I'm thinking about $SL_2(\mathbb{Z})$. Suppose we fix the $2\times1$ matrix $(q_0$ $1)^T$, for $q_0$ a fixed rational number.
My question is: Is it possible to get any $2\times1$ matrix of the form $(q$ $1)^T$ for $q$ rational by (left) multiplying $(q_0$ $1)^T$ by elements of $SL_2(\mathbb{Z})$?
$$\left[ \begin{matrix} a & b \\ c & d \end{matrix}\right] \left[ \begin{matrix} q_0 \\ 1 \end{matrix}\right]=\left[ \begin{matrix} aq_0+b \\ cq_0+d \end{matrix}\right]$$
Suppose we wanted $q=q_0/2$. Then we would need to find $aq_0+b=q_0/2$. But then $b=q_0(a-1/2)$. $b$ is an integer, so that means that $q_0$ must have in its irreducible decomposition a factor of two in the numerator, i.e. $q_0=\frac{2n}{m}$ where $m$ is odd. Then $b=\frac{n}{m}(2a-1)$; $b$ is still an integer, so it must be that $m=\pm 1$, and thus $q_0$ was an integer all along. (note that the fact that the determinant was one didn't even matter to see this)
Assuming I understood your question well, you're asking that given a fixed $q_0$, can we get the rest of the rationals using elements of $SL_2(\mathbb{Z})$, then choosing any non-integral rational number $q$ at this point shows that in general, the answer is no.
Suppose such a matrix exist, say $ M= \left[ {\begin{array}{cc} a & b \\ c & d \\ \end{array} } \right] $ with $ad-bc=1$, and assume that $q\neq 0$, then after multiplying $M$ and ${(q_0,1)}^T$ and equating it to ${(q,1)}^T$ you'll get the matrix $M$ in terms of $a$ as follows $ M= \left[ {\begin{array}{cc} a & q-aq_0-q/a \\ a/q & 1-aq_0/q \\ \end{array} } \right]$ which may or may not belong to $SL_2(\mathbb{Z})$.
For example, if $q_o = 1$ and $q=1$ then $M$ will have entries from $\mathbb{C}$. so the answer is NO.
