Bounded Generation CHAPTER 1

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1 CHAPTER 1 Bounded Generation 1A. Introduction A fundamental theorem of undergraduate linear algebra states that every invertible matrix is a product of elementary matrices. More precisely, it is not difficult to see that every element of SL(l, R) is a product of l 2 such matrices (see Exercise 2). (The same conclusions remain true when R is replaced with any field F.) The situation is more interesting if we replace the field R with the ring Z (or some other ring of integers). This makes it much more difficult to show that there is a number r (analogous to l 2 ), such that every l l matrix of determinant 1 is a product of r elementary matrices. (1.1) Definition. Let A be a commutative ring, and l be a positive integer. 1) An (elementary) column operation in SL(l, A) consists of adding a multiple of some column of a matrix to some other column of the matrix. 2) Any matrix that can be obtained from the identity matrix Id l l by applying a single column operation is called an elementary matrix. In linear algebra, one usually allows additional column operations, such as multiplying a column by an invertible scalar, but we are interested only in matrices of determinant 1, so these other operations are redundant (see Exercise 4). The elementary matrices can also be described more explicitly: (1.2) Notation. For i, j {1,..., l} and a A with i j, let E i,j (a) SL(l, A) be the l l matrix with 1 s on the diagonal, a in the (i, j) entry, and all other entries 0. Then E i,j (a) is an elementary matrix. Conversely, any elementary matrix is of this form, for some choice of i, j, and a (see Exercise 5). It is easy to see that the elementary matrices generate SL(3, Z) (see Exercise 6). Thus, every element of SL(3, Z) is a product of elementary matrices. This means that each matrix γ in SL(3, Z) can be written as a word γ ε 1 ε 2 ε m, where each ε i is an elementary matrix. D. Carter and G. Keller proved that the length of this word can be bounded, independent of the choice of the matrix γ: (1.3) Theorem (Carter-Keller). Every matrix in SL(3, Z) is a product of 48 elementary matrices. For our purposes, the particular constant 48 is not important. All that matters is that we have a uniform bound on the number of elementary matrices needed. This issue can be described in abstract group-theoretic terms: If a set X generates Γ, then every element of Γ can be written as a word in X X 1. We would like the length of the word can be bounded, independent of the particular element of Γ. (1.4) Definition. A subset X of Γ boundedly generates Γ if there is a positive integer r, such that every element of Γ can be written as a word of length r in X X 1. In other words, for each γ Γ, there is a sequence x 1,x 2,..., x l of elements of X X 1, with l r, such that γ x 1 x 2 x l. 1

2 2 1. BOUNDED GENERATION If X X 1 and e X (that is, if X is closed under inverses and contains the identity element), this means that Γ X X X (with r factors in the product). In these terms, we have the following weaker formulation of the Carter-Keller Theorem: (1.3 ) Theorem (Carter-Keller). The elementary matrices boundedly generate SL(3, Z). See 1D and 1E for a nice proof of this fact that does not yield an explicit bound on the number of elementary matrices needed. (1.5) Corollary. For l 3, the elementary matrices boundedly generate SL(l, Z). Proof. The proof is by induction on l, with Thm. 1.3 as the base case. Let γ SL(l, Z), with l> 3. It is not difficult to see that a sequence of less than 2l column operations results in a matrix whose (l, l) entry is 1 (see Exercise 8a). Then less than l additional column operations yield a matrix whose last row is [0, 0,..., 0, 1] (see Exercise 8b). Then the (l 1) (l 1) submatrix obtained by deleting the last row and column must belong to SL(l 1, Z). Thus, by induction on l, a bounded number, say r, of additional column operations results in a matrix that differs from the identity matrix only in the last column. So less than l additional column operations are needed to obtain the identity matrix (see Exercise 8c). Therefore, γ is a product of r + 4l elementary matrices. (1.6) Remark. There is no reason to believe that the constant 48 in Thm. 1.3 is sharp, but a precise bound is probably not particularly interesting. On the other hand, it is interesting to note that the proof of Cor. 1.5 yields a bound that is quadratic in l (see????), and this is the correct order of growth (see Exercise 11). (1.7) Definition. An important special case of Defn. 1.4 is when X is a finite union of cyclic groups: Γ has bounded generation (or has finite width) if (for some r ) there are cyclic subgroups X 1,X 2,..., X r of Γ, such that Γ X 1 X 2 X r. In other words, there exist x 1,..., x r Γ (namely, generators of the cyclic groups X 1,..., X r ), such that each γ Γ can be written in the form γ x e 1 1 x e 2 2 x e r r, for some integers e 1,e 2,..., e r. (1.8) Example. 1) Z n has bounded generation. 2) Nonabelian free groups do not have bounded generation. Suppose, to the contrary, that, for some r, we may write F X 1... X r, with each X i cyclic. By the universality of free groups, this implies that the same r works for any 2-generated group. Hence, if H is any 2-generated finite group, and m is the exponent of H, then #Λ m r. By taking p > r, this contradicts the fact that, for any prime p, there is a 2-generated, finite group P of order p p, such that x p 1 for every x P (see Exercise 12). For any fixed i, j (with i j), the set E i,j (Z) is a cyclic subgroup of SL(l, Z) (see Exercise 13), so the Carter-Keller Theorem has the following consequence (see????): (1.9) Corollary. For l 3, the group SL(l, Z) has bounded generation. It is important to note that the above result is false for l 2 (see Exercise 16). In particular, unlike in SL(l, R), there is no bound on the number of elementary matrices needed to represent a matrix in SL(2, Z) (see Exercises 17 and 18). This is a strengthening of the fact that there are integers a and b, such that the Euclidean Algorithm takes arbitrarily many steps in order to calculate gcd(a, b). A fundamental open question is to determine which lattices have bounded generation. This is independent of commensurability (see Exercise 19). It has been conjectured that, under certain technical conditions, bounded generation is equivalent to the Congruence Subgroup Property (see 1.16 and 1.17). The following is a special case of this: (1.10) Conjecture. Suppose Γ is an irreducible lattice in G, R-rank G 2, and G/Γ is not compact. Then Γ has bounded generation.

