Let $\Gamma = \operatorname{SL}_2(\mathbb{Z})$ be the usual modular group. It is well-known that $\Gamma$ contains infinitely many distinct (non-conjugate even) subgroups which are isomorphic to the infinite cyclic group $(\mathbb{Z}, +)$. Indeed, for any square-free integer $d > 1$, the unit group of the quadratic field $\mathbb{Q}(\sqrt{d})$ will give rise to such a group. More generally, any irreducible, indefinite binary quadratic form $f(x,y) = ax^2 + bxy + cy^2$ will induce such a subgroup in $\Gamma$, with an explicit generator given by

$$\displaystyle \begin{pmatrix} \dfrac{t_f + bu_f}{2} & au_f \\ \\ -cu & \dfrac{t_f - bu_f}{2} \end{pmatrix},$$

where $(t_f, u_f)$ is the fundamental (positive) solution to the Pell equation $x^2 - \Delta(f) y^2 = 4$.

Is this a bijection? That is, each infinite cyclic subgroup of $\Gamma$ must arise from an irreducible binary quadratic form in this way?

Let $\Gamma = \operatorname{SL}_2(\mathbb{Z})$ be the usual modular group. It is well-known that $\Gamma$ contains infinitely many distinct (non-conjugate even) subgroups which are isomorphic to the infinite cyclic group $(\mathbb{Z}, +)$. Indeed, for any square-free integer $d > 1$, the unit group of the quadratic field $\mathbb{Q}(\sqrt{d})$ will give rise to such a group. More generally, any irreducible, indefinite binary quadratic form $f(x,y) = ax^2 + bxy + cy^2$ will induce such a subgroup in $\Gamma$, with an explicit generator given by

$$\displaystyle \begin{pmatrix} \dfrac{t_f + bu_f}{2} & au_f \\ \\ -cu & \dfrac{t_f - bu_f}{2} \end{pmatrix},$$

where $(t_f, u_f)$ is the fundamental (positive) solution to the Pell equation $x^2 - \Delta(f) y^2 = 4$.

Is this a bijection? That is, each infinite cyclic subgroup of $\Gamma$ must arise from an irreducible binary quadratic form in this way (up to conjugacy)?