# Congruence subgroup problem

2010 Mathematics Subject Classification: Primary: 20G30 [MSN][ZBL]

The congruence subgroup problem deals with the following question:

Is every subgroup of finite index in $\def\O{\mathcal{O}}G_\O$, where $\O$ is the ring of integers in an algebraic number field $k$ (cf. Algebraic number theory) and $G$ is a connected linear algebraic group defined over $k$, a congruence subgroup?

This is the classical statement of the congruence subgroup problem. A contemporary version of it is based on the concept of the congruence subgroup kernel, which is a measure of the deviation from a positive solution. Let $\def\G{\Gamma}\G$ denote the group $G_\O$, and let $\hat\G$ and $\bar\G$ be the completions of the group $\G$ in the topologies defined by all its subgroups of finite index and all congruence subgroups of $\G$, respectively. Then there is a surjective continuous homomorphism $\pi:\hat\G \to \bar\G$. The kernel of $\pi$ is called the congruence subgroup kernel and is denoted by $c(G)$. The positive solution of the classical congruence subgroup problem is equivalent to proving $c(G) = 1$. In its modern form, the congruence subgroup problem is that of computing the congruence subgroup kernel $c(G)$.

If $\def\SL{\textrm{SL}}\G=\SL(n,\Z)$, where $\Z$ is the ring of integers, it was known already in the 19th century that the congruence subgroup problem has a negative solution for $n=2$. For $n\ge 3$, it was proved in 1965 that every subgroup of finite index in $\SL(n,\Z)$ is a congruence subgroup (see [BaMiSe]). After this, the congruence subgroup problem was solved [BaMiSe] for $\G = \SL(n,\O)$, $n\ge 3$, and $\def\Sp{\textrm{Sp}}\Sp(2n,\O)$, $n \ge 2$, where $\Sp$ denotes the symplectic group. The results are as follows for these groups; $c(G)\ne 1$ only for totally imaginary fields $k$, in which case the congruence subgroup kernel is isomorphic to the (cyclic) group of roots of unity in $k$. It turned out that the same result holds for simply-connected Chevalley groups other than $\SL(2,k)$ (see [Ma]). The condition of being simply connected is essential, because it follows from the strong approximation theorem (cf. Linear algebraic group) that the congruence subgroup kernel $c(G)$ of a non-simply-connected semi-simple group $G$ is infinite. For every non-semi-simple group $G$ one has $c(G)=c(H)$, where $H$ is a maximal semi-simple subgroup of $G$; in particular, $c(G) = 1$ for a solvable group $G$.

A more general form of the congruence subgroup problem is obtained by replacing $\O$ by the ring

$$\O_S = \{x \in k:v(x) \le 1 \textrm{ for all } v \notin S\}$$ where $S$ is any finite set of inequivalent valuations of the field $k$ containing all Archimedean valuations. In this situation, the congruence subgroup kernel, denoted by $c(G,S)$, depends in an essential way on $S$ (see [Pl], [Ra]).

How to Cite This Entry:
Congruence subgroup problem. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Congruence_subgroup_problem&oldid=21558
This article was adapted from an original article by V.P. Platonov (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article