3 1A. INTRODUCTION 3 The Carter-Keller Theorem (1.3) establishes this conjecture in the case of SL(l, Z). More generally, if l 3, and O is the ring of integers in some number field, then the elementary matrices boundedly generate SL(l, O). The case l 2 is more subtle, but the answer is known, and verifies the conjecture in this case (see Exercises 20 and 21): (1.11) Theorem. Let O be the ring of integers of some number field. The elementary matrices boundedly generate SL(2, O) if and only if O has infinitely many units. (1.12) Remark. Suppose O is the ring of integers of a number field F. The Dirichlet Unit Theorem (??) tells us that O has infinitely many units if and only if F Q and either F : Q 2 or F R. (1.13) Example. 1) The only units of the ring Z are {±1}, so Thm states that the elementary matrices do not boundedly generate SL(2, Z), as we already knew. 2) The only units of the ring Z[i] of Gaussian integers are {±1, ±i}, so Thm states that the elementary matrices do not boundedly generate SL ( 2, Z[i] ). 3) The ring Z[ 2] has infinitely many units (see Exercise 22), so Thm states that the elementary matrices do boundedly generate SL ( 2, Z[ 2] ). This bounded generation is not at all obvious. Generalizing the Carter-Keller Theorem (1.9), Conj has been verified for arithmetic subgroups of many Q-split groups: (1.14) Theorem. Let O be the ring of integers of a number field F, and G be a simple algebraic group over F. If G is F-split, and F-rank G 2, then the arithmetic group G O has bounded generation (by unipotent elements). Exercises for 1A. #1. Show that every element of SL(3, R) is a product of 9 elementary matrices. #2. Let F be a field. Show that every element of SL(l, F) is a product of l 2 elementary matrices. #3. A matrix P is a monomial matrix if P has exactly one nonzero entry in each row, and P has exactly one nonzero entry in each column. Show that if P is any monomial matrix of determinant 1, then P can be reduced to the identity by a sequence of elementary column operations (see Definition 1.1). #4. Let us say that a generalized column operation consists of either: a) adding a multiple of some column of a matrix to some other column of the matrix, b) multiplying some column of a matrix by an invertible scalar, or c) interchanging two columns of a matrix. Show that if a matrix of determinant 1 can be reduced to the identity by a sequence of generalized column operations then it can also be reduced to the identity by a sequence of elementary column operations. [Hint: If a matrix can be reduced to the identity by a sequence of generalized column operations, then it can be written as a product ME, where M is a monomial matrix and E is a product of elementary matrices (because any invertible monomial matrix normalizes the set of elementary matrices).] #5. Verify that every elementary matrix is of the form described in Notation 1.2. #6. Let A be a Euclidean domain. a) Show that the elementary matrices generate SL(2, Z). b) Generalize to SL(l, Z). [Hint: For a b SL(2, Z), either { a, b } {0, 1}, or applying a single column operation can c d make max ( a, b ) smaller.]

4 4 1. BOUNDED GENERATION #7. Suppose r 1 and r 2 are positive integers with r 1 < r 2. Show that if T is product of r 1 elementary matrices, then T is also a product of r 2 elementary matrices. [Hint: Id is an elementary matrix.] #8. Let T SL(l, Z). a) Show that applying some sequence of less than 2l elementary column operations to T results in a matrix whose (l, l) entry is 1. b) Show that if T l,l 1, then applying some sequence of less than l elementary column operations to T results in a matrix whose last row is [0, 0,..., 0, 1]. c) Show that if T differs from the identity matrix only in the last column, then applying some sequence of less than l elementary column operations to T results in the identity matrix. [Hint: (8a) This is easy (using only l 1 operations) if gcd(t l,1,t l,2,..., T l,l 1 ) 1. Apply column operations to reduce to this case.] #9. Assume the conclusion of Thm. 1.3 (but not the specific bound of Thm. 1.3). Show there is a constant C, such that, for l 3, every element of SL(l, Z) is a product of Cl 2 elementary matrices. [Hint: The proof of Cor. 1.5 provides an induction step.] #10. Let U + l be the subgroup of SL(l, Z) consisting of upper-triangular matrices with 1 s on the diagonal, and U l be the subgroup of SL(l, Z) consisting of lower-triangular matrices with 1 s on the diagonal. For l 3, show that if (U + l 1 U l 1 )r SL(l, Z), then (U + l U l )r SL(l + 1, Z). [Hint: It suffices to show that (U + l U l )r is closed under multiplication by E i,i+1 (Z) and E i+1,i (Z), for 1 i l 1. Let V + l E 1,l(Z) E l 1,l (Z) and V l E l,1(z) E l,l 1 (Z). Then, by assumption, and because SL(l 1, Z) normalizes V + l and V l, we have (U + l U l )r SL(l 1, Z)(V + l V l )r. This is closed under multiplication by E 1,2 (Z) and E 2,1 (Z) (on the left).] #11. Find ɛ> 0, such that, for all large l, there exists γ SL(l, Z), such that it γ is not a product of less than ɛn 2 elementary matrices. [Hint: Reduce modulo a large prime p. If r < ɛn 2, then the number of products of elementary matrices (mod p) is less than the number of upper triangular unipotent matrices (mod p).] #12. Define σ GL(m, p) by σ (v 1,..., v m ) (v 2,v 3,..., v m,v 1 ); V {v (F p ) m m i1 v i 0 }; and P σ V, the semidirect product of σ with V. (So (σ i, v) (σ j, w) ( σ i+j,σ j (v)+ w ).) Show that if m p, then a) #P p p, b) x p 1 for every x P; and c) σ, v P, where v (1, 1, 0, 0,..., 0) V. #13. For 1 i, j l with i j, show that the subset { E i,j (n) n Z } of SL(l, Z) is a cyclic subgroup. [Hint: E i,j (a) b E i,j (ab).] #14. Derive Cor. 1.9 from Cor #15. Suppose Γ 1 is a finite-index subgroup of Γ 2. a) Show that if Γ 1 has bounded generation, then Γ 2 has bounded generation. b) Show that if Γ 2 has bounded generation, then Γ 1 has bounded generation. #16. Show that SL(2, Z) does not have bounded generation. [Hint: Use Exer. 15. You may assume (without proof) the fact that some finite-index subgroup of SL(2, Z) is a nonabelian free group.]

