NONABELIAN GROUPS WITH PERFECT ORDER SUBSETS

Transcription

1 NONABELIAN GROUPS WITH PERFECT ORDER SUBSETS CARRIE E. FINCH AND LENNY JONES Abstract. Let G be a finite group and let x G. Define the order subset of G determined by x to be the set of all elements in G that have the same order as x. A group G is said to have perfect order subsets if the number of elements in each order subset of G is a divisor of G. In this article we prove a theorem for a class of nonabelian groups, which is analogous to Theorem 4 in []. We then prove that there are infinitely many nonabelian groups with perfect order subsets. In addition, all values of q are determined such that the special linear group, SL(,q), has perfect order subsets. Next, we give a discussion of some necessary conditions for general nonabelian groups to have perfect order subsets. We conclude by stating some conjectures. 1. INTRODUCTION Definition 1.1. Let G be a finite group and let x G. We define the order subset of G determined by x to be the set of all elements in G that have the same order as x. A group G is said to have perfect order subsets if the number of elements in each order subset of G is a divisor of G. The following is a short list of groups which have perfect order subsets. Note that (Z m ) t is used to indicate the direct product of t copies of Z m. Z n for all n (Z ) Z 3 Z 16 Z 9 Z 15 (Z ) 3 Z 3 Z 5 Z 17 Z 57 Z (Z ) 11 Z 3 Z 5 (Z 11 ) Z 3 Z 89 S 3, the symmetric group on three letters We observe that all but one group on this list is abelian. For a discussion of how the abelian groups on this list were found, and a treatment of the abelian case in general, see []. Until now, S 3 was the only known example of a nonabelian group with perfect order subsets. In this article we prove a nonabelian analog of Theorem 4 from []. Then, utilizing Theorem 1 from [], we prove that there are, in fact, infinitely many such groups. Next, in an attempt to generalize the phenomenon that makes S 3 have perfect order subsets, we determine all values of q such that SL(,q) has perfect order subsets. Finally, 1991 Mathematics Subject Classification. Primary: 0E34, Secondary: 0E45, 11D61. 1

2 CARRIE E. FINCH AND LENNY JONES with the use of certain necessary conditions, we find other nonabelian groups which fail to have perfect order subsets. In particular, we prove the following: Theorem 1.. Let G be a group such that G = S 3 (Z ) t M, where t 0, and M is a cyclic group of odd square-free order not divisible by 3. If G has perfect order subsets then G is isomorphic to one of the following three groups: S 3 S 3 Z Z 7 S 3 (Z ) Z 5 Theorem 1.3. There exist infinitely many nonabelian groups with perfect order subsets. Theorem 1.4. Let SL(,q) denote the group of all matrices of determinant one with entries from the finite field F q of q elements, where q = p n for some prime p. Then SL(,q) has perfect order subsets if and only if q {, 3, 5, 7, 9, 11, 17, 19, 41, 49, 17, 51} Theorem 1.5. Let G be a finite nonabelian group and let p be a prime divisor of G. Suppose that C G (x) = p for every element x in G of order p. If G has more than one conjugacy class of elements of order p, then G does not have perfect order subsets.. The Proofs of Theorems 1. and 1.3 We begin by stating Theorem 4 from [], which we denote here as Theorem A. Theorem A. Let G be a group such that G = (Z ) t M, where t 1, and M is a cyclic group of odd square-free order. If G has perfect order subsets, then G is isomorphic to one of the following nine groups: Z (Z ) Z 3 (Z ) 3 Z 3 Z 7 (Z ) 4 Z 3 Z 5 (Z ) 5 Z 3 Z 5 Z 31 (Z ) 8 Z 3 Z 5 Z 17 (Z ) 16 Z 3 Z 5 Z 17 Z 57 (Z ) 17 Z 3 Z 5 Z 17 Z 57 Z (Z ) 3 Z 3 Z 5 Z 17 Z 57 Z Proof of Theorem 1. First suppose that t = 0. Then the number of elements of order is 3 and the number of elements of order 3 is. Let p be the smallest prime dividing M. Then the number of elements of order p is p 1, and since this must divide G, it follows that p =7. But then the number of elements of order 14 is 18, which does not divide G, since G is square-free. We conclude that M is trivial and G = S 3, which is indeed a POS group.

