# Finite Groups Which Have 20 Elements of Maximal Order

Authors: Zhangjia Han; Longjiang Xie; Pengfei Guo
Journal: Mathematical Problems in Engineering
 (2020)

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## 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 groupsG whose number of elements of maximal order of G is 20.


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## Body

## 1. Introduction

Only finite groups are related in this article and our notation is standard. Furthermore,G always denotes a group, Sp denotes a Sylow p-subgroup of G, and A∗B denotes the center product of groups A and B. For positive integers m and n, Cnm=a1×a2×⋯×am, where ai1≤i≤m is cyclic with ai=n and Cn=Cn1; MG represents the number of elements of maximal order of G, and k=kG is the maximal element order in G.For simplicity, we set symbols in the later:(1)G1=a,b,c,da2=b2=c2=d4=1,a,b=1,a,c=1,b,c=1,a,d=1,dba−1=ab,dcd−1=bc;G2=a,b,ca4=b4=c2=1,a,c=a2,b,c=a2b2,a,b=1;G3=a,b,ca4=b4=c2=1,a,c=a2b2,b,c=a2,a,b=1.This topic here is with respect to one of Thompson’s problems.


### 1.1. Thompson’s Problem

SetGr=x∈Gx=r with r≥1. If a group M is solvable satisfying Gr=Mr, then G is also solvable.For the purpose of solving this famous problem, some authors investigated the solvability of a group by means of a fixedMG and gave some meaningful results (for example, [1–12], etc.). In particular, Chen and Shi [3] classified groups with MG=30. Jiang and Shao [7] classified groups with MG=24. In this article, we give the classification of groups G with MG=20. The result is as follows.Theorem 1.
Assume thatG is satisfied with MG=20 and k=kG as mentioned above. Then,(1)
Ifk=4, then G≅C23⋊C4,D8×C4,Q16×C2,D8∗Q8,G1,G2, or G3(2)
Ifk=6, then(2.1)
G=S5, the symmetric group of degree 5(2.2)
G=2α⋅3β with α∈2,5 and β∈1,2(3)
Ifk=10, then G=C5×C5⋊C2 and CGC2=10(4)
Ifk=12, then(4.1)
22⋅3⋅5G24⋅3⋅5(4.2)
G27⋅32(5)
Ifk=25, then G=C25⋊K is a Frobenius group, where K≤C20(6)
Ifk∈33,44,50,66, then G≅Ck or G is metacyclic and G=a,bak=1,bl=at,bab−1=ar, where l20.Corollary 1.
IfG satisfying MG=20 is A5-free, then G is solvable.


## 1.1. Thompson’s Problem

SetGr=x∈Gx=r with r≥1. If a group M is solvable satisfying Gr=Mr, then G is also solvable.For the purpose of solving this famous problem, some authors investigated the solvability of a group by means of a fixedMG and gave some meaningful results (for example, [1–12], etc.). In particular, Chen and Shi [3] classified groups with MG=30. Jiang and Shao [7] classified groups with MG=24. In this article, we give the classification of groups G with MG=20. The result is as follows.Theorem 1.
Assume thatG is satisfied with MG=20 and k=kG as mentioned above. Then,(1)
Ifk=4, then G≅C23⋊C4,D8×C4,Q16×C2,D8∗Q8,G1,G2, or G3(2)
Ifk=6, then(2.1)
G=S5, the symmetric group of degree 5(2.2)
G=2α⋅3β with α∈2,5 and β∈1,2(3)
Ifk=10, then G=C5×C5⋊C2 and CGC2=10(4)
Ifk=12, then(4.1)
22⋅3⋅5G24⋅3⋅5(4.2)
G27⋅32(5)
Ifk=25, then G=C25⋊K is a Frobenius group, where K≤C20(6)
Ifk∈33,44,50,66, then G≅Ck or G is metacyclic and G=a,bak=1,bl=at,bab−1=ar, where l20.Corollary 1.
IfG satisfying MG=20 is A5-free, then G is solvable.


