2\Bbb Z\cdot \sqrt 2\Bbb Z=2\Bbb Z$ if $k$ is odd, so the sum is also in $\Bbb Z$, or else $i+j$ is odd, in which case one of the two terms is in $\Bbb Z$ and the other is in $\sqrt 2\Bbb Z$ hence the product is in $\sqrt 2\Bbb Z$ and the sum of these is also in $\sqrt 2\Bbb Z$. Thus $AB\in T$.
Note also that $$3a=\begin{pmatrix}3&0&0\\0&1&-2\sqrt2\\0&2\sqrt2&1\end{pmatrix}\in T\qquad3b=\begin{pmatrix}1&-2\sqrt2&0\\2\sqrt2&1&0\\0&0&3\end{pmatrix}\in T,$$
so for any product $M$ of $n$ terms selected from $\{a,b,a^{-1},b^{-1}\}$, we have $3^nM\in T$, and since applying this to $v=(1,0,0)$ extracts the top row, we have $3^nMv=(i,j\sqrt 2,k)$ for some $i,j,k\in\Bbb Z$ as desired.

"Analyzing modulo 3": Consider taking the coefficients of the matrix $\bmod 3$ (treated as elements of $\Bbb Z[\sqrt2]$). This has the effect of simply reducing the coefficient of an element of the form $\Bbb Z\sqrt2$ (there are no mixed terms $a+b\sqrt2$). Dropping the $\sqrt2$ from the $a_{ij}$, $i+j$ odd entries and denoting $-1$ as $\bar 1$, we get:
$$[3a]=\begin{pmatrix}0&0&0\\0&1&1\\0&\bar 1&1\end{pmatrix}\quad
[3a^{-1}]=\begin{pmatrix}0&0&0\\0&1&\bar 1\\0&1&1\end{pmatrix}\quad
[3b]=\begin{pmatrix}1&1&0\\\bar 1&1&0\\0&0&0\end{pmatrix}\quad
[3b^{-1}]=\begin{pmatrix}1&\bar 1&0\\1&1&0\\0&0&0\end{pmatrix}$$
These matrices share the common pattern of having a $2\times2$ block of $1$'s, with a single $\bar 1$ in the block and $0$ outside. Now consider the following four matrices:
$$A=[3a]=\begin{pmatrix}0&0&0\\0&1&1\\0&\bar 1&1\end{pmatrix}\quad
B=\begin{pmatrix}0&0&0\\0&1&1\\0&1&\bar 1\end{pmatrix}\quad
C=\begin{pmatrix}0&1&\bar 1\\0&1&1\\0&0&0\end{pmatrix}\quad
D=\begin{pmatrix}0&\bar 1&1\\0&1&1\\0&0&0\end{pmatrix}$$
It turns out that matrices of this form and their negatives are closed under left multiplication by the generators, which constitutes a proof of the goal because the identity matrix is not of this form. Here is the multiplication table:
\begin{array}{c|cccc}\times&A&B&C&D\\\hline
[3a]=A&-A&0&A&A\\
[3a^{-1}]&0&-B&B&B\\
[3b]&C&C&-C&0\\
[3b^{-1}]&D&D&0&-D
\end{array}
The $0$ elements are because $[3a^{-1}][3a]=[9I]$ is divisible by $3$ and so is equal to $0$, but can only occur if the word $M$ is not reduced (contains adjacent cancelling pairs) - for example $[3a]B=0$, but $\pm B$ only arises from a product $[3a^{-1}]x$ for some $x\in\{A,B,C,D\}$. This proves that if $M$ is any product of $n$ terms selected from $\{a,b,a^{-1},b^{-1}\}$ whose last term is $a$ and with no cancelling pairs, $[3^nM]\in\{\pm A,\pm B,\pm C,\pm D\}$, but $[3^nI]=0$. Thus $M\ne I$. Argumentation by symmetry (or with a different set of matrices) also establishes the result for words that end in $a^{-1},b,b^{-1}$, so we have that $M\ne I$ for any nontrivial word.
A: The essential is thing is that we have two 'independent' rotations of infinite order (i.e. by irrational angle), $a$ and $b$, these will be mapped to the generators of $F_2$, let's call them $x$ and $y$.
It gives a homomorphism $F_2\to SO(3),\ x\mapsto a,\ y\mapsto b$, its image is the subgroup $H$. And we are about to prove that it is injective, which is equivalent to saying that its kernel is zero, that is only the empty word is mapped to $\mathtt{id}$.
The last paragraph you cited aims to finish this technical part of the proof for this particular choice of $a$ and $b$. Using $\cos\theta=\frac13$ and $\sin\theta=\frac{2\sqrt2}3$ you can write up the rotation matrices.
