On the Frobenius Numbers of Symmetric Groups

Transcription

1 Journal of Algebra 221, Article ID jabr , available online at on On the Frobenius Numbers of Symmetric Groups Yugen Takegahara Muroran Institute of Technology, 27-1 Mizumoto, Muroran , Japan Communicated by Walter Feit Received October 6, INTRODUCTION Let Hom A G denote the set of all homomorphisms from a group A to a finite group G. Yoshida proved the following theorem. Theorem A [9]. For a finite abelian group A and for a finite group G, Hom A G 0 mod gcd A G This theorem is a generalization of the following theorem of Frobenius [3]: Theorem A. The number of elements x of a finite group G that satisfy the equation x n = 1 is a multiple of gcd n G. Hereafter, let A be a finitely generated group. Then Hom A S n, where S n is the symmetric group of degree n, is finite; such a number is called a Frobenius number [2]. Hence A contains only a finite number of subgroups of index d for each natural number d [6, p. 66, Exercise 3]; we denote such a number by m A d. Then the Wohlfahrt formula[8] holds: 1 + Hom A S n X n = exp n! d=1 m A d X d d Using this formula and Theorem A, Dress and Yoshida give another proof of Theorem A in the case where G = S n [2]. Let p be a prime integer, and let s be a positive integer. Definition 1.1. A finitely generated group A is said to admit Cp s if the following conditions hold for any positive integer q such that /99 $30.00 Copyright 1999 by Academic Press All rights of reproduction in any form reserved.

2 552 yugen takegahara gcd p q =1: 1 For any integer i with 1 i s + 1 /2, where s + 1 /2 is the greatest integer s + 1 /2, 2 Moreover, m A qp i 1 m A qp i mod p i m A qp s+1 /2 m A qp s+1 /2 +1 mod p s/2 By using a formula in [5], we shall first prove the following result in Section 2: Theorem 1.1. Let A be a finite abelian group. Suppose that the order of a Sylow p-subgroup of A is p s. Then A admits Cp s. This theorem has been essentially proved in [1, 2]. The purpose of this paper is to show the following theorem, which, together with Theorem 1.1, implies Theorem A as a corollary: Theorem 1.2. for any positive integer n. If a finitely generated group A admits Cp s, then Hom A S n 0 mod gcd p s n! Let c = c 1 c 2 c d be any sequence of integers. Let q be any integer such that gcd p q =1. For each nonnegative integer i, define pc q if i = 0, q l c q i = c qp i 1 c qp i otherwise. qp i 1 For any nonzero integer n, let ord p n be the exponent of p in the decomposition of n into prime factors, and let ord p 0 = > 0 ; for any rational number n/m, define ord p n/m to be ord p n ord p m [4]. If there exists i>0 with ord p l c q i 0, we denote the smallest such i by i c q. Definition 1.2. A sequence c = c 1 c 2 c d of integers is said to admit Hp s if the following condition holds for any positive integer q such that gcd p q =1: Either ord p l c q i > 0 for all i>0 or else ord p qp i c q! +ord p l c q i c q s + 1 If a finite group A admits Cp s, the sequence m A 1 m A 2 m A d admits Hp s Lemma 5.1. Hence Theorem 1.2 is a consequence of the following theorem.

3 frobenius numbers of symmetric groups 553 Theorem 1.3. Let c = c 1 c 2 c d be a sequence of integers. Define a sequence u 1 u 2 of rational integers by u 1 + n c n! Xn = exp d d Xd Then u n is an integer for any positive integer n, and a necessary and sufficient condition for the sequence c to admit Hp s is that u n 0 mod gcd p s n! for any positive integer n. Theorem 1.3 is a generalization of [4. p. 97, Exercise 18]. Let p denote the additive group of all p-adic integers. It follows from [2] that Hom p S n 1 + X n 1 = exp n! p k Xpk which is called the Artin Hasse exponential [4, Chap. IV]. To prove Theorem 1.3, we need a result similar to Dwork s lemma [4, Chap. IV, Lemma 3], which yields that the coefficient Hom p S n /n! ofx n in the Artin Hasse exponential is a p-adic integer for each n [4, p. 97, Exercise 17]. d=1 k=1 2. ABELIAN GROUPS Let us prove Theorem 1.1. A sequence λ = λ 1 λ 2 λ r 0 of nonnegative integers satisfying λ 1 λ 2 λ r 0 is called the type of a finite abelian p-group isomorphic to C p λ 1 C p λ 2 C p λr where C p j denotes a cyclic group of order p j.letα λ i p denote the number of subgroups of order p i in a finite abelian p-group of type λ. Then α λ i p is a polynomial in p with nonnegative coefficients [1, 5]. By the duality of abelian p-groups, we have that if λ is the type of an abelian p-group of order p s, then α λ i p =α λ s i p To prove Theorem 1.1, it suffices to show the following. Theorem 2.1. [1, 2]. Let λ = λ 1 λ 2 λ r 0 be the type of a finite abelian p-group of order p s > 1. Then α λ i 1 p α λ i p mod p i for 1 i s + 1 /2. Moreover, α λ s + 1 /2 p αλ s + 1 /2 +1 p mod p s/2