3 NONABELIAN GROUPS WITH PERFECT ORDER SUBSETS 3 Now suppose that t 1. Then the number of elements of order is 3( t )+ t 1 = t+ 1 and the number of elements of order 6 is ( t 1). Thus G, and in particular, 3 M is divisible by both t 1 and t+ 1. If p is an odd prime dividing t +, then p 1 divides t+ 1 and must be square-free. If q is a prime divisor of p 1, then the order of modulo q is p. Hence, if q 1 and q are distinct prime divisors of p 1, it follows that p divides (q 1 1)(q 1), the number of elements of order q 1 q in G. This contradicts the fact that S 3 M is square-free, and therefore p 1 must be prime, with p 1 1 (mod 3). If p 1 and p are two distinct odd primes dividing t +, then from above, both p 1 1 and p 1 are primes congruent to 1 modulo 3. Thus, 9 divides ( p 1 )( p ), the number of elements of order ( p 1 1)( p 1), which again contradicts the fact that S 3 M is square-free. Hence, at most one odd prime divides t +. Now suppose that p divides t+, where p is an odd prime. Then, as above, p 1 is prime, and so 9 divides ( t+ 1)( p ), the number of elements of order ( p 1) in G. We infer that t + is necessarily a power of a single prime. Suppose now that t+ = p b for some odd prime p so that, from above, p 1 is a prime divisor of t+ 1. If b>1, then t+ 1 is not prime, and since t+ 1 is square-free, there exists a prime q p 1 that divides t+ 1. But then p divides (q 1)( p ), the number of elements of order q( p 1) in G, again contradicting the fact that S 3 M is square-free. Thus b =1 and t + is prime. A similar argument allows us to conclude that either t = 1, t is a power of or t is an odd prime. Observe that t and t + are either both even or odd. This gives us three cases to consider. Suppose first that t = 1. Then t + = 3 and G = S 3 Z Z 7 M, since the number of elements of order is 3 1 = 7. But then the number of elements of order 4 is 1 and thus M is trivial. It is easy to check that S 3 Z Z 7 is a POS group. Next, suppose that both t and t + are odd primes. Then t 1 and t+ 1 are both primes and they are both divisors of G. Consequently, 9 divides ( t )( t+ ), the number of elements of order ( t 1)( t+ 1) in G. Therefore this case yields no new POS groups. Finally, suppose that t = a and t + = b for some positive integers a and b. It follows that b 1 a 1 = 1, which implies that a = 1 and b =. Thus, G = S 3 (Z ) Z 5 M since 4 1 = 15, the number of elements of order, divides G. Then the number of elements of order 30 is 4. So if there exists an odd prime p dividing M, it would follow that the number of elements of order 30p is 4(p 1), which is divisible by 16. Since 16 does not divide G,

4 4 CARRIE E. FINCH AND LENNY JONES we conclude that M is trivial. Making the observation that S 3 (Z ) Z 5 is a POS group completes the proof. For the proof of Theorem 1.3, we need Theorem 1 from []. We state it here as Theorem B. Theorem B. Let G (Z p a) t M and Ĝ (Z p a+1)t M, where a and t are positive integers and p is a prime that does not divide M. If G has perfect order subsets, then Ĝ has perfect order subsets. Proof of Theorem 1.3 The proof of Theorem B is combinatorial and does not require that M be abelian. Therefore, Theorem B applies to the groups in Theorem 1. and Theorem 1.3 follows immediately. 3. The Proof of Theorem 1.4. The proof of Theorem 1.4 is fairly straightforward. Given an element x in SL(,q), we count the number of elements in the order subset determined by x using the decomposition of SL(,q) into its conjugacy classes. Some elementary number theoretic facts are used to facilitate this counting. Requiring that the number of elements in each order subset divides the order of SL(,q), places severe restrictions on the values for q, which in turn can be determined by solving a set of Diophantine equations. We begin with some preliminaries PRELIMINARIES. This section is divided into two subsections. In the first subsection, some easily verified facts from number theory are presented without proof. In the second subsection, we use these number-theoretic facts to count the number of elements of a specified order in SL(,q) SOME NUMBER THEORY. Lemma 3.1. Let a and m be integers such that 1 a < m. Then, for any divisor d of m, we have that GCD(a, m) =d if and only if GCD(m a, m) =d. The following four corollaries are immediate consequences of Lemma 3.1. In their statements we use Euler s phi function, φ(r), to indicate the number of positive integers less than r which are relatively prime to r. Corollary 3.. Let n be an integer. Then the exact number of positive integers m such that m n 1 1 with GCD( n 1,m) = 1 is φ(n 1). Corollary 3.3. Let n 1 be an integer. Then the exact number of positive integers m such that m n 1 with GCD( n +1,m) = 1 is φ(n + 1).

