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The Congruence Subgroup Problem for Sln(Z)

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The Congruence Subgroup Problem for Sln(Z) The congruence subgroup problem for SLn(Z) Andrew Putman Abstract 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 ℓ ∈ ( ) 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 { ℓ ∣ ≤ ≤ } 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. 1 Proof of Theorem A. Let G be a finite-index subgroup of Γn. We wish to show that G contains Γn(ℓ) for some ℓ ≥ 2. Passing to a deeper finite-index subgroup, we can assume that G is a normal subgroup of Γn. For all distinct 1 ≤ i, j ≤ n, the fact that ≥ ℓij ∈ G is finite-index in Γn implies that there exists some ℓij 1 such that eij G. Define ℓ ∈ ≤ ≤ ℓ to be the least common multiple of the ℓij, so eij G for all distinct 1 i, j n. Since G is a normal subgroup of Γn, we deduce that G contains the normal closure { ℓ ∣ ≤ ≤ } ( ) of the set eij 1 i, j n distinct , which by Theorem 1.1 is Γn ℓ . Thus G is a congruence subgroup, as desired. 2 Reduction to 2 × 2 matrices The first step in proving Theorem 1.1 is to construct a larger generating set for Γn(ℓ). ( ) { ℓ ∣ ≤ ≤ } < Define EΓn ℓ to be the normal closure in Γn of eij 1 i, j n distinct . For m n, we will regard Γm(ℓ) as a subgroup of Γn(ℓ) via the map A 0 A ↦ ( ) . (2.1) 0 1 The following lemma is the main result of this section. Lemma 2.1. For n ≥ 2 and ℓ ≥ 2, the group Γn(ℓ) is generated by Γ2(ℓ) and EΓn(ℓ). For the proof of Lemma 2.1, we need the following lemma. Lemma 2.2. For n ≥ 3, let a1, . , an ∈ Z satisfy gcd(gcd)(a1, . , an) = 1. Then there exist c2, . , cn ∈ Z such that gcd(a2 + c2a1, . , an + cna1) = 1. Remark. A ring R which satisfies the conclusion of Lemma 2.2 for n ≥ r is said to satisfy Bass’s stable range condition SRr. We remark that in this more general context, the condition gcd(a1, . , an) = 1 should be interpreted as asserting that Ra1 + ⋯ + Ran = R. Lemma 2.2 says that Z satisfies SR3. Proof of Lemma 2.2. If any of the ai are zero then this is trivial, so we can assume that ai ≠ 0 for all 1 ≤ i ≤ n. In this case, it will turn out that we can find a single c ∈ Z such that gcd(a2 + ca1, a3, . , an) = 1. Set b = gcd(a3, . , an), and let p1, . , pk be the distinct primes dividing b. For each 1 ≤ i ≤ k, we know that pi cannot divide both a1 and a2, so there exists some λi ∈ {0, 1} such that a2 + λia1 ≡/ 0 (mod pi). By the Chinese remainder theorem, there exists some c ∈ Z such that c ≡ λi (mod pi) for 1 ≤ i ≤ k, which implies that a2 + ca1 ≡/ 0 (mod pi) for all 1 ≤ i ≤ k. We conclude that gcd(a2 + ca1, b) = 1, and thus that gcd(a2 + ca1, a3, . , an) = 1. 2 Proof of Lemma 2.1. By induction on n, it is enough to show that Γn(ℓ) is generated by Γn−1(ℓ) and EΓn(ℓ). Consider M ∈ Γn(ℓ). Let the bottom row of M be (a1, . , an), so a1 ≡ ⋯ ≡ an−1 ≡ 0 (mod ℓ) and an ≡ 1 (mod ℓ). Also, gcd(a1, . , an) = 1. Since ℓ ∣ a1, we have gcd(ℓa1, a2, . , an) = 1. By Lemma 2.2, we can find c2, . , cn ∈ Z such that gcd(a2 + c2ℓa1, . , an + cnℓa1) = 1. For ≤ ≤ ′ = + 2 i n, set ai ai ciℓa1. By multiplying M on the right by elements of { ℓ ∣ ≤ ≤ } eij 1 i, j n distinct , we can perform column