AlgebraVolume 2013 (2013), Article ID 983186, 5 pageshttp://dx.doi.org/10.1155/2013/983186

Research Article

A Characterization of Projective Special Unitary Group U ₃(7) by nse

Shitian Liu

School of Science, Sichuan University of Science and Engineering, Zigong, Sichuan 643000, China

Received 13 April 2013; Revised 4 July 2013; Accepted 24 July 2013

Academic Editor: Shilin Yang

Copyright © 2013 Shitian Liu. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract.

Let a group and be the set of element orders of . Let and let be the number of elements of order in . Let nse. In Khatami et al. and Liu's works, and are uniquely determined by nse. In this paper, we prove that if is a group such that nse = nse, then .

1. Introduction

A finite group is called a simple -group if is a simple group with . In 1987, J. G. Thompson posed a very interesting problem related to algebraic number fields as follows (see [1]).

Thompson’s Problem. Let and , where is the number of elements with order . Suppose that . If is a finite solvable group, is it true that is also necessarily solvable?

It is easy to see that if and are of the same order type, then

The proof is as follows: let be a group and some simple -group, where ; then if and only if and nse (see [2, 3]). And also the group is characterizable by order and nse (see [4]). Recently, all sporadic simple groups are characterizable by nse and order (see [5]).

Comparing the sizes of elements of same order but disregarding the actual orders of elements in of Thompson’s problem, whether it can characterize finite simple groups? Up to now, some groups especial for , where , can be characterized by only the set nse (see [6, 7]). The author has proved that the group is characterizable by nse (see [8]). In this paper, it is shown that the group also can be characterized by nse.

Here, we introduce some notations which will be used. Let denote the product of integer by integer . If is an integer, then we denote by the set of all prime divisors of . Let be a group. The set of element orders of is denoted by . Let and let be the number of elements of order in . Let nse. Let denote the set of prime such that contains an element of order . denotes the projective special linear group of degree over finite fields of order . denotes the projective special unitary group of degree over finite fields of order . The other notations are standard (see [9]).

2. Some Lemmas

Lemma 1 (see [10]). Let G be a finite group and m a positive integer dividing . If , then .

Lemma 2 (see [11]). Let G be a finite group and let be odd. Suppose that P is a Sylow p-subgroup of G and with . If P is not cyclic and , then the number of elements of order n is always a multiple of .

Lemma 3 (see [7]). Let G be a group containing more than two elements. If the maximal number s of elements of the same order in G is finite, then G is finite and .

Lemma 4 (see [12, Theorem ]). Let G be a finite solvable group and , where , . Let and let be the number of Hall -subgroups of G. Then, satisfies the following conditions for all : (1) () for some ;(2)the order of some chief factor of G is divided by .

To prove , we need the structure of simple -groups.

Lemma 5 (see [13]). Let be a simple -group. Then, is isomorphic to one of the following groups: (1), , , or ; (2), , or ; (3)one of the following: (a), where is a prime and with , , , and is a prime greater than 3, (b), where , with , , are primes, , and , (c), where , or , , with , , are odd primes, , and , (4)one of the following 28 simple groups: , , , , , , , , , , , , , , , , , , , , , , , , , , , or.

Lemma 6. Let be a simple -group and . Then, .

Proof. From Lemma 5, order consideration rules out these cases.So, we consider Lemma 5(3). We will deal with this with the following cases.Case  1. Consider  , where . (i)Let ; then , which contradicts . (ii)Let ; then , which contradicts . (iii)Let ; then , which contradicts . Case  2. Consider  , where . (i)Let ; then and so . But the equation has no solution in , which is a contradiction. (ii)Let ; then and . Thus, and . But , a contradiction. (iii)Let ; then the equation has no solution in , which is a contradiction. Case  3. Consider  .We will consider the case by the following two subcases. Subcase 3.1. Consider   and .We can suppose that .Let ; the equation has no solution.Subcase 3.2. Consider and .We can suppose that .Let ; then the equation has no solution in , which is a contradiction.In review of Lemma 5(4), .This completes the proof of the lemma.

