Research Article | Open Access

Zhangjia Han, Longjiang Xie, Pengfei Guo, "Finite Groups Which Have 20 Elements of Maximal Order", *Mathematical Problems in Engineering*, vol. 2020, Article ID 6585647, 5 pages, 2020. https://doi.org/10.1155/2020/6585647

# Finite Groups Which Have 20 Elements of Maximal Order

**Academic Editor:**Ying Dai

#### Abstract

The structure of finite groups is widely used in various fields and has a great influence on various disciplines. The object of this article is to classify these groups whose number of elements of maximal order of is 20.

#### 1. Introduction

Only finite groups are related in this article and our notation is standard. Furthermore, always denotes a group, denotes a Sylow -subgroup of , and denotes the center product of groups and . For positive integers and , , where is cyclic with and ; represents the number of elements of maximal order of , and is the maximal element order in .

For simplicity, we set symbols in the later:

This topic here is with respect to one of Thompson’s problems.

##### 1.1. Thompson’s Problem

Set with . If a group is solvable satisfying , then is also solvable.

For the purpose of solving this famous problem, some authors investigated the solvability of a group by means of a fixed and gave some meaningful results (for example, [1–12], etc.). In particular, Chen and Shi [3] classified groups with . Jiang and Shao [7] classified groups with . In this article, we give the classification of groups with . The result is as follows.

Theorem 1. *Assume that is satisfied with and as mentioned above. Then,*(1)*If , then , or *(2)*If , then(2.1) , the symmetric group of degree 5(2.2) with and *(3)

*If , then and*(4)

*If , then(4.1)*(5)

*(4.2)**If , then is a Frobenius group, where*(6)

*If , then or is metacyclic and , where .*

Corollary 1. *If satisfying is -free, then is solvable.*

#### 2. Preliminaries

Lemma 1. [12, Lemma 1]. *Assume possesses cyclic subgroups of order . Then, , where and denote the number of elements of order and Euler function of , respectively. Furthermore, if represents the number of cyclic subgroups in of maximal order , then .*

According to Lemma 1, we easily follow the result as given by computations.

Lemma 2. *If , then possible values of , , and are shown as follows:*

Lemma 3. [7, Lemma 4]. *If a nonabelian 2-group has the exponent 4, then contains at least elements with order 4 for with .*

Lemma 4. *If a 2-group has the exponent 4 and , then or .*

*Proof. *Let . If is abelian, then by [8, Lemma 2.5], and so is no more than 5 and is 32. If is nonabelian of the exponent 4 and with , we claim . If , then there is satisfying , because . Clearly, is 32, and it is impossible. Hence, . If is nonabelian with and , we have by Lemma 3. By [13], there is no group of order 64 which is satisfied with the assumption of our lemma. If , then , or by [14].

Lemma 5. *If a 2-group has the exponent 8 and , then 8 divides .*

*Proof. *Obviously, is noncyclic. Setting for two different maximal subgroups in , then and . So and , where is a 3-maximal subgroup in satisfying . Since divides , (mod 8).

We show the result by induction on .

If , then is divisible by 8 by applying a result in [13]. If , then 8 divides , , and by induction, and so 8 divides .

#### 3. Proof of Theorem

*Proof. *Since , we have that and 11 by [3, Lemma 6] and by [3, Corollary 2]. In the following, we discuss the cases for the other values of .

*Case 1. *. is either a 3-group or {2, 3}-group. If the former holds, then its exponent is 3. Using [15, Theorem 3.8.8], is ( a integer), this contradicts the hypothesis. If the later holds, then , the set of prime divisors of . By applying a result in [1, Theorem], is a Frobenius group, where or . Suppose . Then, is elementary abelian. It follows a contradiction by using [15, Theorem 3.8.8]. If is isomorphic to , then it has two elements with order 3, this contradicts the hypothesis. Thus, .

*Case 2. *. Now is a 2-group. Using Lemma 4, , or . Thus, (1) holds.

*Case 3. *. By a result of [3, Lemma 8], , where and . If , then is a -group (its centralizer of every 5-element is a 5-group). If is nonsolvable, then there exists satisfying ; are -groups, is simple, and must divide using [4, Theorem 2.1]. So , , or . But processes elements of order 6, and their indices of its maximal subgroups are 27, 36, 40, and 45. Hence, the index of normalizer in of the arbitrary cyclic subgroup in of order 6 is not less than 27. So has more than 15 subgroups of order 6. It leads to that has more than 20 elements of order 6, and so it is impossible. Thus, . Hence, may be or . Now, we have . Thus, , or . However, contains elements with order 6 surpassing 20; this induces that is not isomorphic to . If , then since has 20 elements of order 6. Thus, (2.1) follows.