## 2. Preliminaries

Lemma 1. [12, Lemma1].
AssumeG possesses n cyclic subgroups of order r. Then, nrG=nϕr, where nrG and ϕr denote the number of elements of order r and Euler function of r, respectively. Furthermore, if n represents the number of cyclic subgroups in G of maximal order k, then MG=nϕk.According to Lemma1, we easily follow the result as given by computations.Lemma 2.
IfMG=20, then possible values of n, k=kG, and ϕk are shown as follows:Lemma 3. [7, Lemma 4].
If a nonabelian 2-groupG has the exponent 4, then G contains at least G−CGx/2 elements with order 4 for x∈G\ZG with x=2.Lemma 4.
If a 2-groupG has the exponent 4 and MG=20, then G≅C23⋊C4,D8×C4,Q16×C2,D8∗Q8,G1,G2, or G3.Proof.
LetG=2t. If G is abelian, then 2t−1≤20 by [8, Lemma 2.5], and so t is no more than 5 and G is 32. If G is nonabelian of the exponent 4 and x∈ZG with x=2, we claim G≤64. If G≥128, then there is K<G satisfying K≅C2×C2×C2×C2×C4, because x∈ZG. Clearly, n4K is 32, and it is impossible. Hence, G≤64. If G is nonabelian with y∈G\ZG and y=2, we have G≤64 by Lemma 3. By [13], there is no group of order 64 which is satisfied with the assumption of our lemma. If G=32, then G≅C23⋊C4,D8×C4,Q16×C2,D8∗Q8,G1,G2, or G3 by [14].Lemma 5.
If a 2-groupG has the exponent 8 and G≥64, then 8 divides n8G.Proof.
Obviously,G is noncyclic. Setting H=A∩B for two different maximal subgroups AandB in G, then H⊴G and G/H≅C2×C2. So A∪B∪C=G and A∩B∩C=H, where C is a 3-maximal subgroup in G satisfying H≤C. Since ϕ8=4 divides n8H, n8G=n8A\H+n8B\H+n8C\H+n8H≡n8A+n8B+n8C (mod 8).
We show the result by induction onG.
IfG=64, then n8G is divisible by 8 by applying a result in [13]. If G>64, then 8 divides n8A, n8B, and n8C by induction, and so 8 divides n8G.