5 NONABELIAN GROUPS WITH PERFECT ORDER SUBSETS 5 Corollary 3.4. Let n 1 be an integer and let p be an odd prime. Then the exact number of positive integers m such that m pn 3 with GCD(p n 1,m) = 1 is φ(pn 1). Corollary 3.5. Let n 1 be an integer and let p be an odd prime. Then the exact number of positive integers m such that m pn 1 with GCD(p n + 1,m) = 1 is φ(pn + 1). Although the next two lemmas are presented in a number-theoretic setting, we need them in the next subsection to count the number of elements of a particular order in SL(,q). As above, φ is Euler s phi function. Lemma 3.6. Let m>1 be an integer and suppose that d divides m. ( Then m ) the number of positive integers k < m such that GCD(m, k) =d is φ d ( m ) Lemma 3.7. If d divides m and φ(m) divides m, then φ divides m. d SOME GROUP THEORY. Recall that the cardinality of SL(,q) is q(q 1)(q + 1) and that the exact number of conjugacy classes in SL(,q) is q + 4 when q is odd, and q + 1 when q is even. The following tables indicate the structure of the conjugacy classes of SL(,q). Tables I and II list the conjugacy classes when q is odd and q is even respectively. [3]

6 6 CARRIE E. FINCH AND LENNY JONES Conjugacy Class Rep Order of Rep Cardinality of Class z 1 c p q 1 d p q 1 zc p q 1 zd p q 1 a l q 1 GCD(l, q 1) 1 l q 3 q(q + 1) b m q +1 GCD(m, q + 1) 1 m q 1 q(q 1) Table I (q is odd)

7 NONABELIAN GROUPS WITH PERFECT ORDER SUBSETS 7 Conjugacy Class Rep Order of Rep Cardinality of Class c q 1 a l q 1 GCD(l, q 1) 1 l q q(q + 1) b m q +1 GCD(m, q + 1) q(q 1) 1 m q Table II (q is even) We see from Table I that the number of elements of each of the orders 1,, p and p divides the order of SL(,q) for any value of q. So the restrictions on q are imposed by the order subsets determined by the elements a l and b m. The same is true when q is even. We do not, however, need to count the number of elements in every such order subset. The reason for this is the following. Let d be a divisor of q 1 with d not equal to d 1 or q 1. By Lemma 3.6, the ( ) q 1 number of positive integers l q 1 such that GCD(l, q 1) = d is φ. d By Lemma 3.1, the number of such integers l< q 1 is φ ( ) q 1 d. Therefore, the total number of elements in SL(,q) of order q 1 is φ ( ) q 1 d q(q + 1). It is d then clear from Lemma 3.7 that we need consider only the case when d =1. A similar argument handles the situation for b m. Utilizing Corollaries 3.4 and 3.5, we summarize this counting for d = 1 in Proposition 3.8. Proposition 3.8. The total number of elements in SL(,q) of order q 1 is φ(q 1) q(q + 1) and the total number of elements in SL(,q) of order q +1 is φ(q + 1) q(q + 1).