3. Proof of Theorem

Let be a group such that , and let be the number of elements of order . By Lemma 3, we have that is finite. We note that , where is the number of cyclic subgroups of order . Also we note that if , then is even. If , then by Lemma 1 and the above discussion, we have

Theorem 7. Let be a group with , , , , , , , , , , , , ,  where is the projective special unitary group of degree 3 over field of order 7. Then, .

Proof. We prove the theorem by first proving that , second showing that , and so .By (2), . If , then is even, and then , .In the following, we prove that . If  , then by (2), . If , then . Therefore, . Now we consider Sylow 17-subgroup acts fixed-point-freely on the set of elements of order 2 of ; then , which is a contradiction. Hence, we have . Furthermore, by (2) , , and . If , then , and so . If , then by (2), , and so . If , then . If , then . If , then . If , then by (2),   nse. So, . If , then . If , then by (2), . Thus, .In the following, we first prove that , then consider the proper set of , and so . Finally, we get the desired result by [2].To remove the “bad” prime 5, we should prove that .Since and exp, 4, 8, 16, 32, 64, 128, 256, then by Lemma 1, and (this case occurs only when , , , and ).If , then exp.(i)If , then . It follows that nse. So we have and so, .(ii)If , then or 809088.(a)Let ; then , and so . Then, , and so . (b)Let ; then , but . Then, . (iii)If , then , and so .If and exp, then by Lemma 1, , and so .If and exp, then by Lemma 1, . By (2),   = 101136, 202272, 809088, 1843968. (i)  = 101136, 202272 or 809088, and then , but . Then, .(ii)  = 1843968, and then . So, .If and exp, then by Lemma 1, , and so .Assume that . We deal with it with the following cases.Case A. Let 7. We know that exp.(i)Let exp  = 7. Then by Lemma 1, , and so . If =7, then since , 43, which is a contradiction. If , then we can assume that , , where , , and are nonnegative integers and . Since , then since is at most 11, the equation has no solution in . If , then similarly, , 3, 1), (8, 3, 1), (9, 3, 1), (10, 3, 1), (11, 3, 1), (9, 2, 1), (10, 2, 1), (11, 2, 1). Let . Then by Sylow’s theorem, the number of Sylow 5-subgroups of is 1, 6, 16, 21, 36, 56, 96, 126, 196, 216, 336, 441, 576, 686, 756, 896, 1176, 2016, 2646, 3136, 3456, 3528, 4116, 7056, 9261, 10976, 12096, 18816, 24696, 42336, 65856, and 395136, and so the number of elements of order 5 is at most 4, 24, 64, 84, 144, 224, 384, 504, 784, 864, 1344, 1764, 2304, 2744, 3024, 3584, 4704, 8064, 10584, 12544, 13824, 14112, 16464, 28224, 37044, 43904, 48384, 75264, 98784, 169344, 263424, and 1580544, but none of which belongs to nse, which is a contradiction. Similarly, we can rule out the other cases “8, 3, 1), (9, 3, 1), (9, 2, 1).” Similarly, for , we can get a contradiction as the case “."(ii)Let exp. Then by Lemma 1, for 101136, 202272, 809088, 1843968}, and so . Since for 101136, 202272, 809088}, , which is a contradiction. If , by Sylow’s theorem, for some integer , but the equation has no solution in , which is a contradiction.(iii) Let exp. Then by Lemma 1, and so . If , then since , by Sylow’s theorem, for some integer , but the equation has no solution in , which is a contradiction. If , then we can assume that , , where , , and are nonnegative integers and . Since , then , 3, 1), (6, 3, 1), (7, 3, 1), (8, 3, 1), (9, 3, 1), (6, 2, 1), (7, 2, 1), (8, 2, 1), (9, 2, 1), (10, 2, 1), (11, 2, 1), (8, 1, 1), (9, 1, 1), (10, 1, 1), and (10, 1, 1). Let . Then by Sylow’s theorem, the number of Sylow 5-subgroups of is 1, 6, 16, 21, 36, 56, 96, 126, 196, 216, 336, 441, 686, 756, 1176, 2401, 2016, 2646, 4116, 7056, 9261, 10976, 14406, 18816, 24696, 28416, 42336, 86436, 148176, and 230496. It follows that the number of elements of order 5 is 4, 24, 64, 84, 144, 224, 384, 504, 784, 864, 1344, 1764, 2744, 3024, 4704, 9604, 8064, 10584, 16464, 28224, 37044, 43904, 57624, 75264, 98784, 113664, 169344, 345744, 592704, and 921984, but none of which belongs to nse. Similarly, we can rule the other cases.