5 1B. APPLICATIONS OF BOUNDED GENERATION 5 #17. Show that the elementary matrices do not boundedly generate SL(2, Z). [Hint: Use Exer. 16.] #18. Define the Fibonacci sequence {F n } inductively, by F n F n 1 + F n 2 (with F 0 0 and F [ 1 1). As an explicit example ] of the conclusion of Exer. 17, show that the matrix F n F n 1 ( 1) n 1 F n 1 ( 1) n 1 SL(2, Z) is not a product of less than n elementary matrices. F n [Hint: Let x and y, and let x and ȳ be their images in PSL(2, Z). Then xy n 1 1 and x 2 y are elementary matrices, (xyx 2 y) n F2n+1 F 2n, and PSL(2, Z) is the 1 0 F 2n F 2n 1 free product of x and ȳ (see??)**. ] #19. Suppose Γ 1 is commensurable with Γ 2. Show that Γ 1 has bounded generation if and only if Γ 2 has bounded generation. #20. Let O be the ring of integers in some number field. a) Show that if the elementary matrices boundedly generate SL(2, O), then SL(2, O) has bounded generation. b) Generalize to SL(l, O). [Hint: You may assume, without proof, that O is finitely generated as an abelian group. That is, there are finitely many elements ω 1,..., ω k of O, such that OZω Zω k.] #21. Suppose O is the ring of integers of a number field F, S is the set of all (infinite) places of F (see??), and G v S SL(2, F v ), so SL(2, O) is an irreducible lattice in G (see??). Show R-rank G 2 if and only if O has infinitely many units. [Hint: You may assume (without proof) the consequence of the Dirichlet Unit Theorem stated in Rem ] #22. Show that ( 2 1 ) n is a unit in Z[ 2], for every n Z +. 1B. Applications of bounded generation We briefly mention some of the consequences of bounded generation. (1.15) Theorem. Assume Γ has bounded generation, and Γ /[Γ, Γ ] is finite, for every finite-index subgroup Γ of Γ. Then Γ is superrigid, in the sense that it has only finitely many irreducible representations of degree d, for each natural number d. Sketch of proof. It suffices to show, for each γ Γ, that there are only countably many possibilities for the trace of ρ(γ), as ρ ranges over all irreducible representations ρ : Γ GL(d, C) (see??). One can show, more precisely, that all the eigenvalues of ρ(γ) are algebraic over Q. To illustrate some of the ideas that are involved, we show how the Carter-Keller Theorem (1.3 ) can be used to prove that Γ SL(3, Z) is superrigid. Note, first, that ρ(e) is unipotent, for every elementary matrix E (see??)**. This means that ( ρ(e) Id ) d 0, so the Q-subalgebra Q [ ρ(e) ] of Mat d d (C) spanned by the powers of ρ(e) is finite dimensional (see Exercise 1), and the same goes for Q [ ρ(e) 1]. From the Carter-Keller Theorem (1.3 ), we know that Γ E 1 E r, for some elementary matrices E 1,..., E r Γ. Because ρ(γ ) is closed under multiplication, we know that Q [ ρ(γ ) ] is simply the Q-span of the elements of ρ(γ ). In other words, it is the Q-span of { ρ(e n 1 1 E n r r ) n i Z } {ρ(e 1 ) n 1 ρ(e 1 1 )m 1 ρ(e 2 ) n 2 ρ(e 1 2 )m2 ρ(e r ) n r ρ(e 1 r )m r m i,n i Z + }.

6 6 1. BOUNDED GENERATION Therefore Q [ ρ(γ ) ] Q [ ρ(e 1 ) ] Q [ ρ(e 1 ) 1] Q [ ρ(e r ) ] Q [ ρ(e r ) 1] is finite dimensional over Q. Hence, for each γ Γ, the powers of ρ(γ) span a finite-dimensional Q-subspace of Mat l l (C), so ρ(γ) satisfies a nontrivial polynomial with rational coefficients. The eigenvalues of ρ(γ) must all satisfy this same polynomial, so they are algebraic over Q. (1.16) Theorem. If Γ has bounded generation, Γ is an irreducible lattice in G, G is simply connected, and G/Γ is not compact (or a certain more general technical condition is satisfied), then Γ has the Congruence Subgroup Property (see Definition 2.3). Idea of proof. The known proofs are quite difficult. One approach uses bounded generation to show, for every ɛ> 0, that the number of subgroups of index n in Γ is no more than Cn ɛ log n (for some constant C C(ɛ)). It can be shown that this implies the Congruence Subgroup Property (see 2.4). The converse is conjectured to be true: (1.17) Conjecture (Rapinchuk). If Γ has the Congruence Subgroup Property, then it has bounded generation. (1.18) Remark. In light of the above conjecture, it is interesting to note that the proof of the Carter- Keller Theorem (1.3 ) presented in 1E below is based on Mennicke symbols and other ideas that were developed to prove the Congruence Subgroup Property for SL(3, Z). (1.19) Theorem. Suppose OZ [ 2 ] (or, more generally, O is the ring of integers of an algebraic number field, such that O has infinitely many units), and Γ is a subgroup of finite index in SL(2, O), then Γ has no nontrivial orientation-preserving action on the real line R. Idea of proof. Suppose Γ acts nontrivially on R, and let F be the set of fixed points of the action. This is a closed set, so its complement is a union of open intervals. By restricting to a single interval (which is homeomorphic to R), we may assume F. One can show that if U is any unipotent subgroup of Γ, then every U-orbit is a bounded subset of R. The elementary matrices boundedly generate SL(2, O) (see 1.11), so, by passing to a finite-index subgroup, we may assume that there are unipotent subgroups U 1,..., U r, such that Γ U 1 U 2 U r (see Exercise 2). Since every orbit of each U i is bounded, this implies that every Γ -orbit is bounded (see Exercise 3). Thus, the each orbit has a finite supremum. This supremum is fixed by Γ (see Exercise 4). This contradicts the conclusion of the preceding paragraph. (1.20) Theorem. If the elementary matrices boundedly generate SL ( 3, Z[x] ), then SL ( 3, Z[x] ) has Kazhdan s property T (see??). Idea of proof. Suppose we have an action of SL ( 3, Z[x] ) on a Hilbert space by affine isometries. We wish to show that the action has a fixed point (see??). For any i, j with i j, one can prove that every orbit of E i,j ( Z[x] ) is bounded. By bounded generation, we conclude that every orbit of SL ( 3, Z[x] ) is bounded (see Exercise 5). Thus, the closed convex hull of any orbit has a well-defined centroid (see Exercise 6). This centroid is fixed by SL ( 3, Z[x] ).

