The congruence subgroup problem for SLn(Z)

Following Bass–Milnor–Serre, we prove that SLn(Z) has the congruence subgroup property for n ≥ 3. This was originally proved by Mennicke and Bass–Lazard–Serre. Let Γn = SLn(Z). The congruence subgroup problem for Γn (solved independently by Mennicke [Me] and Bass–Lazard–Serre [BLS]) seeks to classify all finite-index subgroups of Γn. For l ≥ 2, the level l principal congruence subgroup of Γn, denoted Γn(l), is the kernel of the homomorphism Γn → SLn(Z/l) that reduces the entries in matrices modulo l. Clearly Γn(l) is finite-index in Γn. A subgroup G of Γn is a congruence subgroup if there exists some l ≥ 2 such that Γn(l) ⊂ G. Mennicke and Bass–Lazard–Serre proved the following theorem. Theorem A. For n ≥ 3, every finite-index subgroup of Γn is a congruence subgroup. Remark. This is false for n = 2. Indeed, SL2(Z) ≅ (Z/4) ∗Z/2 (Z/6) contains a free subgroup of finite index, and thus contains a veritable zoo of finite-index subgroups. Most of these are not congruence subgroups. Bass–Milnor–Serre [BMiS] later generalized Theorem A to deal with finite-index subgroups of SLn(O) for number rings O; they proved that SLn(O) satisfies a version of the congruence subgroup problem if and only if O has a real embedding. In this note, we will describe Bass–Milnor–Serre’s proof, specialized to just prove Theorem A. 1 Reduction to a generating set It turns out that the key to proving Theorem A is to construct (normal) generating sets for Γn(l). For distinct 1 ≤ i, j ≤ n, let eij ∈ Γn denote the elementary matrix with 1’s along the diagonal and at position (i, j), and 0’s elsewhere. Observe that eij ∈ Γn(l). Bass–Milnor–Serre proved the following theorem. Theorem 1.1. For n ≥ 3 and l ≥ 2, the group Γn(l) is normally generated (as a subgroup of Γn) by {eij ∣ 1 ≤ i, j ≤ n distinct}. The remaining sections of this note will be devoted to the proof of Theorem 1.1, which will be completed in §5. Before we get to that, we will show how to derive Theorem A from it.