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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.

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    $\begingroup$ They are super-rigid by Margulis, hence have finite abelianization. This also follows from Margulis normal subgroup theorem. $\endgroup$ – Misha Feb 7 '15 at 1:04
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    $\begingroup$ The finiteness of abelianization is a consequence of Serre's proof of the congruence subgroup property, see the corollary to Theorem 3 (bottom of p. 500) in J.-P. Serre: Le Problème des groupes de congruence par $SL_2$. Ann. Math. 92 (1970) pp 489--527. I guess, a closer inspection of the arguments would allow for more precise description of the group structure as well... $\endgroup$ – Matthias Wendt Feb 7 '15 at 8:25
  • $\begingroup$ Any finite index subgroup of the Hilbert modular group is an irreducible lattice in a higher rank group; though these do not have property $T$, they do have (as Misha said) the Margulis normal subgroup property that any infinite normal subgroup is of finite index; this was also proved for the Hilber modular case by Serre (Mathias WEndt's comment). For either of these reasons, the abelianisation of a finite index subgroup is finite. $\endgroup$ – Venkataramana Feb 7 '15 at 8:52
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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.

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