7 1C. COUNTING THE ELEMENTARY MATRICES IN A PRODUCT 7 Exercises for 1B. #1. Show that if a matrix T satisfies a nontrivial polynomial with rational coefficients, then the Q-span of the powers Id, T, T 2,... of T is finite dimensional. #2. Suppose Γ X 1 X 2 X r, where each X i is a subgroup of Γ, and Γ is a subgroup of finite index in Γ. Show there exist a subgroup Γ of finite index in Γ, and finitely many subgroups Y 1,Y 2,..., Y s of Γ, such that a) Γ Y 1 Y 2 Y s, and b) each Y i is conjugate (in Γ ) to a subgroup of some X j. [Hint: Let F i be a set of coset representatives for Γ X i in X i. For any γ Γ, we may write γ x 1 f 1 x 2 f 2 x r f r, with each x i X i and f i F i. Then γ x 1x 2 x r f 1 f 2 f r, where x i belongs to a conjugate of X i. We obtain a subgroup of finite index, because there are only finitely many possibilities for the product f 1 f 2 f r.] #3. Suppose Γ acts on R (by orientation-preserving homeomorphisms), Γ X 1 X 2 X r, where each X i is a subgroup of Γ, and for each i, every X i -orbit is a bounded subset of R. Show that every Γ -orbit is a bounded subset of R. #4. Suppose Γ acts on R (by orientation-preserving homeomorphisms), and C is a bounded subset of R that is Γ -invariant. Show that the supremum of C is fixed by every element of Γ. [Hint: Anything defined from an invariant set is invariant.] #5. Suppose Λ is a group, Λ acts by isometries of a metric space (Ω, d), E 1,..., E r are subgroups of Λ with Λ E 1 E 2 E r, and for each i, every E i -orbit is bounded. Show that every Λ-orbit is bounded. #6. A point b in a metric space B is the centroid of a subset Ω of B if Ω is contained in the closed ball of some radius r centered at b, but Ω is not contained in any closed ball of radius less than r (centered at any point of B). a) Let Ω be a (nonempty) bounded subset of R l. Show that Ω has a unique centroid. b) Let Ω be a (nonempty) bounded subset of a Hilbert space B. Show that Ω has a unique centroid. 1C. Counting the elementary matrices in a product The most straightforward way to prove that a particular generating set X boundedly generates a group Γ is to provide an algorithm that writes each element of Γ as a product of elements of X, and find a bound on the number of elements of X that are used by the algorithm. For example, this approach can quite easily be used to prove that every element of SL(l, R) is a product of l 2 elementary matrices (see Exercise 1A#2). In other words, any element of SL(l, R) can be reduced to the identity by a sequence of no more than l 2 elementary column operations. This is the method that was used by Carter and Keller to prove Thm. 1.3, but the calculations are complicated. To illustrate the idea in a very simple setting, let us show how a strong version of the following famous number-theoretic conjecture of E. Artin easily implies that the elementary matrices boundedly generate the group SL ( 2, Z[1/m] ), for any m>1.

8 8 1. BOUNDED GENERATION (1.21) Definition. We say that an integer m is a primitive root modulo a natural number n if the powers of m include a complete set of nonzero residues modulo n. (1.22) Example. 2 is a primitive root modulo 13, because { 2 k 0 k 11 } { 1, 2, 4, 8, 3, 6, 12, 11, 9, 5, 10, 7} (mod 13). This set includes all the numbers from 1 to 12; a complete set of nonzero residues modulo 13. Artin conjectured that there are infinitely many primes p, such that 2 is a primitive root modulo p. In fact, the same should be true, not only for the number 2, but for any integer, other than some obvious exceptions (see Exercise 1). (1.23) Conjecture (Artin). Let m be any nonzero integer, such that m ±1, and m is not a perfect square. Then there are infinitely many primes p, such that m is a primitive root modulo p. We will assume the stronger conjecture that the prime p can be chosen to lie in an arithmetic progression. This requires a stronger assumption on m (see Exercise 2). (1.24) Conjecture. Let m be any nonzero integer, such that m ±1, and m is not a perfect power, and a and b be nonzero integers, such that gcd(a, b) 1. Then there are infinitely many primes p, such that i) p a(mod b), and ii) m is a primitive root modulo p. (1.25) Proposition. If Conjecture 1.24 is true, and m>1, then every element of SL ( 2, Z[1/m] ) is a product of 5 elementary matrices. a b Proof. Given a matrix SL ( 2, Z[1/m] ), it suffices to show that the matrix can be reduced c d to the identity by a sequence of 5 elementary column operations. We may assume m is square free (see Exercise 3), so m is not a perfect power. Then, from Conj. 1.24, we have a prime p, such that (i) and (ii) hold. p b 1) From (i), applying a single column operation yields. d 2) From (ii), there is a positive [ integer ] k, such that m k b(mod p). Thus, applying another p m k column operation yields. 3) The element m k is obviously a unit in Z[1/m], so every element of Z[1/m] is a multiple of m k. Thus, [ a column ] operation can add any element of Z[1/m] to p, so another 1 m k operation yields ) One more column operation yields. Since the determinant is 1, the matrix we have 1 0 now must be ) One more column operation yields. 0 1 Exercises for 1C. #1. Suppose m is an integer that is either ±1 or a perfect square. Show there are only finitely many primes p, such that m is a primitive root modulo p. #2. Suppose a and b are relatively prime integers, k is a divisor of gcd(a 1, b) (with k>1), and

9 1D. CALCULATING IN PRODUCT RINGS 9 m is a k th power. Show there are only finitely many primes p a(mod b), such that m is a primitive root modulo p. #3. Let m be a positive integer. Show there is a square-free positive integer m, such that Z[1/m] Z[1/m ]. 1D. Calculating in product rings Here is an alternate approach to bounded generation that does not require one to keep track of the number of elementary matrices used in every calculation. It is based on the following observation that converts the question of bounded generation to simply the question of generation, without bounds (see Exercise 1): (1.26) Definition. Let Z be the direct product of infinitely many copies of the ring Z. That is, Z is the set of all sequences (a k ) k1, with each a k Z. It is an uncountable ring. (1.27) Lemma. The elementary matrices boundedly generate SL(3, Z) if and only if the elementary matrices generate SL(3, Z ). By introducing notation for the subgroup generated by elementary matrices, the lemma can be restated as follows: (1.28) Definition. For any ring A, let E(l, A) elementary matrices in SL(l, A) be the subgroup of SL(l, A) generated by the elementary matrices. (1.27 ) Lemma. The elementary matrices boundedly generate SL(3, Z) if and only if E(3, Z ) SL(3, Z ). Unfortunately, Z has zero divisors (see Exercise 2), so this lemma replaces the nice ring Z with a ring that is not an integral domain. To remedy this, we mod out a prime ideal in Z. (1.29) Definition. 1) Let p be a prime ideal in Z, such that (a) if (a k ) is an element of Z in which only finitely many components are nonzero, then (a k ) p and (b) if (a k ) is an element of Z in which only finitely many components are zero, then (a k ) p. Such an ideal exists by Zorn s Lemma (see Exercise 3). 2) Let Z Z /p. Then Z is an integral domain (because p is prime), and Z embeds (diagonally) as a subring (see Exercise 4). (Similarly, we may define A for any integral domain A.) (1.30) Remark. The ideal p cannot be constructed explicitly (but exists by the Axiom of Choice), so it is merely a technical device that avoids the need to deal with zero divisors. Thus, a novice reader may wish to simply ignore the distinction between Z and Z in a first reading. One may replace Z with Z in Lem (see Exercise 7). Furthermore, it is not necessary to prove that the elementary matrices generate all of SL(3, Z) it suffices to obtain a subgroup of finite index (see Exercise 8): (1.31) Lemma. The following are equivalent: 1) the elementary matrices boundedly generate SL(3, Z); 2) E(3, Z) SL(3, Z); 3) E(3, Z) has finite index in SL(3, Z); 4) the elementary matrices boundedly generate SL(3, Z). It is important to realize not only that Z is an integral domain, but that many other nice properties of Z will be inherited by Z (cf. 10 and 11).