4 554 yugen takegahara Proof. Let λ = λ 1 λ k 1 λ k 1 λ k+1, where k is the largest number satisfying λ k = λ 1, and let ˆλ = λ 2 λ 3 λ r 0. It follows from [5, Theorem 1] that α λ i p =α λ i p +p s i αˆλ s i p which is, by the duality, equivalent to the following equation [5, Corollary]: Combining these equations, we have α λ i p =α λ i 1 p +p i αˆλ i p α λ i p α λ i 1 p =p i αˆλ i p p s i+1 αˆλ s i + 1 p This formula was used in [7]. The theorem follows from this equation. To show the second congruence of the theorem, use the equation s = s + 1 /2 + s/2 3. A GENERALIZATION OF DWORK S LEMMA Throughout this section, let u = u 1 u 2 u n be a sequence of integers. Put E u X =1 + Define the series v 0 = 1 v 1 v 2 by E u X p E u X p = u n n! Xn n=0 v n n! Xn Then the following Dwork s lemma [4, Chap. IV, Lemma 3] holds: u n /n! p for any n 1 if and only if v n u /n! p p for any n 1. For each rational number g and a nonnegative integer k, we write g 0 mod p k if g p k p. We shall refine Dwork s result as follows. Proposition 3.1. The following conditions are equivalent : 1 v n 0 mod gcd p s n! for any positive integer n. 2 v n 0 mod p gcd p s n! for any positive integer n.

5 frobenius numbers of symmetric groups 555 We denote by the set of all nonnegative integers. For each positive integer n, let J n be the set of all elements a 0 a 1 a n 1 j of n+1 satisfying the following conditions: n 1 i=0 a i = p a n/p <p n 1 j<n ia i + j = n if p divides n Note that J 1 is empty. For each a = a 0 a 1 a n 1 j J n, define a rational number ϕ u a by ϕ u a = { p 1! n 1 a 0!a 1! a n 1! ui i! ai } vj j! Lemma 3.1. We have u n + v n p + n! n! u n/p ϕ u a = p n/p! n! p un/p if p divides n, p n/p! a J n 0 otherwise. In particular, u 1 + v 1 /p = 0. Proof. By the definition of v n, we have u i i! Xpi = u i i! Xi p 1 + j=1 v j j! Xj Hence u i i! Xpi = p un n! + v n pn! + a J n ϕ u a X n + ui i! p X pi Comparing the coefficients of X n in the equation above, we obtain the result. We need the lemmas below. Lemma 3.2. Assume that u i v i /p 0 mod gcd p s i! for any i with 1 i<n. Then n!ϕ u a 0 mod gcd p s n! for any a = a 0 a 1 a n 1 j J n. Proof. Let a = a 0 a 1 a n 1 j J n. There are two cases.