(iv)Let exp. Then by Lemma 1, , and so . Since , by Sylow’s theorem, for some integer , but the equation has no solution in , which is a contradiction. Case B. Let 5. We know that exp   = 5. Then by Lemma 1, , and so   = 5. Since , then , which is a contradiction.Case C. Let . We know that exp. (i)Let exp. Then by Lemma 1, and so . Since , , which is a contradiction. Therefore, . From cases A and B, and by assumption “", we can assume that . Then, + , where and are nonnegative integers and . Since , then since is at most 11, the equation has no solution in .(ii)Let exp. Then by Lemma 1, and so . If , then 43, which is a contradiction. If , 81, then similarly as the case “exp( and ," the equation has no solution in since is at most 11.(iii)Let exp(. Then by Lemma 1, , and so . Since , then , which is a contradiction.Therefore, . We prove that .Suppose to the contrary that . If , sets and are Sylow 43-subgroups of ; then and are conjugate in , and so and are also conjugate in . Therefore, we have , where is the number of cyclic subgroups of order 5 in . Since , and so for some integer , the equation has no solution in since . Therefore, ; it follows that the Sylow 5-subgroup of acts fixed-point-freely on the set of elements of order 43, and so , which is a contradiction. Thus, .Therefore, we have that .Case A (). Therefore, + , where and are nonnegative integers and . Since , then since is at most 11, the equation has no solution in .Case B (). As exp, by Lemma 1, and so . Since , , which is a contradiction.Case C (). Since exp, then by Lemma 1, , and so . Since , , which is a contradiction.Case D   (). In the following, we first show that , , , , , or . and second prove that .Step  1  ( or ). We have known that .If , sets and are Sylow 43-subgroups of ; then and are conjugate in , and so and are also conjugate in . Therefore, we have , where is the number of cyclic subgroups of order 3 in . Since ,, so for some integer . Since , the equation has no solution. Therefore, ; it follows that the Sylow 3-subgroup of acts fixed-point-freely on the set of elements of order 43, , and so . Similarly, and , .Therefore, we can assume that . Since , then , or .Step  2   (). First prove that there is no group such that or and nse. Then by [2], .Let and nse.If is soluble, set is a Hall -subgroup of , and all the Hall -subgroups of are conjugate, so the number of Hall -subgroups of is . Let by Lemma 4, (mod 43), (mod 43), ,, ,, where , are nonnegative integers and ,  . Hence, . So and , but , which is a contradiction. So is insoluble.Therefore, has a normal series such that is isomorphic to a simple -group with , 4 as 9 and 1849 do not divide the order of .If is isomorphic to a simple -group, from [14], . Then, . Let . Then, . It is easy to see that , and so . Since , . Therefore, , and so or . That is, or . By Sylow’s theorem, . Since , we have that , and so , which contradicts .Hence, is isomorphic to a simple -group; then by Lemma 6, . So . Therefore, or .If , then order consideration . It follows that is a normal subgroup generated by a 2-central element of . So, there exists an element of order , which is a contradiction.If , then . But nsense, which is a contradiction.Therefore, , and by assumption, nse = nse; then by [2], .This completes the proof of the theorem.

Acknowledgments

The author is supported by the Department of Education of Sichuan Province (Grants nos. 12ZB085, 12ZB291, and 13ZA0119). The author is very grateful for the helpful suggestions of the referee.

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