10 10 1. BOUNDED GENERATION (1.32) Remark. Readers familiar with ultraproducts will realize that the ring Z can also be constructed as an ultraproduct of infinitely many copies of Z (see Exercise 18). In this approach, a theorem of logic states that any first-order" property that is true of Z will also be true of Z. Exercises for 1D. #1. Prove Lem [Hint: SL(3, Z ) SL(3, Z). If there exists g r SL(n, Z), such that g r is not a product of r elementary matrices, then (g r ) r 1 SL(3, Z) is not a product of elementary matrices.] #2. Find uncountably many zero divisors in Z. #3. Show Z has a prime ideal p as described in Defn [Hint: The set of all sequences as in (1a) is an ideal Z of Z. Let p be maximal in the class of all ideals that contain Z, but do not contain any sequence as described in (1b). The maximality implies that p is prime.] #4. Show that the map n (n, n, n,...) from Z to Z induces an (ring) embedding of Z in Z. #5. Let (a k ), (b k ) Z. Show that if (a k ) and (b k ) represent the same element of Z, then a k b k for infinitely many k. #6. Show that if a is any nonzero element of Z, then a can be represented by a sequence (a k ) Z, such that every a k is nonzero. #7. Show that the elementary matrices boundedly generate SL(3, Z) if and only if E(3, Z) SL(3, Z). #8. Prove (3 1) of Lem #9. Let A be an integral domain. Show that if the elementary matrices generate SL(3, A), then the boundedly generate SL(3, A). [Hint: You may assume (without proof) that Lem remains valid with A in the place of Z.] #10. Assume A is an integral domain. For each of the following statements, show that if the statement is true for R A, then it is also true for (i) R A and (ii) R A. a) Every 2-generated ideal of R is principal. b) Every finitely generated ideal of R is principal. c) Every element of R is a sum of 4 squares. d) Every element of SL(l, R) is a product of r elementary matrices. e) If a, b, c are elements of R that are relatively prime, then there exist a a(mod c) and b b(mod c), such that a and b are relatively prime. #11. For each of the following statements, show that the statement is (i) true for R Z, (ii) not true for R Z, and (iii) true for R Z. a) The only units of R are +1 and 1. b) If a, b, c are elements of R that are relatively prime, and a 0, then there exists b b(mod c), such that a and b are relatively prime. [Hint: (11b(i)) If a 0, then Z/aZ has only finitely many maximal ideals; choose t Z, such that a + tc does not lie in any of them.] #12. For each of the following statements, show that the statement is (i) true for R Z, but (ii) not true for R Z. a) Every ideal of R is principal. b) Every nonzero element of R can be written as a product of (finitely many) primes and units. c) R is a Euclidean domain. #13. Show that if a and b are elements of Z that are relatively prime, and b 0, then there is a prime element p a(mod b). [Hint: You may assume (without proof) that the statement is true when Z is replaced with Z.]

11 1E. MENNICKE SYMBOLS AND A PROOF OF BOUNDED GENERATION 11 #14. Assume A is an integral domain, such that E(2, A) is a normal subgroup of SL(2, A). Show there is a natural number r, such that if g SL(2, A), and ε is an elementary matrix in SL(2, A), then g 1 εg is a product of r elementary matrices. #15. Show that if F is a field, then F is a field. #16. Show that A is finite if and only if A is finite. #17. Let F be an infinite field. Show there is a natural number r, such that if g SL(2, F), and g ± Id, then every element of SL(2, F) is a product of r elements, each of which is conjugate to either g or g 1. [Hint: You may assume, without proof, the fact (true for every infinite field) that if N is any normal subgroup of SL(2, F), and N {± Id}, then N SL(2, F).] #18. (Requires familiarity with ultraproducts) Show that Z is isomorphic to an ultraproduct U Z, where U is a nonprincipal ultrafilter on N. 1E. Mennicke symbols and a proof of bounded generation (1.33) Notation. For convenience, let 1) C Z SL(3, Z)/ E(3, Z) and 0 2) SL(2, Z) 0 be the image of SL(2, Z) under the natural embedding in the top left corner of SL(3, Z). (We define C A similarly, for any commutative ring A. All rings are assumed to have a multiplicative identity 1.) In order to show that SL(3, Z) has bounded generation, it suffices to show that C Z is finite (see 1.31). There are 5 steps in our proof of this finiteness: 1. We have SL(3, Z) SL(2, Z) E(3, Z) (see Notn. 1.33(2)). 2. E(3, Z) is normal in SL(3, Z), so C Z is a group. 3. C Z is abelian. 4. C Z has exponent dividing 24 (i.e. x 24 e). 5. C Z is cyclic. From these facts, it is easy to show that C Z is finite (see Exercise 1); in fact, #C Z 24. (1.34) Definition. Recall that elements a 1,..., a m of a ring A are relatively prime if a 1 A + + a m A A (or, equivalently, if there exist t 1,..., t m A, such that a 1 t a m t m 1). Step 1. We have SL(3, Z) SL(2, Z) E(3, Z). We wish to show that each element of SL(3, Z) can be reduced, by a finite sequence of elementary column operations, to a matrix in SL(2, Z). To accomplish this, we note that the following property of the ring Z carries over to Z (see Exercise 1D#10e): if a, b, c are relatively prime, then there exist a a(mod c) and b b(mod c), (SR2) such that a and b are relatively prime. Given SL(3, Z), we know that a3,b 3,c 3 are relatively prime, so (SR2) implies a 3 b 3 c 3 there exist a 3 a 3 (mod c 3 ) and b 3 b 3 (mod c 3 ), such that a 3 and b 3 are relatively prime. Thus, by applying two column operations (adding appropriate multiples of the last column to the first