6 556 yugen takegahara Case 1. Suppose that j 1. Since n 1 i=0 a i = p, p! a 0!a 1! a n 1! p Hence, if u i /i! p for any i with 1 i<nand v j / pj! p, then ϕ u a p, and therefore n!ϕ u a 0mod p ordp n!. So we may assume that either u i 0 mod p s for some i with 1 i<nor v j /p 0 mod p s. Then n!ϕ u a 0 mod p s, because n! n 1 p i! a i j! Thus we have either n!ϕ u a 0 mod p ord p n! or n!ϕ u a 0 mod p s, and hence n!ϕ u a 0 mod gcd p s n!, as desired. Case 2. Suppose that j = 0. In this case, we have a i <pfor any i. Then p 1! a 0!a 1! a n 1! p Therefore the proof is completely analogous to that in Case 1. Lemma 3.3. Let t be an integer. Suppose that t 0 mod gcd p s k! for a positive integer k. Then pk! t p pk! t p k! p k! mod gcd ps pk! Proof. If t/k! p, then t p t mod p k! k! Furthermore, if t 0 mod p s, then t p pk! pk! t k! k! 0 mod ps+1 because pk! p k! p = jk! p k! jk k! = p! jk 1! k 1! jk k! j=1 and jk 1! k 1! jk k! = This completes the proof. j=1 jk 1 k 1 p Proof of Proposition 3.1. Assume that condition 1 holds. Let us show that condition 2 holds. We use induction on n. By Lemma 3.1, v 1 = pu 1 p p, and v 1 0 mod p. Suppose that n>1. It follows from the inductive assumption that v i 0 mod p gcd p s i! for any i with 1 i<n. Then, by using Lemmas 3.1, 3.2, and 3.3, we have v n /p u n 0 mod gcd p s n!, as desired. Likewise, condition 2 implies condition 1.

7 frobenius numbers of symmetric groups THE PROOF OF THEOREM 1.3 Let us prove Theorem 1.3. Let denote the set all elements q i of 2 such that gcd p q =1. For each positive integer n, we denote by n the set of all mappings ν from to that satisfy ν q i qp i = n q i Definition 4.1. Define Let c = c 1 c 2 c d be a sequence of integers. c ν = q i l c q i ν q i ν q i! for each ν n. Proposition 4.1. Let c = c 1 c 2 c d be a sequence of integers. Define a sequence u 1 u 2 of rational integers by u 1 + n c n! Xn = exp d d Xd Then u n is an integer for any positive integer n, and the following conditions are equivalent: d=1 1 u n mod gcd p s n! for any positive integer n. 2 n! ν n c ν 0 mod p gcd p s n! for any positive integer n. Proof. First, we have u n = µ n! 1 µ 1 µ1!2 µ 2 µ2! cµ 1 1 cµ 2 2 where µ = 1 µ 1 2 µ 2 runs over all partitions of n. Since the coefficient n 1 µ 1 µ1!2 µ 2 µ2! of c µ 1 1 cµ 2 2 is the number of permutations of type 1µ 1 2 µ 2, it follows that u n is an integer for any positive integer n. Letu = u 1 u 2 u n. We obtain E u X p E u X p pc =exp d c d Xd + d d Xpd d=1 = exp gcd p q =1 d=1 pc q q Xq + c qp i 1 c qp i qp i 1 X qpi

8 558 yugen takegahara where the summation gcd p q =1 is over all positive integers q such that gcd p q =1. Therefore we have E u X p E u X p =1 + c ν X n ν n Now, by virtue of Proposition 3.1, Proposition 4.1 holds. Let c = c 1 c 2 c d be a sequence of integers. We denote by c the set of all elements q i of with ord p l c q i 0. Lemma 4.1. Let c = c 1 c 2 c d be a sequence of integers, and let ν n. Assume that for every q i c with ν q i > 0, Then ord p qp i! +ord p l c q i s + 1 n! c ν 0 mod p gcd p s n! Proof. If ν q i = 0 for any q i c, then by the definition, ord p c ν 1 see, e.g., [4, p. 7, Exercise 14], and so the assertion holds. Suppose that ν q i > 0 for some q i c. Since ord p n! = j=1 n/pj, it follows that [ ν q i qp i ] ord p n! = ord p j=1 q i j p ν q i qp i! c q i c Here, for any q i c with ν q i > 0, we have because and ord p ν q i qp i! ord p ν q i! +ν q i ord p qp i! ν q i qp i! qp i! ν q i ν q i jqp i 1! = ν q i! qp i 1! jqp i qp i! j=1 jqp i 1! qp i 1! jqp i qp i! = jqp i 1 qp i 1 p Thus ord p n! c ν ν q i ord p qp i! +ord p l c q i q i c because ord p c ν ν q i ord p l c q i ord p ν q i! q i c Now, by assumption, ord p n! c ν s + 1, and the assertion holds. We have thus proved the lemma.