8 8 CARRIE E. FINCH AND LENNY JONES Consolidating and summarizing the work in this section, Theorem 3.9 provides the link to perfect order subsets. Theorem 3.9. The group SL(,q) has perfect order subsets if and only if are both integers. (q 1) φ(q 1) and (q + 1) φ(q + 1) 3.. THE PROOF OF THEOREM 1.4. Theorem 3.9 drastically reduces the possibilities for q in terms of the prime factors of q 1 and q+1. Specifically, we have Theorem Let m = p a m i i. Then φ(m) i=1 exactly one of the following is true: t t =1, p 1 =; so that m = a t =1, p 1 =3; so that m =3 b t =, p 1 =and p =3; so that m = a 3 b t =, p 1 =and p =5; so that m = a 5 b t =3, p 1 =, p =3and p 3 =7; so that m = a 3 b 7 c with all exponents positive. is an integer if and only if In our situation m = q ± 1, and so one possibility would be q 1 = a 5 b and q + 1 = c 3 d 7 e. This example gives us = (q + 1) (q 1) = c 3 d 7 e a 5 b, or equivalently ( ) c 1 3 d 7 e a 1 5 b =1. Considering every possibility arising from the conditions in Theorem 3.10, we get a total of seventeen exponential Diophantine equations, only one of which corresponds to the case when q is even. Some of these equations, however, can be seen immediately to have no solutions since GCD(q 1,q + 1) = 1 or. For example, if q 1 = a 3 b and q + 1 = c 3 d 7 e, we get the equation c 1 3 d 7 e a 1 3 b = 1, which has no solution if we insist that all variables are positive. Eliminating equations of this particular type reduces the number of equations to twelve. When q is odd, we are concerned with only the solutions of these equations where the variables are positive, since positive exponents distinguish the cases from one another. Also, while any particular equation reflects necessary conditions for q, a solution to such an equation might not actually produce a value of q that is a power of a prime. Therefore we need to check each solution to guarantee sufficiency. All of these equations can be solved using a combination of divisibility, congruence and Pell equation arguments. To illustrate the techniques involved, we outline the solution for equation ( ). It is clear that either a = 1 or c = 1 in ( ). So suppose first that c = 1. If a 4, then reduction of ( ) modulo 3, 5, 7 and 8 imply that

9 NONABELIAN GROUPS WITH PERFECT ORDER SUBSETS 9 any solution of ( ) is also a solution to the Pell equation x 5y =1. Since the fundamental solution of x 5y = 1 is , any solution (u, v) to ( ) with both u and v positive is given by u + v 5 = ( ) n for some positive integer n. It is easy to show, however, that for any n, u is never divisible by 1. Thus, a 3. Reduction of ( ) modulo 7 reveals that a must be odd. Since a is clearly not equal to 1, we have a =3. Reduction of ( ) modulo 7 with a = 3 shows us that b 1 is divisible by 6. If d>1, then reduction of ( ) modulo 9 implies however that b + 1 is divisible by 6, a contradiction. Consequently, d =1. If b>1, ( ) is impossible modulo 5. Thus, a = 3, b = d = e = 1 is the only solution to ( ) with c =1. Now suppose that a = 1. Then ( ) is impossible modulo 4 if c 3, and impossible modulo 6 if c = 1. Thus, c =. If e, then ( ) has no solutions modulo 141, and so e = 1. Now, ( ) has no solutions if d = 1 and since ( ) is impossible modulo 1971 if d 3, we conclude that the only solution with a = 1 is b = 3, c = d = and e = 1. We can now recover the corresponding values for q to determine whether these solutions to ( ) actually do produce groups. Indeed, in both cases we get legitimate values of q. The first solution gives us q = 41 while the second solution yields q = 51. Table III is a compilation of the twelve nontrivial equations, in no particular order, together with their solutions and the corresponding values of q. The solutions are indicated as ordered tuples where the entries are the values of the variables in alphabetical order. There are also columns for q 1 and q + 1, which indicate how each equation arises. Equation q 1 q + 1 Solutions Corresponding Values of q 3 b 3 a = 3 a 3 b (0,1) c 1 a 1 5 b =1 a 5 b c none none b 1 a 1 =1 a b (1,) 3 b 1 3 c a 1 =1 a b 3 c (,1,1) 5 (4,1,) 17 c 1 a 1 3 b =1 a 3 b c (1,1,3) 7 c 1 5 d a 1 3 b =1 a 3 b c 5 d (1,,,1) 19 (4,1,1,) 49 c 1 3 d a 1 5 b =1 a 5 b c 3 d (1,1,,1) 11 (5,1,1,4) 161 b 1 5 c a 1 =1 a b 5 c (3,1,1) 9 b 1 3 c 7 d a 1 =1 a b 3 c 7 d none none d 1 a 1 3 b 7 c =1 a 3 b 7 c d (1,,1,7) 17 c 1 3 d 7 e a 1 5 b =1 a 5 b c 3 d 7 e (3,1,1,1,1) 41 (1,3,,,1) 51 d 1 5 e a 1 3 b 7 e =1 a 3 b 7 e d 5 e none none Table III

