The abelianization of $\operatorname{SL}_2(\mathbb{Z}[\frac{1}{m}])$

For all $m \geq 1$, we prove that the abelianization of $\operatorname{SL}_2(\mathbb{Z}[\frac{1}{m}])$ is (1) trivial if $6 \mid m$; (2) $\mathbb{Z} / 3\mathbb{Z}$ if $2 \mid m$ and $\gcd(3,m)=1$; (3) $\mathbb{Z} / 4 \mathbb{Z}$ if $3 \mid m$ and $\gcd(2,m)=1$; and (4) $\mathbb{Z} / {12}\mathbb{Z} \cong \mathbb{Z} / 3\mathbb{Z} \times \mathbb{Z} / 4\mathbb{Z}$ if $\gcd(6,m)=1$. This completes known computational results of Bui Anh & Ellis for $m \leq 50$. The proof is completely elementary, and in particular does not use the congruence subgroup property. We also find a new presentation for $\operatorname{SL}_2(\mathbb{Z}[\frac{1}{2}])$. This presentation has two generators and three relators. Thus, $\operatorname{SL}_2(\mathbb{Z}[\frac{1}{2}])$ admits a presentation with deficiency equal to the rank of its Schur multiplier. This also gives new and very simple presentations for the finite groups $\operatorname{SL}_2(\mathbb{Z} / m \mathbb{Z})$, where $m$ is odd.