# Infinite cyclic subgroups of $\text{SL}_2(\mathbb{Z})$

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)?

Edited: Oops, I missed that $$\mathrm{SL}_2(\mathbb{Z})$$ also has infinite cyclic subgroups generated by parabolic elements, but those are also stabilizers of quadratic forms of discriminant $$0$$ (that is, that are squares).