12 12 1. BOUNDED GENERATION two columns), we may assume that a 3 and b 3 are relatively prime. From here, it is not difficult to reduce the matrix to the desired form (see Exercise 2). (1.35) Remark. In the literature, the axiom (SR2) is known as a Stable Range Condition. The subscript 2 refers to the fact that the natural map SL(l, A)/ E(l, A) SL(l + 1, A)/ E(l + 1, A) is surjective for l 2 (cf. Exercise 4d). Step 2. E(3, Z) is normal in SL(3, Z). It suffices to show that g 1 E i,j (t) g E(3, Z), for every g SL(3, Z) and every elementary matrix E i,j (t). For simplicity, let us assume (i, j) (1, 3) (see Exercise 5). From Step 1, there is some h E(3, Z), such that gh SL(2, Z). Since h (obviously) normalizes E(3, Z), we may assume g SL(2, Z). Then, letting H , we have E i,j (t) H E(3, Z), and g normalizes H (see Exercise 6). So g 1 E i,j (t) g g 1 Hg H E(3, Z). (1.36) Remark. By appealing to Step 1, we have implicitly relied on axiom (SR2) in the proof of Step 2. It can actually be shown that E(3, A) is a normal subgroup of SL(3, A) for any commutative ring A, without assuming (SR2), but this is much more difficult. For the remaining steps of the proof, it is very helpful to have a shorthand notation for elements of C Z : (1.37) Notation. Let W W Z { (a, b) Z 2 a and b are relatively prime }. It is important to note that W is precisely the set of first rows of elements of SL(2, Z). That is, [ for ] (a, b) Z 2, we have (a, b) W if and only if there exists (c, d) Z 2, such that a b SL(2, Z) (see Exercise 7). c d For (a, b) W, let ba be the element of C Z with a b 0 b c d 0 E(3, Z) CZ, a a b where c and d are elements of Z chosen so that SL(2, Z). Because E(3, Z) is c d normal in SL(3, Z), it is easy to see that is a well-defined function from W to C Z (see Exercise 9). We need to establish some basic properties of the function. Properties (MS1) and (MS2a) below are what it means to be a Mennicke symbol. The property (MS1) is rather obvious (since we mod out by elementary column operations in the definition of C Z ), but the multiplicativity expressed in (MS2a) (and MS2b) is quite surprising. See Exer. 13 for additional interesting properties that follow from these axioms. (1.38) Proposition. (onto) is onto. b + ta ba b (MS1) a a + tb, for all (a, b) W and all t Z. b1 b (MS2a) 2 b1 b2 a a a.

13 (MS2b) ba1 ba2 1E. MENNICKE SYMBOLS AND A PROOF OF BOUNDED GENERATION 13 [ b a 1 a 2 ]. Proof. (onto) Given α C Z, we have α γ E(3, Z), for some γ SL(3, Z). By Step 1, we may a b a b assume γ, for some SL(2, Z). Then α ba. Since α is an arbitrary element c d c d of C Z, this implies that is onto. (MS1) It is easy to find ε 1,ε 2 E(3, Z), such that a b 0 a b+ ta 0 a b 0 a + tb b 0 c d 0 ɛ1 0 and c d 0 ɛ (see Exercise 11). (MS2a) The proof is clever, but not long or difficult, and uses, in an essential way, the fact that we are working in SL(3, Z), rather than SL(2, Z). See Exer. 12. (MS2b) This follows from (MS1) and (MS2a); see Exer. 13f. The remaining steps of the proof are based on the Mennicke axioms presented in Prop ba The basic idea is that these axioms are so strong that they almost imply that is trivial for all a and b. More precisely, they imply the conclusions expressed in Steps 3, 4, and 5. We will use the following strengthening of?? SR2 (see Exercise 1D#11b): if a, b, c are relatively prime, and a 0, then there exists b b(mod c), (SRhalf) such that a and b are relatively prime. Step 3. C Z is abelian. Let b1 a and b2 1 a be any two nontrivial elements of C Z. Then b (see Exercise 15), so, from (SRhalf), we know there exists b 1 b 1 mod a 1, such that [ b 1 is relatively prime to b 2. From (MS1), we know that replacing b 1 with b 1 does not affect b1 a ], so we may 1 assume b 1 is relatively prime to b 2. Then, by the Chinese Remainder Theorem (16), there exists a Z, such that a a i mod b i for i 1, 2. Therefore, applying, in order, (MS1) twice, (MS2a), commutativity of Z, (MS2a), and (MS1) twice, we have b1 b2 b1 b2 b1 b 2 b2 b 1 b2 b1 b2 b1. a 1 a 2 a a a a a a a 2 a 1 Since b1 a and b2 1 a are arbitrary nontrivial elements of C Z (see 1.38(onto)), we conclude that 2 C Z is abelian. Step 4. C Z has exponent dividing 24. It may be advisable to read Step 5 first, because that part of the argument is more straightforward. The calculations here are not difficult to verify, but may seem rather arbitrary. They are adapted from a proof for C Z (see Exercises 17 and 19), but that argument does not apply verbatim, because raising each component of a sequence to a power cannot be realized by raising the sequence to a power (if the sequence of exponents is unbounded). This is remedied by noting that every power of a 2 2 matrix T can be expressed in the form ft + g Id. The following axiom is satisfied by the ring A Z (see Exercise 21), and carries over to A Z (see Exercise 22). (1.39) Assumption. If a and b are relatively prime, then there exist c and f i,g i,b i,d i for i 1, 2, with b 1 b, such that a bi a) f i Id +g i SL(2, A), c d i b) a 12 (f 1 + g 1 a)(f 2 + g 2 a) (mod c), c) f 2 i 1 (mod g i ), and d) f i + g i a 1 (mod b i ).

