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)? 



Infinite cyclic subgroups of $\text{SL}_2(\mathbb{Z})$
The maximal infinite cyclic subgroups are of the form you mentioned (to get all infinite cyclic subgroups you also need to include their finite index subgroups). This follows from Theorem 1.4 in [Sarnak, Peter, Class numbers of indefinite binary quadratic forms. J. Number Theory 15 (1982), no. 2, 229--247](https://www.sciencedirect.com/science/article/pii/0022314X82900282) since the generator of such a subgroup must be a primitive hyperbolic element.


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).


