Matrix groups generated by translation and inversion in the unit sphere Let $\alpha$ be algebraic over $\mathbb{Q}$, and consider the subgroup $G$ of $\mathrm{SL}_2(\mathbb{C})$ generated by inversion in the unit sphere and translation by $\alpha$.  That is, consider $G=\langle S, T \rangle$, in which
$$ S:= \left(\begin{smallmatrix} 0 & -1 \\ 1 & 0 \end{smallmatrix}\right) \quad \text{and} \quad T:= \left( \begin{smallmatrix} 1 & \alpha \\ 0 & 1 \end{smallmatrix}\right).$$
For which $\alpha$ is the structure of $G$ known?  Of particular interest to me is the following: for which $\alpha$ do we obtain $\mathrm{SL}_2(\mathbb{Z}) \subset G$?
Perhaps this occurs iff $1/\alpha$ is an algebraic integer?
 A: Let $K$ be the number field generated by $\alpha$. Consider the prime factorization of the fractional ideal $(\alpha)$. If this prime factorization contains a prime ideal $P$ to a positive exponent, then $\alpha$ lies in the localization $S^{-1}\mathcal{O}_K$ away from all of the prime ideals in the prime factorization which occur with negative exponents. $P$ is a prime ideal in this localization, so we can take the quotient by $P$, which gives rise to a map
$$\text{SL}_2(S^{-1} \mathcal{O}_K) \to \text{SL}_2(S^{-1} \mathcal{O}_K/P).$$
$G$ is contained in the domain of this map, as is $\text{SL}_2(\mathbb{Z})$, but the image of $T$ under this map is the identity since $\alpha \equiv 0 \bmod P$, so the image of $G$ is generated by $S$ and in particular does not contain the image of $\text{SL}_2(\mathbb{Z})$.
Thus a necessary condition for the desired property to hold is that the prime factorization of $(\alpha)$ contains only negative exponents. This is equivalent to $\frac{1}{\alpha}$ being an algebraic integer. However, I don't know how to prove that the property holds unless $\frac{1}{\alpha}$ is in fact an integer (in which case it's straightforward). 
Edit: I believe your conjecture is false. If $\text{SL}_2(\mathbb{Z}) \subset G$ and $\alpha$ is real, then it follows that a fundamental domain of $\text{SL}_2(\mathbb{Z})$ acting on the upper half plane $\mathbb{H}$ is a union of fundamental domains of $G$ acting on $\mathbb{H}$; in particular, the latter have smaller hyperbolic area than the former. But if $\alpha$ is an algebraic number with a real conjugate of absolute value greater than $1$, then WLOG $\alpha$ is this conjugate (the property of containing $\text{SL}_2(\mathbb{Z})$ is invariant under Galois action), and it is straightforward to write down a fundamental domain which has larger hyperbolic area than a fundamental domain of $\text{SL}_2(\mathbb{Z})$, which is a contradiction.
As an explicit example where $\frac{1}{\alpha}$ is an algebraic integer, take $\alpha = \frac{1 + \sqrt{5}}{2}$. 