operations to convert its bottom ( ′ ′ ) ∣ row into a1, a2, . , an . Next, since ℓ a1 we can multiply our matrix on the right { ℓ ∣ ≤ ≤ } by elements of eij 1 i, j n distinct to perform column operations and convert ( ′ ′ ) its bottom row into ℓ, a2, . , an . The next step is the most subtle and is the { ℓ ∣ ≤ ≤ } reason why we need to take the normal closure of eij 1 i, j n distinct . Write ′ = + k an 1 kℓ. Multiplying our matrix on the right by e1n (which does not necessarily lie { ℓ ∣ ≤ ≤ } ( ′ ′ ) in eij 1 i, j n distinct ) converts its last row to ℓ, a2, . , an−1, 1 . Multiplying { ℓ ∣ ≤ ≤ } our matrix on the right by elements of eij 1 i, j n distinct , we can then perform column operations and convert its last row into (0,..., 0, 1). Multiplying our matrix −k ( ) on the right by e1n then does not change its last row, and using the fact that EΓn ℓ { ℓ ∣ ≤ ≤ } is the normal closure of eij 1 i, j n distinct we see that the resulting matrix is the result of multiplying our original matrix M on the right by an element of EΓn(ℓ). { ℓ ∣ ≤ ≤ } We now can multiply our matrix on the left by elements of eij 1 i, j n distinct to perform row operations and convert its last column to (0,..., 0, 1). Our matrix now lies in Γn−1(ℓ) ⊂ Γn(ℓ), as desired. 3 The nature of the quotient Define Qn(ℓ) = Γn(ℓ)/EΓn(ℓ). The main result of this section is as follows. Lemma 3.1. For n ≥ 3, the group Qn(ℓ) is a finitely generated abelian group. More- over, the action of Γn on Qn(ℓ) induced by the conjugation action of Γn on Γn(ℓ) is trivial. Remark. Lemma 2.1 implies that elements of Qn(ℓ) can be represented by 2 × 2 matrices; however, the condition n ≥ 3 in Lemma 3.1 is necessary. Proof of Lemma 3.1. Since Qn(ℓ) is a quotient of the finitely generated group Γn(ℓ) (we remark that the group Γn(ℓ) is finitely generated since it is a finite-index sub- group of the finitely generated group Γn), it is itself finitely generated. The fact that Qn(ℓ) is abelian will follow from the fact that the Γn-action on it is trivial since Γn(ℓ) ⊂ Γn, so it is enough to prove this. The group Γn is generated by elemen- tary matrices eij. Moreover, we have the easily-verified matrix identity eik = [eij, ejk] when 1 ≤ i, j, k ≤ n are distinct. Since n ≥ 3, we see that Γn is actually generated by {ein, eni ∣ 1 ≤ i ≤ n − 1}, so it is enough to prove that these generators act trivially on Qn(ℓ). Lemma 2.1 implies that Qn(ℓ) is generated by the image of Γ2(ℓ), so it is enough to prove that for 1 ≤ i ≤ n − 1 and M ∈ Γ2(ℓ), we have [ein,M] ∈ EΓn(ℓ) and [eni,M] ∈ EΓn(ℓ). For i ≥ 3, we have [ein,M] = [eni,M] = 1, so we just have to deal 3 with i = 1 and i = 2. All four of the necessary calculations are similar; we will give the details for e1n, and in fact to simplify our notation we will assume that n = 3 (the general case will be clear from this). Write ⎛ a b 0 ⎞ M = ⎜ c d 0 ⎟ . ⎝ 0 0 1 ⎠ We then have ⎛ 1 0 1 ⎞ ⎛ a b 0 ⎞ ⎛ 1 0 −1 ⎞ ⎛ d −b 0 ⎞ [e13,M] = ⎜ 0 1 0 ⎟ ⎜ c d 0 ⎟ ⎜ 0 1 0 ⎟ ⎜ −c a 0 ⎟ ⎝ 0 0 1 ⎠ ⎝ 0 0 1 ⎠ ⎝ 0 0 1 ⎠ ⎝ 0 0 1 ⎠ ⎛ a b 1 ⎞ ⎛ d −b −1 ⎞ = ⎜ c d 0 ⎟ ⎜ −c a 0 ⎟ ⎝ 0 0 1 ⎠ ⎝ 0 0 1 ⎠ ⎛ 1 0 1 − a ⎞ = ⎜ 0 1 −c ⎟ ⎝ 0 0 1 ⎠ This clearly lies in EΓn(ℓ), as desired.
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