## 3. Proof of Theorem

Proof.
SinceMG=20, we have that k≠2,5 and 11 by [3, Lemma 6] and k≠22 by [3, Corollary 2]. In the following, we discuss the cases for the other values of k.Case 1.
k=3. G is either a 3-group or {2, 3}-group. If the former holds, then its exponent is 3. Using [15, Theorem 3.8.8], MG is 9s+4 (s a integer), this contradicts the hypothesis. If the later holds, then πeG=1,2,3, the set of prime divisors of G. By applying a result in [1, Theorem], G=N⋊Q is a Frobenius group, where N≅C3t,Q≅C2 or N≅C22t,Q≅C3. Suppose Q≅C2. Then, N is elementary abelian. It follows a contradiction by using [15, Theorem 3.8.8]. If Q is isomorphic to C3, then it has two elements with order 3, this contradicts the hypothesis. Thus, k≠3.Case 2.
k=4. Now G is a 2-group. Using Lemma 4, G≅E8⋊C4,D8×C4,Q16×C2,D8∗Q8,G1,G2, or G3. Thus, (1) holds.Case 3.
k=6. By a result of [3, Lemma 8], G=2α3β5γ, where α,β≠0 and 0≤γ≤1. If γ=1, then G is a C55-group (its centralizer of every 5-element is a 5-group). If G is nonsolvable, then there exists 1≤H⊴K⊴G satisfying H⊴G; G/K are π1-groups, K/H is simple, and G/K must divide OutK/H using [4, Theorem 2.1]. So K/H≅A5, A6, or U42. But U42 processes elements of order 6, and their indices of its maximal subgroups are 27, 36, 40, and 45. Hence, the index of normalizer in U42 of the arbitrary cyclic subgroup in U42 of order 6 is not less than 27. So U42 has more than 15 subgroups of order 6. It leads to G that has more than 20 elements of order 6, and so it is impossible. Thus, K/H≠U42. Hence, K/H may be A5 or A6. Now, we have G/K≤2. Thus, G/H≅A5,A6,S5, or S6. However, S6 contains elements with order 6 surpassing 20; this induces that G/H is not isomorphic to S6. If H=1, then G≅S5 since S5 has 20 elements of order 6. Thus, (2.1) follows.
Assume thatG/H≅A5,A6, or S5 and H≠1. Let x∈G with x=6. Then, 5∤NGx. Otherwise, NGx has an element of order 30 since k=6, a contradiction. Hence, 5G:NGx. As G has only 10 subgroups with order 6, it makes G:NGx=5 or 10. If 6H and H=2u3v, then any 2-element in H is of order 2 and any 3-element in H is of order 3. Hence, H has 2u−13v−1 elements of order 6. Note that H⊴G and G is a C55-group, and it follows easily its number whose elements with orders 2, 3, or 6 in H are multiples of 5. Hence, 2u−1=5s and 3v−1=5t for sandt, and so H has 25st elements of order 6. However, MG=20, and this is impossible. Thus, we may assume H=2u or 3v.
Suppose firstH=2u. Then, G=2u+2⋅3⋅5 or 2u+3⋅32⋅5 or 2u+3⋅3⋅5 since G/H≅A5 or A6 or S5. Since CGx has at most 20 elements with order 6, CGx=2r⋅3 or 2r⋅32. Hence, S2 is 2r, 2r+1, or 2r+2, where S2∈Syl2G. As CGx has no element with order 4, we have r≤4. Otherwise, CGx contains more than 20 elements with order 6, and it is impossible. So H≤24, ZH is 2t (1≤t≤4). Furthermore, 5H−1 and 5ZH−1. Hence, H and ZH are 16 and G=26⋅3⋅5 or 27⋅32⋅5 or 27⋅3⋅5 and CGx=24⋅3 or 24⋅32 or 24⋅3. Thus, CGx has at least 30 elements with order 6, and it contradicts the hypothesis.
Suppose nextH=3v. Then, G=22⋅3v+1⋅5 or 23⋅3v+2⋅5 or 23⋅3v+1⋅5 since G/H≅A5 or A6 or S5. It is well known that A6 and S5 have elements of order 4, so CGx has no Sylow 2-subgroup of G when G≅A6 or S5. Thus, we can let CGx=2⋅3r or 4⋅3r. Since G:NGx=5 or 10, we know that S3 is 3r and 3r+1, where S3∈Syl3G. It means that v≤r. By hypothesis, r≤3 and so v≤3. Thus, H|≡1 (mod 5), a contradiction.
