BOUNDS ON ORDERS OF FINITE SUBGROUPS OF P GL n (K) Eli Aljadeff and Jack Sonn

BOUNDS ON ORDERS OF FINITE SUBGROUPS OF P GL n (K) Eli Aljadeff and Jack Sonn Abstract. We characterize the pairs (K, n), K a field, n a positive integer, for which there is a bound on the orders of finite subgroups of P GL n (K). Explicit bounds are given in important cases. An application is made to the analogous problem for central simple K-algebras of degree n. 1. Introduction Let K be a field, n an integer > 1, GL n (K) the group of invertible n n matrices over K, P GL n (K) = GL n (K)/K. Is there a bound on the orders of the finite subgroups of GL n (K)? of P GL n (K)? Obviously the answer depends on K, since the answer is no if K contains infinitely many roots of unity. The answer is also no if K is an infinite field of characteristic p 0, as one sees by looking at the unitriangular matrices. Thus we must restrict the field K or consider only nonmodular finite groups, i.e. finite groups whose order is not divisible by char(k). The answer also depends on n as we shall see below. Let B(K, n) (B (K, n) resp.) denote the statement that there is a bound on the orders of the finite nonmodular subgroups of GL n (K) (P GL n (K) resp.). A theorem of Jordan [4, Thm. 15.7] states that for any K and n, if G is a finite nonmodular subgroup of GL n (K), then G contains an abelian normal subgroup of index J(n) := (49n) n2. By Jordan s theorem, B(K, n) holds if and only if there is a bound on the orders of the finite nonmodular abelian subgroups of GL n (K). W. Feit has communicated to us an unpublished result of B. Weisfeiler, which replaces Jordan s bound J(n) by the bound W (n) = (n + 2)! for n > 63, W (n) = n 4 (n + 2)! for n 63. Based on Weisfeiler s work, Feit [5] has determined finite subgroups of GL(n, K) of maximal order, for all n, where K is any cyclotomic field. Recently, S. Friedland [6] has determined precise bounds on the finite abelian subgroups of GL(n, Q), and using Weisfeiler s bound, obtains Feit s result for K = Q and all large n. In Section 2, we consider B(K, n) and give a characterization of B(K, n) in terms of cyclotomic extensions of K. If µ denotes the group of all roots of unity, B(K, n) B(K k(µ), n) where k denotes the prime field of K. For every n, there is a field K for which B(K, n) holds but B(K, n + 1) does not hold. For fields K such that [K k(µ) : k] <, we give an explicit bound in terms of n and [K k(µ) : k]. Research supported by the Fund for the Promotion of Research at the Technion and by the Technion VPR Fund 1 Typeset by AMS-TEX

2 ELI ALJADEFF AND JACK SONN In Section 3, we consider P GL n (K) and B (K, n). It is easy to show that B (K, n) holds for all n if and only if B(K, n) holds for all n. The main result is that for every n, B(K, n) is equivalent to B (K, n). As in the case of GL n (K), one can give explicit bounds when [K k(µ) : k] <. In the last short section, the results above are applied to multiplicative subgroups of finite dimensional central simple K-algebras. 2. GL n As defined above, B(K, n) is the assertion that there is a bound on the orders of finite nonmodular subgroups of GL n (K). Clearly B(K, n) B(K, n 1). 2.1. Proposition. Let n be an integer > 1, K a field. B(K, n) holds if and only if [K(µ m ) : K] > n for all sufficiently large m, i.e. there exists m 0 such that m m 0 [K(µ m ) : K] > n. (If char(k) is p 0, we exclude from the requirement those m divisible by p.) In this case, if A is a finite abelian nonmodular subgroup of GL n (K), then A m 0 n, hence if G is a finite nonmodular subgroup of GL n (K), then G J(n)m 0 n, where J(n) is Jordan s bound (on the index of a maximal abelian normal subgroup) (49n) n2. Proof. Suppose there are arbitrarily large m (p m as needed) such that [K(µ m ) : K] n. The field K(µ m ) can be represented in the algebra M n (K) of n n matrices over K, hence µ m is embedded in GL n (K) for arbitrarily large m. Now conversely assume there exists m 0 such that m m 0 [K(µ m ) : K] > n. Let G be a finite nonmodular subgroup of GL n (K). By Jordan s theorem, G has an abelian normal subgroup A of index J(n), which reduces the proof to the case G = A abelian. Consider the commutative subalgebra K[A] of M n (K) generated by A over K. Since A is nonmodular, K[A] is semisimple, hence a direct sum of fields r i=1 L i. If K is an extension of K for which A is diagonalizable, we have dim K K[A] = dim K K[A] K K n. Hence if d i = [L i : K], we have r 1 d i n. The projection K[A] L i maps A onto a finite group A i of roots of unity in L i, L i = K[A i ] is a cyclotomic extension, and A is embedded in r 1 A i. Let m i = A i, so L i = K(µ mi ). By hypothesis there exists m 0 such that m m 0 [K(µ m ) : K] > n. Since d i = [L i : K] n, we have m i < m 0. It follows that A m i < m n 0. 2.2. Proposition. Let k be the prime field of K. Then B(K, n) B(K k(µ), n). In fact, [K(µ m ) : K] > n for all m m 0 [(K k(µ))(µ m ) : K k(µ)] > n for all m m 0. Proof. Immediate since [(K k(µ))(µ m ) : K k(µ)] = [K(µ m ) : K]. By Proposition 2.2, we may assume K k(µ).