9 frobenius numbers of symmetric groups 559 Lemma 4.2. Let c = c 1 c 2 c d be a sequence of integers. Let q i c. Suppose that ord p qp ic q! +ord p l c q i c q s + 1 Then ord p qp i! +ord p l c q i s + 1 Proof. We may assume that i>i c q. Then it is clear that ord p qp i! ord p qp i c q! +i 1 Since ord p l c q i i 1 and ord p l c q i c q 0, it follows that ord p qp i! +ord p l c q i ord p qp i! i 1 and this yields the lemma. ord p qp i c q! +ord p l c q i c q Using these lemmas, we can prove Theorem 1.3. Proof of Theorem 1.3. By Proposition 4.1, u n is an integer for any positive integer n. Assume that c admits Hp s. Then it follows from Lemmas 4.1 and 4.2 that for any positive integer n, n! c ν 0 mod p gcd p s n! ν n Hence Proposition 4.1 yields that u n 0 mod gcd p s n! for any positive integer n. Conversely, assume that u n 0 mod gcd p s n! for any positive integer n. Then it follows from Proposition 4.1 that for any q i c q c, qp ic q! c ν 0 mod p gcd p s qp ic q! ν qp i c q We show that ord p qp i c q! +ord p l c q i c q s + 1 for any q i c q c. We use induction on qp i c q where q i c q c. Suppose that q 0 p i c q 0 where q 0 i c q 0 c is minimal, i.e., q 0 p i c q 0 qp i c q for any q i c q c.ifqp i <q 0 p i c q 0 for q i, then clearly i<i c q, and therefore ord p l c q i > 0. So, for any ν q 0 p i c q 0 such that ν q 0 i c q 0 = 0, c ν 0 mod p Hence q 0 p i c q 0!l c q 0 i c q 0 0 mod p s+1 therefore by the formula, and ord p q 0 p i c q 0! +ord p l c q i c q 0 s + 1

10 560 yugen takegahara Let q i c q c, and suppose that qp i c q >q 0 p i c q 0. By the inductive assumption, for any q i c q c such that q p i c q <qp i c q, ord p q p i c q! +ord p l c q i c q s + 1 We have the following lemma under these hypotheses. Let ν qp i c q. Suppose that ν q i c q = 0. Then qp i c q! c ν 0 mod p gcd p s qp i c q! Let us prove this lemma. If ν q i > 0 for some q i c, then because ord p q p i c q! +ord p l c q i c q s + 1 q p i c q q p i <qp i c q Therefore Lemma 4.2 implies that for every q i c with ν q i > 0, ord p q p i! +ord p l c q i s + 1 Then the result follows from Lemma 4.1. By this lemma and the formula, qp i c q!l c q i c q 0 mod p s+1, which yields that ord p qp i c q! +ord p l c q i c q s + 1 Thus c admits Hp s, thereby completing the proof of Theorem A LEMMA Theorem 1.2 is a consequence of Theorem 1.3, because the following holds. Lemma 5.1. Let A be a finite group, and let c = m A 1 m A 2 m A d. Suppose that A admits Cp s. Then c admits Hp s. Proof. Let q i c q c. Then i c q s + 1 /2 +1, and { [ ord p lc q s+1 ] 0 if s is even if s is odd.

11 frobenius numbers of symmetric groups 561 Hence, if i c q = s + 1 /2 +1, then ord p qp i c q! + ord p l c q i c q s+1 /2 j=0 2 [ s+1 2 s + 1 If i c q > s + 1 /2 +1, then ord p qp i c q! + ord p lc q i c q p j [ + ord p lc q s+1 ] ] [ ordp lc q s+1 ] i c q 1 j=0 s+1 /2 j=0 s + 1 p j + ord p lc q i c q because ord p l c q i c q i c q +1. Thus c admits Hp s. p j ACKNOWLEDGMENT The author would like to thank the referee for important comments. REFERENCES 1. L. M. Butler, A unimodality result in the enumeration of subgroups of a finite abelian group, Proc. Amer. Math. Soc , A. W. M. Dress and T. Yoshida, On p-divisibility of the Frobenius numbers of symmetric groups, preprint. 3. P. Hall, On a theorem of Frobenius, Proc. London Math. Soc , N. Koblitz, p-adic Numbers, p-adic Analysis, and Zeta-Functions, 2nd ed., Springer Verlag, New York, T. Stehling, On computing the number of subgroups of a finite abelian group, Combinatorica , M. Suzuki, Group Theory I, Springer Verlag, New York, Y. Takegahara, On Butler s unimodality result, Combinatorica , K. Wohlfahrt, Über einen Satz von Dey und die Modulgruppe, Arch. Math. Basel , T. Yoshida, Hom A G, J. Algebra ,