LetG be solvable with G=2α⋅3β⋅5γ, where α,β≥1 and γ≥0. If γ≥1, then G 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 H is a 5-Hall subgroup, then G contains a normal subgroup with order 5. Hence, G contains an element with order 15, and it contradicts the hypothesis. If the former holds and H is a 2,3-Hall subgroup, then G has 2u−13v−1 elements of order 6 by the nilpotency of H. Hence, we have that u=4 and v=1, and it contradicts the hypothesis. If the later holds, then there is a series H⊴K⊴G satisfying K/H is a 5-group by [4, Theorem 2.3]; G/H and K are both Frobenius groups whose kernels are K/H and H, respectively. Therefore, G/H contains an element with order 15 when 3 divides G/H, and it contradicts the hypothesis. Hence 3H, N⊴G with N=3. Choosing y∈G with y=5 and considering y acts on N by conjugation, then G has an element with order 15, and it contradicts the hypothesis. This implies that G cannot be a 2-Frobenius group. So γ=0 and G=2u⋅3v. By Lemma 2, it has 10 subgroups with order 6, and G:NGx=1, 2, 3, 4, 6, 8, or 9 for any x with order 6. Let CGx=2r⋅3s. Since G has no element with order 9, any 3-element y in CGx which commute x satisfies y=3, and it leads to xy=6. Thus, s≤2. In addition, it follows from CGx has no element with order 4 and arbitrary 2-element z satisfying z,x=1 that r≤3 by MG=20.
IfG:NGx=8 or 9, then it has y∈G with y=6 satisfying G:NGx being 1 or 2. In total, G has 10 subgroups with order 6, which implies G has an element with order 6 satisfying 1≤G:NGx≤4. Furthermore, we obtain that G is a divisor of 2CGx or 4CGx. Hence, G25⋅32. Thus, (2.2) follows.Case 4.
k=8. If G=32, then n8G=20 is divisible by 8 by [14], a contradiction. If G is a 2-group with G≥64, then there is no group which satisfies our assumption using Lemma 5. If G is a 2,5-group, then G23α+3⋅5. Choosing x∈G with x=8, we get G:NGx=1,2,4, or 5, as G has precisely 5 cyclic subgroups with order 8. If y∈G with y=8 satisfying G:NGy=2 or 4, then G has another element z with z=8 satisfying G:NGz=1. It implies that G must have an element x with x=8 satisfying G:NGx=1 or 5. If G:NGz=1, then 5NGx. As Autx=4, we have 5CGx, and it leads to G that has an element with order 40, which contradicts the hypothesis. Thus, G:NGx=5. So all cyclic subgroups with order 8 are conjugate in G and so do their centralizers. Let C=CGx. Then, C=8,16, or 32 by [8, Lemma 2.5]. If C=32, then C has precisely 16 elements of order 8 by [8, Lemma 2.5]. Choosing y∈G\C with y=8, then CGy∩C has 4 or 8 elements of order 8 and so CGy∪C has 28 or 24 elements of order 8. Assume that CGy∩C has 4 elements with order 8, and let z∈CGy∩C with z=8. Then, z is not in x, z is not in y, and CGz is conjugate to C. As C∩CGy is abelian, CGy∩C≤CGz. Hence, C∪CGy∪CGz contains 16×3−8=40 elements of order 8, and it contradicts the hypothesis. Assume that CGy∩C has 8 elements with order 8, and let z∈CGy∩C be any element with z=8. Then, C∪CGy∪CGz has 16×3−16=32 elements with order 8. Let t∈G\C∪CGy∪CGz be any element with t=8. Then, we get that C∪CGy∪CGz∪CGt contains 16×4−8×6+8×4−8=40 elements with order 8 using the same arguments, and it contradicts the hypothesis. If C=16, then C is abelian and it contains 8 elements with order 8. In fact, we get easily that CGt∩C contains no element of order 8 for any t∈G\C with t=8. Thus, its number whose elements in G with order 8 are divided by 8, and it is impossible. If C=8, then C≅C8 and C contains 4 elements with order 8. As all centralizers of elements with order 8 are conjugate, G:CGx=5=G:NGx. By using a theorem of Burnside, G≅C5⋊C8, which is impossible since k=8.Case 5.
k=10. By [3, Lemma 8], we can set G=2α⋅5β, where α,β>0. As G has 5 cyclic subgroups with order 10, we have G:NGx is 1,2,4, or 5 if x∈G with x=8.
IfG:NGx=2 or 4, then G has y with y=8 satisfying G:NGy=1 or 2. Let CGx=2u⋅5v. Then, u≤2 and v=1 as CGx has at most 20 elements with order 10. Hence, the order of any 5-element of G is 5. It is easy to see that G cannot have 20 elements with order 10, and it is impossible.