BOUNDS ON ORDERS OF FINITE SUBGROUPS OF P GL n (K) 3 Suppose k is F p. Then if [K : k] =, K contains infinitely many roots of unity, so B(K, n) does not hold. Hence B(K, n) K finite, which is a trivial case. We therefore assume for the rest of this section that char(k)=0, k = Q. B(K, n) can be localized: 2.3. Proposition. B(K, n) is equivalent to the following assertion: (i) For almost all primes p, [Q(µ p ) : K Q(µ p )] > n, and (ii) for all primes p, if [Q(µ p ) : K Q(µ p )] n, then [K Q(µ p ) : Q] <. Observe that [K(µ m ) : K] = [Q(µ m ) : K Q(µ m )] for every m. Proof. Assume B(K, n). Then (i) certainly holds. Assume for some p, [Q(µ p ) : K Q(µ p )] n, and [K Q(µ p ) : Q] is infinite. Set K = K Q(µ p ). Then K contains the (unique) Z p -extension Q p, since if not, K Q p is a proper subfield of Q p, hence finite over Q, which implies that [K : K Q p ] is infinite. But [K : K Q p ] = [K Q p : Q p ] [Q(µ p ) : Q p ] = p 1 <, contradiction. Thus K Q p, so [K(µ p ) : K] = [Q(µ p ) : K ] = [Q(µ p ) : K Q(µ p )] = [Q(µ p ) : K Q(µ p )] n, contradicting B(K, n). Conversely, assume B(K, n) does not hold. Then there are infinitely many m such that [K(µ m ) : K] n. We show that (i), (ii) do not both hold. Assume (i) holds. Then for some m 0, [K(µ p ) : K] > n for all p > m 0. It follows then that if m is divisible by a prime p > m 0, [K(µ m ) : K] [K(µ p ) : K] > n, hence the infinite set of m for which [K(µ m ) : K] n consists of m divisible only by (the finitely many) primes m 0. By the pigeonhole principle, there is a prime p and an infinite sequence of m s such that [K(µ m ) : K] n and the powers of p that divide these m are unbounded. Hence n [K(µ p ) : K] = [Q(µ p ) : K Q(µ p )]. In particular, [K Q(µ p ) : Q] is infinite, and [K(µ p ) : K] n, violating (ii). 2.4. Corollary. B(K, n) holds for all n for every prime p, [K Q(µ p ) : Q] < and lim p [K(µ p ) : K] =. 2.5. Example. Let K be the composite of all quadratic extensions of Q. K is an infinite extension of Q for which B(K, n) holds for all n. Of course if [K : Q] is finite, then B(K, n) holds for all n. 2.6. Example. We will show that for every n > 1, there are fields K for which B(K, n 1) holds, but B(K, n) does not hold. By Dirichlet s theorem there is a prime p 1 (mod n). Let Q p K Q(µ p ) with [Q(µ p ) : K] = n. Then [K(µ p ) : K] = n, violating B(K, n), but it is easily checked that B(K, n 1) holds, using Proposition 2.3. A horizontal family of examples can be constructed as well, where K is a composite of fields K p, with p running through an infinite set of primes 1 (mod n), and Q K p Q(µ p ), [Q(µ p ) : K p ] = n. When [K : Q] is finite, and in particular B(K, n) holds for all n, it is of interest to compute a more explicit bound than m n 0 in Proposition 2.1. In this case, let A be a finite abelian subgroup of GL n (K), K[A] = r i=1 L i, [L i : K] = d i, d i n, L i = K(µ mi ), m i = A i as in the proof of Proposition 2.1. Now d i = [K(µ mi ) : K] = [Q(µ mi ) : K Q(µ mi )] = [Q(µ mi ) : Q]/[K Q(µ mi ) : Q] φ(m i )/[K : Q] = φ(m i )/c K

