Bass–Milnor–Serre theorem

Remarks

The idea of the proof: every subgroup of finite index contains a normal subgroup $N$ of finite index $m$. Then $N$ contains the powers $e_{ij}(r)^m = e_{ij}(mr)$ of the elementary matrices. Hence $N$ contains the normal subgroup $E_n(\mathcal{O}, \mathfrak{q})$ generated by the $e_{ij}(q)$ with $q \in \mathfrak{q}$, for $\mathfrak{q} = m\mathcal{O}$. The quotient $\Gamma(\mathfrak{q})/E_n(\mathcal{O}, \mathfrak{q})$ is a relative $SK_1$ group, which Bass, Milnor and Serre computed with Mennicke symbols and reciprocity laws for power residue symbols. When $K$ has a real embedding this quotient vanishes, so $N \supseteq \Gamma(\mathfrak{q})$. Example: for $K = \Q$, every finite-index subgroup of $SL_3(\Z)$ contains the kernel of reduction modulo some integer $m \ge 1$. The hypothesis $n \ge 3$ is necessary: $SL_2(\Z)$ has subgroups of finite index that are not congruence subgroups. When $K$ is totally imaginary the conclusion fails, but the obstruction is small: it is a finite cyclic group whose order is the number of roots of unity in $K$. The same paper treats the symplectic groups $Sp_{2n}(\mathcal{O})$ for $n \ge 2$.

Going deeper

For a student

The theorem solved the congruence subgroup problem for $SL_n$, $n \ge 3$, over rings of integers, and the solution is phrased in algebraic K-theory. It also gives $SK_1(\mathcal{O}) = 0$ for the ring of integers of every number field, so $K_1(\mathcal{O}) \cong \mathcal{O}^\times$. The method, via relative $K_1$ groups and Mennicke symbols, became a model for later work on other arithmetic groups. In terms of profinite completions, it says the congruence kernel of $SL_n(\mathcal{O})$ is trivial when $K$ has a real place.