10 10 CARRIE E. FINCH AND LENNY JONES The only non-legitimate value of q that is obtained is q = 161 = (7)(3). Hence, Theorem 1.4 is established. 4. Some Non-Examples As mentioned before, the special linear group SL(,q) is, in some sense, a natural generalization of S 3. A more obvious generalization is, of course, S n. However, S 4 contains exactly 9 elements of order, so that S 4 does not have perfect order subsets. In fact, computer evidence suggests that the number of elements of order in S n, for n 4, is never a divisor of n!. We can eliminate infinitely many alternating groups from the list of possible groups with perfect order subsets by the following easy proposition. Proposition 4.1. Let A n, with n 4, be the alternating group on n letters, where n or n 1 is prime. Then A n does not have perfect order subsets. Proof: First suppose that n = p is prime. Counting the number of elements of order p in A p gives (p 1)! which does not divide A p = p!. Similarly, if (p + 1)! n 1 =p is prime, the number of elements of order p in A p+1 is which p (p + 1)! does not divide A p+1 =. Theorem 1.5 gives us a criterion to eliminate some more nonabelian groups The Proof of Theorem 1.5. Let N be the number of conjugacy classes of elements in G of order p. Since C G (x) = p, for every element x in G of order p, each such conjugacy class has order G. Thus, the total number of p elements of order p in G is N G. Note that N < p. If G has perfect order p subsets, then N G k = G q for some positive integer k, which implies that N = 1, contradicting the fact that N. A slight modification of Theorem 1.5 given below in Corollary 4. can be used to generate another infinite collection of groups that do not have perfect order subsets. Corollary 4.. Let p be a prime such that p divides G but p does not divide G and let S be a Sylow p subgroup of G. If C G (S) =S and G has more than one conjugacy class of elements of order p, then G does not have perfect order subsets.

11 NONABELIAN GROUPS WITH PERFECT ORDER SUBSETS 11 Proposition 4.3. The simple groups PSL(,q), where q>3 is prime, do not have perfect order subsets. Proof: Let S be a Sylow q subgroup of G =PSL(,q). Since G = (q 1)q(q + 1), we have S = q and it is easy to check that C G (S) =S. Assume, by way of contradiction, that G contains exactly one conjugacy class of elements of order q. We count the number N of elements of order q in two ways. First, we see that N = G S subgroups of G. Then N = k(q 1), and so k = q +1 since we know that k 1 (mod q). (q 1)(q + 1) =. Next, let k be the number of Sylow q. But this is impossible 5. Conjectures We conclude by stating three conjectures. If G has perfect order subsets, we know from [] that G is even. Conjecture 5.1. If G has perfect order subsets and G is not a power of, then 3 divides G. From computer evidence and the examination of the character tables in [1], we conjecture the following. Conjecture 5.. For n 4, S n and A n do not have perfect order subsets. Conjecture 5.3. If G has perfect order subsets and G is nonabelian, then G is not simple. References [1] J. H. Conway, R. T. Curtis, S. P. Norton,, R. A. Parker, R. A. Wilson, Atlas of finite groups, Clarendon Press, Oxford, 1985 [] Carrie E. Finch and Lenny Jones, A curious connection between Fermat numbers and finite groups, Amer. Math. Monthly 109 (00) [3] Larry Dornhoff, Group Representation Theory, Part A, Marcel Dekker, Inc., New York, 1971 Addendum Walter Feit has recently communicated to us that Conjecture 5.1 is not true. He gives the following counterexample. Let p>3 be a Fermat prime. Let H be the cyclic group of order p 1. Then H is a power of and H has perfect order subsets. Let G be the Frobenius group of order p(p 1) with Frobenius

12 1 CARRIE E. FINCH AND LENNY JONES complement H and Frobenius kernel of order p. Then G is nonabelian and has perfect order subsets, but 3 does not divide G. Department of Mathematics University of South Carolina Columbia, SC 908 address: cfinch@math.sc.edu Department of Mathematics, Shippensburg University, Shippensburg, PA address: lkjone@ship.edu