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

For all $m\ge 1$, we prove that the abelianization of ${\mathrm{SL}}_2(\mathbb{Z}[\frac{1}{m}])$ is (1) trivial if $6\mid m$; (2) $\mathbb{Z}\mathrm{/}3\mathbb{Z}$ if $2\mid m$ and $\gcd (3,m)=1$; (3) $\mathbb{Z}\mathrm{/}4\mathbb{Z}$ if $3\mid m$ and $\gcd (2,m)=1$; and (4) $\mathbb{Z}\mathrm{/}12\mathbb{Z}\cong \mathbb{Z}\mathrm{/}3\mathbb{Z}\times \mathbb{Z}\mathrm{/}4\mathbb{Z}$ if $\gcd (6,m)=1$. This completes known computational results of Bui Anh & Ellis for $m\le 50$. The proof is completely elementary, and in particular does not use the congruence subgroup property. We also find a new presentation for ${\mathrm{SL}}_2(\mathbb{Z}[\frac{1}{2}])$. This presentation has two generators and three relators. Thus, ${\mathrm{SL}}_2(\mathbb{Z}[\frac{1}{2}])$ 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 ${\mathrm{SL}}_2(\mathbb{Z}\mathrm{/}m\mathbb{Z})$, where $m$ is odd.