TITLE: Abelianization of Hilbert modular group
QUESTION [6 upvotes]: Let $d>0$ be a square free positive integer and let $\mathcal{O}_d$ be the ring of integers in $\mathbb{Q}[\sqrt{d}]$.  What is the abelianization of the Hilbert modular group $\text{SL}_2(\mathcal{O}_d)$?  If this is too hard, is at least the rank of the abelianization known?
I'd also be interested in knowing this for finite-index subgroups of $\text{SL}_2(\mathcal{O}_d)$.
These groups are lattices in $\text{SL}_2(\mathbb{R}) \times \text{SL}_2(\mathbb{R})$.  I believe that this implies that they don't have property (T), so there isn't a cheap way of seeing that the rank of the abelianization is $0$.  But they do have some higher-rank behavior; for instance, Serre proved that they do have the congruence subgroup property.

REPLY [6 votes]: The question you ask (at least for ${\rm SL}_{2}(\mathcal{O}_{d})$) is the subject of a paper by Hatice Boylan and Nils-Peter Skoruppa (available on arXiv here). Basically, all the abelian quotients of ${\rm SL}_{2}(\mathcal{O}_{d})$ come from prime ideals above $2$ and $3$ in $\mathcal{O}_{d}$. In cases when there are none (like $d = 5$), the group ${\rm SL}_{2}(\mathcal{O}_{d})$ is perfect.