Assume that , or and . Let with . Then, . Otherwise, has an element of order 30 since , a contradiction. Hence, . As has only 10 subgroups with order 6, it makes or 10. If and , then any 2-element in is of order 2 and any 3-element in is of order 3. Hence, has elements of order 6. Note that and is a -group, and it follows easily its number whose elements with orders 2, 3, or 6 in are multiples of 5. Hence, and for , and so has elements of order 6. However, , and this is impossible. Thus, we may assume or .

Suppose first . Then, or or since or or . Since has at most 20 elements with order 6, or . Hence, is , , or , where . As has no element with order 4, we have . Otherwise, contains more than 20 elements with order 6, and it is impossible. So , is (). Furthermore, and . Hence, and are 16 and or or and or or . Thus, has at least 30 elements with order 6, and it contradicts the hypothesis.

Suppose next . Then, or or since or or . It is well known that and have elements of order 4, so has no Sylow 2-subgroup of when or . Thus, we can let or . Since or 10, we know that is and , where . It means that . By hypothesis, and so . Thus, (mod 5), a contradiction.

Let be solvable with , where and . If , then is sure to be a Frobenius group or a 2-Frobenius group by [4, Theorem 2.1]. If the former holds and the Frobenius kernel is a 5-Hall subgroup, then contains a normal subgroup with order 5. Hence, contains an element with order 15, and it contradicts the hypothesis. If the former holds and is a -Hall subgroup, then has elements of order 6 by the nilpotency of . Hence, we have that and , and it contradicts the hypothesis. If the later holds, then there is a series satisfying is a 5-group by [4, Theorem 2.3]; and are both Frobenius groups whose kernels are and , respectively. Therefore, contains an element with order 15 when 3 divides , and it contradicts the hypothesis. Hence , with . Choosing with and considering acts on by conjugation, then has an element with order 15, and it contradicts the hypothesis. This implies that cannot be a 2-Frobenius group. So and . By Lemma 2, it has 10 subgroups with order 6, and , 2, 3, 4, 6, 8, or 9 for any with order 6. Let . Since has no element with order 9, any 3-element in which commute satisfies , and it leads to . Thus, . In addition, it follows from has no element with order 4 and arbitrary 2-element satisfying that by .

If or 9, then it has with satisfying being 1 or 2. In total, has 10 subgroups with order 6, which implies has an element with order 6 satisfying . Furthermore, we obtain that is a divisor of or . Hence, . Thus, (2.2) follows.

*Case 4. *. If , then is divisible by 8 by [14], a contradiction. If is a 2-group with , then there is no group which satisfies our assumption using Lemma 5. If is a -group, then . Choosing with , we get , or 5, as has precisely 5 cyclic subgroups with order 8. If with satisfying or 4, then has another element with satisfying . It implies that must have an element with satisfying or 5. If , then . As , we have , and it leads to that has an element with order 40, which contradicts the hypothesis. Thus, . So all cyclic subgroups with order 8 are conjugate in and so do their centralizers. Let . Then, , or 32 by [8, Lemma 2.5]. If , then has precisely 16 elements of order 8 by [8, Lemma 2.5]. Choosing with , then has 4 or 8 elements of order 8 and so has 28 or 24 elements of order 8. Assume that has 4 elements with order 8, and let with . Then, is not in , is not in , and is conjugate to . As is abelian, . Hence, contains elements of order 8, and it contradicts the hypothesis. Assume that has 8 elements with order 8, and let be any element with . Then, has elements with order 8. Let be any element with . Then, we get that contains elements with order 8 using the same arguments, and it contradicts the hypothesis. If , then is abelian and it contains 8 elements with order 8. In fact, we get easily that contains no element of order 8 for any with . Thus, its number whose elements in with order 8 are divided by 8, and it is impossible. If , then and contains 4 elements with order 8. As all centralizers of elements with order 8 are conjugate, . By using a theorem of Burnside, , which is impossible since .

*Case 5. *. By [3, Lemma 8], we can set , where . As has 5 cyclic subgroups with order 10, we have is , or 5 if with .