14 14 1. BOUNDED GENERATION have Given ba C Z, we choose c, f i,g i,b i,d i as in (1.39). Then, employing (MS2a) and (MS2b), we b c (Exer. 10) a a 2 2 c (1.39(b)) f i1 i + g i a 2 2 cgi (1.39(c)) f i1 i + g i a 2 2 bi g i (Exer. 10 and 1.39(a)) f i1 i + g i a 2 2 b i (1.39(c)) f i + g i a i1 2 [ bi i1 1 ] 2 (1.39(d)) e. Since ba is an arbitrary element of C Z (see 1.38(onto)), we conclude that the exponent of C Z divides 24. b1 b2 Step 5. C Z is cyclic. Given a, 1 a C Z, it suffices to show that these two elements are in the 2 same cyclic subgroup of C Z (see Exercise 23). To do this, we repeatedly apply Dirichlet s Theorem on primes in arithmetic progressions (1.40). Abusing notation, let us think of a i and b i as elements of Z, instead of Z. Because a i and b i are relatively prime, (we may assume that) each coordinate of a i is relatively prime to the corresponding coordinate of b i (see Exercise 25). Then, by Dirichlet s Theorem on primes in arithmetic progressions, there exists p Z, such [ that ] p[ a] 1 (mod b 1 ), and every coordinate of p is prime (see Exercise 26). By (MS1), we have b1 a b1 1 p. Thus, replacing a 1 with p does not change anything, so we may assume a 1 p, which means that every coordinate of a 1 is prime. In fact, by repeating this argument, we may assume that every coordinate of each of a 1, a 2, b 1, and b 2 is prime. Also, because Dirichlet s Theorem provides infinitely many primes in each arithmetic progression, we may assume no coordinate of b 1 is equal to the corresponding coordinate of b 2, so b 1 and b 2 are relatively prime (see Exercise 25). Thus, the Chinese Remainder Theorem (16) implies there is some q Z, such that q a i (mod b i ) for i 1, 2. Furthermore, by Dirichlet s Theorem (1.40), we may assume every coordinate of q is prime. By (MS1), we have [ bi a i ] a 1 q a 2. [ bi q ], so we may assume In Z, the group of units modulo any prime is cyclic. Since every coordinate of q is prime, let us pretend that the group (Z/qZ) of units modulo q is cyclic. (This is not quite true, but see Exer. 28 for a way to deal with this issue.) Thus, letting b be a generator of this group, there exist integers e 1 and e 2, such that Then we have [ bi a i ] b i b e i (mod q) for i 1, 2. [ bi ] b e i q q ei b q b. q

15 1E. MENNICKE SYMBOLS AND A PROOF OF BOUNDED GENERATION 15 So b1 a and b2 1 a both belong to the cyclic subgroup generated by bq. 2 This completes the proof of Thm Exercises for 1E. #1. Show that any cyclic group of finite exponent is finite. #2. Complete the proof of Step 1. #3. Use the Stable Range Condition SR 1 1 to complete the proof of Prop. 1.38(onto). 2 [Hint: When a 3 and b 3 are relatively prime, applying two column operations yields (a 3,b 3, 1) as the last row.] #4. We say that a commutative ring A satisfies the Stable Range Condition SR k if, for all relatively prime a 1,a 2,..., a k+1 A, there exist a i a i (mod a k+1 ) for 1 i k, such that a 1,a 2,..., a k are relatively prime. a) Show Z does not satisfy SR 1. b) Show that if A satisfies SR k, and I is any ideal of A, then A/I also satisfies SR k. c) Show SR k implies SR k+1 for k 1. d) Show that if A satisfies SR k, then, for all l k, then we have SL(l, A) SL(k, A) E(l, A), where SL(k, A) denotes the image of SL(k, A) under the natural embedding in the top left corner of SL(l, A). [Hint: (4b) If a 1 + I, a 2 + I, a 3 + I are relatively prime in A/I, then there exist a i a i (mod I), such that a 1,a 2,a 3 are relatively prime.] #5. Eliminate the assumption that (i, j) (1, 3) from Step 2. [Hint: Any elementary matrix can be conjugated to E1, 2( ) by a permutation matrix.] #6. In the notation of the proof of Step 2, show that SL(2, Z) normalizes H. #7. For elements a and b of a commutative [ ring ] A, show a and b are relatively prime if and a b only if there exist c, d A, such that SL(2, A). c d #8. Suppose A is an integral domain, and γ 1 and γ 2 are two elements of SL(2, A) that have the same first row. Show there exists ε E(2, A), such that eγ 1 γ 2. [Hint: If ad bc 1 and ad bc 1, then a(d d ) b(c c ).] ab a b #9. Show that does not depend on the choice of c, d Z, such that SL(2, Z). c d [Hint: Use Exer. 8 and the fact that E(3, Z) is normal.] a b 1 #10. Show that if SL(2, Z), then ba ca. c d #11. Prove (MS1). #12. Prove (MS2a) a b 2 0 [Hint: Conjugating by the matrix E(3, Z) shows that the matrices c 2 d 2 0 and d 2 0 c b1 represent the same element of C Z. Multiplying the latter by a (on the left) and b 2 0 a applying row operations yields the desired conclusion.] #13. Prove the following Mennicke symbol identities, by using only (MS1), (MS2a) and SR 2. (Furthermore, the scalars a and b belong to a commutative ring A, and writing ba implies that a and b are relatively prime.)

16 16 1. BOUNDED GENERATION a) 01 e (the identity element of C Z ). ba b(1 a) b) a. ba c) 1 if there is a unit u in A, such that either a u(mod b) or b u(mod a). [ d) ba b a ]. e) (Kervaire Reciprocity) ba ab. [ f) (MS2b) ba1 ba2 b a 1 a ]. 2 [Hint: (13c) If a u (mod b), then ba bu 1u 1. (13e) For d b a, we have ba da db a ab b. (13f) Immediate from (MS1) and (13e).] #14. Show: a) SR 1 implies SR b) SR 1 1 implies SR 2. 2 #15. Show that if a and b are relatively prime, and either a 0 or b 0, ba then e. [Hint: If a 0, then b is a unit.] #16. (Chinese Remainder Theorem) Suppose a 1,b 1,a 2,b 2 are elements of a ring A, and b 1 is relatively prime to b 2. Show there exists a A, such that a a 1 (mod b 1 ) and a a 2 (mod b 2 ). #17. For the case A Z, provide a short proof that every Mennicke symbol ba has finite order as an element of C Z. [Hint: For e φ(b), use (MS2b) to show [ ba ] e 1.] #18. Suppose a, b Z, such that gcd(a, b) 1. Show there exists b b(mod a), such that gcd ( φ(b), φ(b ) ) 12. [Hint: If p k is a prime power, and p k 5, then φ(p k ) p k 1 (p 1) >2, so we we may write b x p y p (mod p k ), where x p,y p 1 (mod p). Let b xy, where x and y are primes, such that x x p (mod p k ) and y y p (mod p k ), for every maximal prime divisor p k of φ(b).] 12 #19. For the case A Z, use Exer. 18 to provide a short proof that ba 1 for every (a, b) W. 12 [Hint: There exist t, t Z, such that et + e t 12, where e φ(b) and e φ(b ). Then ba et ba b e t a 1.] #20. Suppose a, b Z, such that gcd(a, b) 1. Show there exists b b(mod a), such that gcd ( φ(b), φ(b ) ) 12. [Hint: Exer. 18.] #21. Given relatively prime a, b Z, let b 1 b. b 2 b(mod a), such that gcd ( φ(b 1 ), φ(b 2 ) ) 12. α i exponent of a modulo b i (so gcd(α 1,α 2 ) 12).