4 ELI ALJADEFF AND JACK SONN where we set c K := [K : Q]. Now m i φ(m i ) 2 c 2 K d2 i. Then A m i i c2 K d2 i = c2r K ( d i ) 2 c 2r K ( 1 r di ) 2r c 2r K (n/r)2r (using the geometric-arithmetic inequality). 2.7. Lemma. max{(n/r) r : 1 r n} e n/e. Proof. To maximize (n/r) r, replace r by a continuous variable x, and maximize log(n/x) x = x log n x log x. The derivative vanishes at log x = log n 1, or x = n/e, giving (n/x) x = ( n n/e )n/e = e n/e. It follows that A c 2n K e2n/e = (c K e 1/e ) 2n. This proves 2.8. Proposition. Let [K : Q] be finite (K Q(µ)). Then for any n, if A is a finite abelian subgroup of GL n (K), then A ([K : Q]e 1/e ) 2n. Hence if G is a finite subgroup of GL n (K), G J(n)([K : Q]e 1/e ) 2n. Remark. When B(K, n) holds in the characteristic p case, it is also of interest to write down an explicit value for m 0. Here we may assume K finite, and m 0 can be taken as the smallest positive integer such that [K(µ m ) : K] > n for all m m 0. If K = q K, then the extension of degree n of K is the field of qk n elements, containing the qk n 1th roots of unity. so if m 0 = qk n, then m m 0, p m K(µ m ) has more than qk n 1 roots of unity, hence K(µ m) > qk n, so [K(µ m) : K] > n. We therefore have 2.9. Proposition. Suppose char(k)=p and K k(µ) is finite. If A is a finite abelian nonmodular subgroup of GL n (K), then A qk n2, q K = K k(µ). Hence if G is a finite nonmodular subgroup of GL n (K), G J(n)qK n2 = (49nq K) n2. Remark. Let K be an infinite field of finite characteristic p. There is no bound on finite subgroups or even indices of maximal abelian normal subgroups of finite subgroups of GL n (K) for n 3 as one sees by considering unitriangular matrices. A theorem of Brauer and Feit [3] states that there is a bound on the index of maximal abelian normal subgroups in finite subgroups G of GL n (K) depending on n and the order of a p-sylow subgroup G p of G. On the other hand, exp G p < pn. (Proof: Let a be an element of p-power order p m in GL n (K). Since a pm = 1, a pm 1 = (a 1) pm = 0 so a 1 is nilpotent, hence (a 1) n = 0 which implies that p m the first power of p n, which is < pn.) By Zelmanov s theorem (Restricted Burnside Problem) [9] there is a bound on the order of a p-sylow subgroup G p of G depending on rank(g p ) (and n), hence in the Brauer-Feit theorem, order can be replaced by rank. 3. PGL Recall that B (K, n) denotes the statement that there is a bound on finite nonmodular subgroups of P GL n (K).