If or 4, then has with satisfying or . Let . Then, and as has at most 20 elements with order 10. Hence, the order of any 5-element of is 5. It is easy to see that cannot have 20 elements with order 10, and it is impossible.

If , then all cyclic subgroups with order 10 are conjugate in . Let . Then, and . We always have that divides 4 and is a -group. Hence, . Let . If , then using Sylow’s Theorem. Thus, , , and . Clearly, . If , then , a contradiction. Hence, . If , then contains at least 128 elements with order 10, and it contradicts the hypothesis. If , then has at least 64 elements with order 10, and it is impossible as well. Hence and . If , then contains at least 24 elements with order 10, and it is impossible. Hence, and . If , then it has 5 Sylow 2-subgroups in using Sylow’s Theorem. Thus, has 60 elements with order 10, and it contradicts the hypothesis. Therefore . If , where and , then , and has 20 elements with order 10. Thus, (3) holds.

*Case 6. *. Let with . Then, . By [8, Lemma 2.5], has at least elements with order 4. In addition, its 3-elements in are of order 3. So using [8, Lemma 2.5] and the hypothesis. Thus, and or and . Using Lemma 2, has 5 cyclic subgroups with order 12, and so .

If , we let , and . Clearly, . Assume . Then, . If is abelian, then and has precisely 16 elements with order 12. If we choose in with , then has 16 elements with order 12. And has at least 28 elements with order 12, and it contradicts the hypothesis. Assume that is nonabelian with . Furthermore, we get that . Since and , we know that is normal in . Then, since is solvable. Let with . Then, acts fixed-point-freely on . Therefore, , and it contradicts the hypothesis. Hence, does not divide , , and . By the hypothesis, has at most 20 elements of order 12. Hence, has at most 10 elements with order 4. If , then and has 8 elements with order 12. Let and . Then, has no element with order 12. If not, let with . Then, and as and are both abelian. Hence, , a contradiction. If and is abelian with type (4, 2, 2), then contains 8 elements of order 4. It also gets a contradiction similar to case . If is abelian with type (4, 4), then contains 12 elements with order 4 and must have 24 elements with order 12, and it contradicts the hypothesis. If and is nonabelian, then . Now has 8 elements with order 4. Let with . Then, has at least 4 elements with order 12. If has exactly 4 elements with order 12, then must contain 28 elements of order 12, and it contradicts the hypothesis. So has at least 8 elements with order 12. Thus, is 24, and . Let with . Then, is neither in nor in and is conjugate to . As is abelian, . Hence, contains 32 elements with order 12, and it also contradicts the hypothesis. Thus, and . Therefore, and . It is noted that is solvable since the simple groups and have no element with order 12. Thus, (4.1) holds.

If or 4, then there exists with satisfying or 2 as has 5 cyclic subgroups with order 12 and by using Lemma 2. Noting , if there is with satisfying , then is normal in , and hence both and are normal in . Let . Then, . If , then , and hence, . Thus, 20 = = = = 2. So , which contradicts the fact that its number whose elements with order 4 in is always divisible by 4 by using [3, Lemma 9]. If and , then has at least 33 elements with order 3, so has at least 66 elements with order 12, a contradiction. If and , then , which is impossible. Hence, there must exist an element of with satisfying . Therefore, . Thus, (4.2) holds.

*Case 7. *. Let . Then, . If , then may be noncyclic, and its number whose elements with order 25 exceeds 20, a contradiction. Thus, and . Since , we have (5) holds.

*Case 8. *. Let with . Then , , which implies that or is metacyclic. Hence, or , where . Thus, (6) holds.

#### 4. Conclusion

The structure of finite groups is widely used in various fields and has applications in various branches of mathematics. In particular, the classification of finite simple groups is a ground-breaking and amazing piece of contemporary mathematics. It is of great help in solving related problems by checking the specific structure of finite groups. In this paper, we determined those groups in which its number whose elements of maximal order of is 20. The main result enriches the research of finite group theory and provides theoretical guidance for the development of various disciplines.

#### Data Availability

The results have been proved by rigorous mathematical theory, and the data are available.

#### Conflicts of Interest

The authors declare that they have no conflicts of interest.

#### Acknowledgments

This work was supported by the National Scientific Foundation of China (Grant no. 11661031) and Hainan Provincial Natural Science Foundation of China (Grant no. 119MS039).

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#### Copyright

Copyright © 2020 Zhangjia Han et al. 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.