17 NOTES 17 t 1,t 2 Z with α 1 t 1 + α 2 t f i,g i Z with T α it i i f i Id +g i T i, where T i Show that a bi a) f i Id +g i T α it i c i SL(2, Z). b) a 12 a α 1t 1 a α 2t 2 (f 1 + g 1 a)(f 2 + g 2 a) (mod c). c) 1 det(t α it i i ) det(f i Id) f 2 i (mod g i ). d) f i + g i a a α it i 1 t i 1 (mod b i ). #22. Show that (1.39) is satisfied for A Z. a bi SL(2, Z). c d i #23. Suppose H is an abelian group, the exponent of H is finite, and every pair of elements of H is contained in a cyclic subgroup of H. Show that H is cyclic. #24. Let a, b, c Z. Show that a b(mod c) if and only if there exist representatives â, ˆb and ĉ of a, b and c in Z, such that â i ˆb i (mod c i ) for every i. #25. Let a, b Z. Show that a and b are relatively prime if and only if there exist representatives â and ˆb of a and b in Z, such that â i is relatively prime to ˆb i for every i. #26. Dirichlet s Theorem on primes in arithmetic progressions states that if a and b are relatively prime elements of Z (with b 0), then there are infinitely many primes p, such that p a(mod b). In other words, if a and b are relatively prime, with b 0, then there exist p 1,p 2,p 3,..., such that (1.40) p i a(mod b), p i A is a maximal ideal, and p i is relatively prime to p j whenever i j. Show that this remains true with Z in the place of Z. #27. Show there is a (prime) element q of Z, such that qz is a maximal ideal, but the group (Z/qZ) of units modulo q is not cyclic. [Hint: Let q i be a sequence of primes tending to infinity.] #28. The argument in the last paragraph of the proof of Step 5 (on page 14) pretends that (Z/qZ) is cyclic. That is not true (see Exercise 27). Correct this error. [Hint: The axiom if q is a prime in A, then (A/qA) is cyclic modulo 24th powers (1.41) is true for A Z, so it is also true for A Z.] Notes 1A. The Carter-Keller Theorem (1.3 ) was first proved in [CK1] (in a more general form that allows Z to be replaced with the ring of integers of any number field.) Elementary proofs appear in [CK1] and [AM]. Theorem 1.11 is due to D. Carter, G. Keller, and E. Paige [CKP, Mo]. It had previously been proved by B. Liehl [Li] under some restrictions on O. Even earlier, G. Cooke and P. Weinberger [CW] had shown that the general case (with an excellent bound on the number of elementary matrices required) follows from a certain generalization of the Riemann Hypothesis. Theorem 1.14 is due to O. I. Tavgen [Ta1, Ta2] (in a more general form that also applies to many quasi-split groups). Additional arithmetic groups have been proved to have bounded generation by I. Erovenko and A. Rapinchuk [ER1, ER2].

18 18 1. BOUNDED GENERATION 1B. Theorem 1.15 is due to A. Rapinchuk [R2]. (A sketch of the proof appears in [PR2, App. A.2].) Theorem 1.16 was proved by A. Lubotzky [L] and by V. Platonov and A. Rapinchuk [PR1]. (A sketch of Lubotzky s proof appears in [LS, 12.7].) Rapinchuk s Conjecture (1.17) first appeared in [R1]. Theorem 1.19 is due to L. Lifschitz and D. Morris [LM]. Theorem 1.20 is due to Y. Shalom [Sh]. 1C. not written yet 1D. not written yet 1E. This proof is due to D. Carter, G. Keller, and E. Paige [CKP, Mo]. It generalizes to prove the bounded generation of SL(2, O) when O has infinitely many units, but much additional work is required, because the Mennicke axioms (MS1), (MS2a), and (MS2b) of Prop are not obvious in this setting. The Mennicke axiom (MS2a) is implicit in the proof of J. Mennicke [Me, Lem. 3.2] that SL(3, Z) has the Congruence Subgroup Property. Mennicke symbols were then studied by H. Bass [Ba1, Ba2] and others, and played a fundamental role in the development of Algebraic K-Theory. References [AM] S. I. Adian and J. Mennicke: On bounded generation of SL n (Z), Internat. J. Algebra Comput. 2 (1992), no. 4, [Ba1] H. Bass, K-theory and stable algebra, Inst. Hautes Études Sci. Publ. Math. 22 (1964), _0 [Ba2] H. Bass, Algebraic K-theory, Benjamin, New York, [CK1] D. Carter and G. Keller: Bounded elementary generation of SL n (O), Amer. J. Math. 105 (1983), [CK2] D. Carter and G. Keller: Elementary expressions for unimodular matrices, Comm. Algebra 12 (1984), [CKP] D. Carter, G. Keller, and E. Paige: Bounded expressions in SL(n, A) (unpublished). [CW] G. Cooke and P. Weinberger: On the construction of division chains in algebraic number rings, with applications to SL 2, Comm. Algebra 3(6) (1975), [ER1] V. Erovenko and A. Rapinchuk: Bounded generation of some S-arithmetic orthogonal groups, C. R. Acad. Sci. Paris Sér. I Math. 333 (2001), no. 5, [ER2] V. Erovenko and A. S. Rapinchuk: Bounded generation of S-arithmetic subgroups of isotropic orthogonal groups over number fields, preprint (2005). [GS] F. Grunewald and J. Schwermer: Free non-abelian quotients of SL 2 over orders of imaginary quadratic numberfields, J. Algebra 69 (1981), [Li] B. Liehl: Beschränkte Wortlänge in SL 2, Math. Z. 186 (1984), [LM] L. Lifschitz and D. W. Morris: Isotropic nonarchimedean S-arithmetic groups are not left orderable, C. R. Math. Acad. Sci. Paris 339 (2004), no. 6, MR (2005f:11061) [L] A. Lubotzky: Subgroup growth and congruence subgroups, Invent. Math. 119 (1995), no. 2, MR (95m:20054) [LS] A. Lubotzky and D. Segal: Subgroup growth, Progress in Mathematics, 212. Birkhäuser Verlag, Basel, ISBN MR (2004k:20055) [Me] J. L. Mennicke: Finite factor groups of the unimodular group, Ann. Math. 81 (1965) [Mo] D. W. Morris: Bounded generation of SL(n, A) (after D. Carter, G. Keller, and E. Paige), (preprint). [PR1] V. Platonov and A. Rapinchuk: Abstract characterizations of arithmetic groups with the congruence property (Russian), Dokl. Akad. Nauk SSSR 319 (1991), no. 6, ; translation in Soviet Math. Dokl. 44 (1992), no. 1, MR (93e:11056)