# For $G$ an adjoint Chevalley group, are all of $G(\mathbb Z)$'s finite-index subgroups congruence subgroups?

Let $$G$$ be an adjoint Chevalley group. Are all of $$G(\mathbb Z)$$'s finite-index subgroups congruence subgroups?
I read a theorem that states: When $$G$$ is the universal Chevalley group and it's not of type $$\operatorname{SL}_{2}$$ every finite index subgroup of $$G(\mathbb Z)$$ is a congruence subgroup (in Matsumoto - Sur les sous-groupes arithmétiques des groupes semi-simple déployés).

I've been looking for sources and results on that topic that I can learn and quote form, unfortunately, all I could find was in French.

• You are asking if a Chevally group of adjoint type has the congruence subgroup property. The answer is no, unless the Chevalley group is simply connected. The precise reference is Serre's article in Seminaire Bourbaki (volume 10) numdam.org/book-part/SB_1966-1968__10__275_0. See section 1.2 part c) of that paper. – Venkataramana Jul 11 at 0:38
• I forgot to mention that this is also in French, but the expository paper is very easy to read (and there is google translate) – Venkataramana Jul 11 at 0:55
• Venkataramana has given a concise answer known for a long time (with a reference to Serre) to the question raised in the header, though I've tried to answer another question raised about sources in English. – Jim Humphreys Jul 15 at 18:56

Chapter VI of my old Springer Lecture Notes in Mathematics 789 Arithmetic Groups (1980) is in English and gives a version of H. Matsumoto's and C. Moore's arguments for the subgroups of SL$$_n (\mathbb{Z})$$ of finite index when $$n \geq 3$$. But this account also gives some clues to the general case, where even for simply connected groups there are exceptions for imaginary fields.