$SO(\mathbb{Q})$ is not a finitely generated group for $n \geq 2$ 
Suppose that $SO(\mathbb{Q}) = \{ A \in M_n{\mathbb{(Q)}}: A^TA=I,
\det A = 1 \}$ is a subgroup of $n\times n$ matrices with rational
  entries under matrix multiplication. Show that for $n \geq 2$ this
  group is not finitely generated.

The cardinality of $SO(\mathbb{Q})$ is countable, so it's possible that it is finitely generated. I think I need to find an algorithmic approach that takes a set of matrices in $SO(\mathbb{Q})$ and produces another matrix that can't be generated by that set. But this seems like a bit complicated. Any ideas?
 A: Hint. If $SO(n;\mathbb Q)$ is finitely generated, then there exist a finite set of primes $P$ (including $1$ here by convention), such that when an element $a_{ij}$ of any $A\in SO(n;\mathbb Q)$ is written as an irreducible fraction, its denominator is a product of powers of primes from $P$. Now note that $(m^2-n^2,\ 2mn,\ m^2+n^2)$ is a Pythagorean triple and there are infinitely many primes of the form $m^2+n^2$.
A: Assume you have a finite set of generators $A_1,\ldots, A_k$. For suitable $a_i\in \mathbb N$ (the common denominator), we have $a_iA_i\in M_n(\mathbb Z)$ and also $a_iA_i^{-1}\in M_n(\mathbb Z)$. Let $p$ be a prime with $p\not\mid \prod a_i$ and $p\equiv 1\pmod 4$. Then there exists a solution of $p=u^2+v^2$ with $u,v\in\mathbb Z$. Let
$$A=\begin{pmatrix}\frac{u^2-v^2}{u^2+v^2}&\frac{2uv}{u^2+v^2}&0&\cdots&0\\
\frac{-2uv}{u^2+v^2}&\frac{u^2-v^2}{u^2+v^2}&0&\cdots &0\\
0&0&1&\cdots&0\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
0&0&0&\cdots &1\end{pmatrix} $$
If $A$ wree a product of $A_i$s and $A_i^{-1}$s, multiplying it with a power of $\prod a_i$ would make it an integer matrix, but such a factor cannot get rid of the $p$ in the denominator!