BOUNDS ON ORDERS OF FINITE SUBGROUPS OF P GL n (K) 5 3.1. Observation. ) The embedding GL n (K) P GL n+1 (K), a mod K, a GL n (K) shows that B (K, n + 1) B(K, n). ( a 0 0 1 3.2. Observation. The embedding P GL n (K) Aut M n (K) GL n 2(K), a mod K conjugation by a on M n (K) shows that B(K, n 2 ) B (K, n). Thus: 3.3. Proposition. There is a bound on finite nonmodular subgroups of GL n (K) for all n if and only if there is a bound on finite nonmodular subgroups of P GL n (K) for all n, i.e. B(K, n) holds for all n if and only if B (K, n) holds for all n. Let k be the prime field of K. If char(k) =p 0, i.e. k = F p, then since by Proposition 2.2, B(K, n) B(K k(µ), n) K k(µ) is a finite field, B(K, n) is independent of n, so B (K, n) holds K k(µ) is finite. We therefore assume from now on that char(k)=0. We now prove 3.4. Theorem. B(K, n) is equivalent to B (K, n) for all n 2. Proof. One direction is easy. Suppose B(K, n) does not hold. Case 1. K contains infinitely many roots of unity. Then the diagonal subgroup of P GL n (K) has unbounded torsion, violating B (K, n). Case 2. K contains finitely many roots of unity. Then for infinitely many m, [K(µ m ) : K] n, so µ m embeds into GL n (K). It follows that µ m mod scalars is unbounded. For the opposite direction, we assume B(K, n) and consider a finite subgroup G of P GL n (K). 3.5. Claim. exp G (e 1/e m 0 /m K ) n where m K = µ K =the number of roots of unity in K. Proof. Let u GL n (K) have order h mod K, so u h = a K. The commutative subalgebra K[u] of M n (K) is semisimple and of dimension n over K by Cayley- Hamilton. Hence K[u] = n 1 L i, with L i a finite extension field of K of degree d i, i = 1,..., r. If u i is the projection of u onto L i, then L = K[u i ], and u h i = a. Thus the order h i of u i mod K divides h, and h = lcm{h i } r 1. Let us temporarily fix i and drop the subscript, so L = K[u] is a field extension of degree d over K. We wish to bound h=order of u mod K in terms of d. If d = 1 then h = 1 and there is nothing to do, so we assume d > 1. Let µ K, µ L denote the group of roots of unity in K, L, respectively, m K = µ K, m L = µ L. Since K(µ L ) L, [K(µ L ) : K] [L : K] d n. By the hypothesis B(K, n), there exists m 0 such that m m 0 [K(µ m ) : K] > n, hence m L < m 0. By a result of Risman [8], h dm L /m K < dm 0 /m K. Restoring subscripts, h i < d i m 0 /m K, hence h r i=1 d i(m 0 /m K ) r ( 1 r di ) r (m 0 /m K ) r (n/r) r (m 0 /m K ) r (e 1/e m 0 /m K ) r (e 1/e m 0 /m K ) n.

6 ELI ALJADEFF AND JACK SONN 3.6. Claim. exp H 2 (G, K ) divides lcm(exp G, m K ). Proof. By the universal coefficient theorem [7, Thm. sequence 15.1], we have a split exact 0 Ext 1 Z(G ab, K ) H 2 (G, K ) Hom(M(G), K ) 0 where M(G) is the Schur multiplier of G and G ab = G/G, G the commutator subgroup of G. If G ab = C e1 C et with C ei cyclic of order e i, then Ext 1 Z(G ab, K ) = Ext 1 Z(C e1, K ) Ext 1 Z(C et, K ) which has exponent dividing exp G. Moreover exp Hom(M(G), K ) divides m K. We proceed with the proof of Theorem 3.4. Lift G back to a subgroup Ĝ of GL n(k) to obtain an exact sequence 1 K Ĝ G 1 and let α H 2 (G, K ) be the corresponding cocycle class. Let m = exp H 2 (G, K ) and let f α be a representative cocycle. then f m = g B 2 (G, K ), i.e. for every σ, τ G, f(σ, τ) m = g(σ)g(τ)g(στ) 1 with g : G K a mapping. Let F = K(µ m, {g(σ) 1/m : σ G}). Extending coefficients to F, we may view Ĝ as a subgroup of GL n (F ). The cocycle class of f in H 2 (G, F ) is equivalent to f 1 := f( g) 1/m, and f1 m = 1, i.e. f 1 takes values in µ m. We then obtain a finite subgroup G of GL n (F ) in an exact sequence 1 µ m G G 1. By Jordan s theorem, G has an abelian subgroup A of index J(n) = (49n) n 2 (or W (n) if we use Weisfeiler s result). (At this point the temptation is to use B(F, n) to bound A ; however B(K, n) does not necessarily imply B(F, n).) Consider the semisimple commutative subalgebra F [A] of M n (F ). F [A] = r 1L i, where L i is a field extension of F of degree d i, and dim F F [A] n as in the proof of Proposition 2.1. The projection of A on L i is a finite subgroup A i of L i hence contained in µ Li, and L i = F [A i ] is a cyclotomic extension. We have A i = exp A i exp A exp G exp H 2 (G, K ) exp G m K (exp G) 2. It follows that A r 1 A i m n K (e1/e m 0 /m K ) 2n2. It now follows that G G J(n) A (49n) n2 m n K (e1/e m 0 /m K ) 2n2, hence B (K, n) holds. As in the case of GL n (K), we can give a more explicit bound on G if [K k(µ) : k] is finite, with k the prime field. If k = F p, then we merely take m 0 = qk n as in section 1, so G (49n) n2 (e 1/e qk) n 2n2 /(q K 1) 2n2 n. Let k = Q, and assume [K Q(µ) : Q] <. We begin with the bound on exp G in Claim 3.5. In the proof we had L = K[u] with [L : K] = d, h dm L /m K. Now m L φ(m L ) 2 = [Q(µ L ) : Q] 2 = [Q(µ L ) : K Q(µ L )] 2 [K Q(µ L ) : Q] 2 d 2 c 2 K, where

BOUNDS ON ORDERS OF FINITE SUBGROUPS OF P GL n (K) 7 c K = [K Q(µ) : Q]. Hence h d 3 c 2 K /m K. Then restoring subscripts and proceeding as in the proof of Claim 3.5, we get h ( d 3 i )(c2 K /m K) r (e 3/e c 2 K /m K) n, hence exp G (e 3/e c 2 K/m K ) n. In the proof of Theorem 3.4, we obtain A i m K (exp G) 2, hence A A i m n K (exp G)2n m n K (e3/e c 2 K /m K) 2n2, whence G J(n) A (49n) n2 (e 3/e c 2 K) 2n2 /m 2n2 n K. 4. Central simple algebras Let B be a finite dimensional central simple K-algebra (with center K). Let Γ be a subgroup of the multiplicative group B of invertible elements of B, such that [Γ : K Γ] <. (If Γ also spans B over K, B is called a projective Schur algebra over K (see [1]).) Let L be a splitting field of B, so that B K L = M n (L). We have Γ L = Γ K (Γ identified with Γ 1), so Γ/Γ K embeds into P GL n (L), where n = deg K B. Thus by Proposition 2.1, if B(L, n) holds, then there is a bound on the orders of (nonmodular) Γ/Γ K. We can say more: 4.1. Proposition. If B(K, n) holds, there is a bound on the orders of (nonmodular) Γ/Γ K for all central simple K-algebras of degree n over K. Proof. There exists a splitting field L of B which is a regular extension of K (K is algebraically closed in L), for example a generic splitting field of B has this property [2, Theorem 9.1, p. 26]. Then L k(µ) = K k(µ), hence by Proposition 2.2, B(K, n) B(K k(µ), n) B(L k(µ), n) B(L, n). Note that the explicit bounds given above for P GL n (K) will hold for all central simple K-algebras of degree n. References [1] E. Aljadeff and J. Sonn, On the projective Schur group of a field, J. Alg. 178 (1995), 530 540. [2] S. Amitsur, Generic splitting fields of central simple algebras, Ann. Math. 62 (1955), 8 43. [3] R. Brauer and W. Feit, An analogue of Jordan s theorem in characteristic p, Ann. Math. 84 (1966), 119 131. [4] J. D. Dixon, The Structure of Linear Groups, Van Nostrand Reinhold, London, 1971. [5] W. Feit, Orders of finite linear groups, preprint (1995).

8 ELI ALJADEFF AND JACK SONN [6] S. Friedland, The maximal orders of finite subgroups in GL n (K), Proc. A.M.S. 125 (1997), 3519 3526. [7] P. Hilton and U. Stambach, A Course in Homological Algebra, Springer-Verlag, Berlin, 1977. [8] L. Risman, On the order and degree of solutions to pure equations, Proc. AMS 55 (1976), 261 266. [9] E. Zelmanov, On the restricted Burnside problem, Proc. Int. Cong. Math. Kyoto (1991), Math. Soc. Japan, 395 402. Department of Mathematics, Technion, 32000 Haifa, Israel E-mail address: aljadeff@tx.technion.ac.il, sonn@math.technion.ac.il