Let $\Gamma(2)$ be the congruence group:
$$ \Gamma(2) = \left\{ \left( \begin{array}{cc} a & b \\ c & d \end{array} \right) \in \text{SL}_2(\mathbb{Z}) : \left( \begin{array}{cc} a & b \\ c & d \end{array} \right) \equiv \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right) \pmod 2\right\} $$
We know by some means, this group $\Gamma(2)$ is finitely generated.
Clearly, $\Gamma(2) \subset \text{GL}_2(\mathbb{Q})$ the set of invertible $2 \times 2$ matrices with rational coefficients. What are the cosets? Let's consider the double-coset of a single element:
$$ \Gamma_0(2) \, \left( \begin{array}{cc} 1 & \frac{1}{5} \\ 0 & 1\end{array} \right) \, \Gamma_0(2) = \bigsqcup_k \Gamma(2) \, x_k $$
for some other coset representatives $x_k \in \text{GL}_2(\mathbb{Q})$. Maybe this is in a paper somewhere, but I'm hoping this could be solved from first principles. Certainly if we write out all the matrices and put the relevant conditions we might hope to solve it.
The answer should be a disjoint union of left- (or right-) cosets of $\Gamma(2)$.
This is related to modular forms:
In number theory, the Hecke algebra corresponding to a congruence subgroup $\Gamma$ of the modular group is spanned by elements of the double coset space ${\displaystyle \Gamma \backslash \mathrm {GL} _{2}^{+}(\mathbb {Q} )/\Gamma }$ ; the algebra structure is that acquired from the multiplication of double cosets described above. Of particular importance are the Hecke operators ${\displaystyle T_{m}}$ corresponding to the double cosets ${\displaystyle \Gamma _{0}(N)g\Gamma _{0}(N)}$ or ${\displaystyle \Gamma _{1}(N)g\Gamma _{1}(N)}$, where ${\displaystyle g=\left({\begin{smallmatrix}1&0\\0&m\end{smallmatrix}}\right)}$ (these have different properties depending on whether m and N are coprime or not), and the diamond operators ${\displaystyle \langle d\rangle }$ given by the double cosets $${\displaystyle \Gamma _{1}(N)\left({\begin{smallmatrix}a&b\\c&d\end{smallmatrix}}\right)\Gamma _{1}(N)}$$ where ${\displaystyle d\in (\mathbb {Z} /N\mathbb {Z} )^{\times }}$ and we require ${\displaystyle \left({\begin{smallmatrix}a&b\\c&d\end{smallmatrix}}\right)\in \Gamma _{0}(N)}$ (the choice of $a$, $b$, $c$ does not affect the answer).
However, at the moment all of this modular forms is a distract, since I can't even compute group orbits, or double-cosets in this single example.