IfG:NGx=5, then all cyclic subgroups with order 10 are conjugate in G. Let CGx=2u⋅5v. Then, u≤2 and v=1. We always have that NG〈x〉/CG〈x〉 divides 4 and CGx is a 2,5-group. Hence, G24⋅52. Let S5∈Syl5G. If S5⋬G, then G:NGS5=16 using Sylow’s Theorem. Thus, u=2, G=24⋅52, and NGx/CGx=4. Clearly, O5G=5. If O2G=1, then CGO5G≤O5G, a contradiction. Hence, O2G≠1. If O2G=4, then G contains at least 128 elements with order 10, and it contradicts the hypothesis. If O2G=2, then G has at least 64 elements with order 10, and it is impossible as well. Hence S5⊴G and G/CGS54. If 2CGS5, then G contains at least 24 elements with order 10, and it is impossible. Hence, CGS5=S5 and S2≤4. If G=22⋅52, then it has 5 Sylow 2-subgroups in G using Sylow’s Theorem. Thus, G has 60 elements with order 10, and it contradicts the hypothesis. Therefore G=2⋅52. If G=x×y⋊z, where x=y=5 and z=2, then CGz=10, and G has 20 elements with order 10. Thus, (3) holds.Case 6.
k=12. Let x∈G with x=12. Then, CGx=2u⋅3v. By [8, Lemma 2.5], CGx has at least 2u−1 elements with order 4. In addition, its 3-elements in CGx are of order 3. So 2⋅2u−1+23v−1−4≤20 using [8, Lemma 2.5] and the hypothesis. Thus, v=2 and 2≤u≤3 or v=1 and 2≤u≤4. Using Lemma 2, G has 5 cyclic subgroups with order 12, and so 1≤G:NGx≤5.
IfG:NGx=5, we let C=CGx,S2∈Syl2C, and S3∈Syl3C. Clearly, x≤ZC. Assume 32C. Then, x<C. If C is abelian, then C=22⋅32 and C has precisely 16 elements with order 12. If we choose y in G∖C with y=12, then CGy has 16 elements with order 12. And C∪CGy has at least 28 elements with order 12, and it contradicts the hypothesis. Assume that C is nonabelian with C>36. Furthermore, we get that C=72. Since S3≤NCS3 and x3≤NCS3, we know that S3 is normal in C. Then, CCFC=S3×O2C since C is solvable. Let y∈S2\ZC with y=4. Then, y acts fixed-point-freely on O3C\ZC. Therefore, 4O3C, and it contradicts the hypothesis. Hence, 32 does not divide C, S3=3, and CGx=S2×S3. By the hypothesis, CGx has at most 20 elements of order 12. Hence, S2 has at most 10 elements with order 4. If u=3, then S2=C4×C2 and CGx has 8 elements with order 12. Let y∈CGx and y=12. Then, C∩CGy has no element with order 12. If not, let z∈C∩CGy with z|=12. Then, CGy≤CGz and C≤CGz as C and CGy are both abelian. Hence, CGy=CGz=C, a contradiction. If u=4 and S2 is abelian with type (4, 2, 2), then S2 contains 8 elements of order 4. It also gets a contradiction similar to case u=3. If S2 is abelian with type (4, 4), then S2 contains 12 elements with order 4 and CGx must have 24 elements with order 12, and it contradicts the hypothesis. If u=4 and S2 is nonabelian, then S2≅C4×C2⋊C2. Now S2 has 8 elements with order 4. Let y∈G∖C with y|=12. Then, C∩CGy has at least 4 elements with order 12. If C∩CGy has exactly 4 elements with order 12, then C∩CGy must contain 28 elements of order 12, and it contradicts the hypothesis. So C∩CGy has at least 8 elements with order 12. Thus, C∩CGy is 24, and C∩CGy≅C12×C2. Let z∈C∩CGy with z=12. Then, z is neither in x nor in y and CGz is conjugate to C. As C∩CGy is abelian, C∩CGy≤CGz. Hence, C∪CGy∪CGz contains 32 elements with order 12, and it also contradicts the hypothesis. Thus, u=2 and C=〈x. Therefore, 22⋅3⋅5nG and G24⋅3⋅5. It is noted that G is solvable since the simple groups A5 and A6 have no element with order 12. Thus, (4.1) holds.
IfG:NGx=3 or 4, then there exists y∈G with y=12 satisfying G:NGx=1 or 2 as G has 5 cyclic subgroups with order 12 and by using Lemma 2. Noting 5∤G, if there is a∈G with a=12 satisfying G:NGa=1, then a is normal in G, and hence both a3 and a4 are normal in G. Let S3′∈Syl3G. Then, S3′≤CGa. If S3′=3, then S3′⊲¯G, and hence, n3G=2. Thus, 20 = MG = n12G = n3G⋅n4CGS3′ = 2n4CGS3′. So n4CGS3′=10, which contradicts the fact that its number whose elements with order 4 in CGS3′ is always divisible by 4 by using [3, Lemma 9]. If S3′=32 and P3⋬G, then G has at least 33 elements with order 3, so G has at least 66 elements with order 12, a contradiction. If S3′=32 and S3′⊲¯G, then 20=MG=n12G=n3G⋅n4CGS3′=8n4CGS3′, which is impossible. Hence, there must exist an element a of G with a=12 satisfying G:NGa=2. Therefore, G27⋅32. Thus, (4.2) holds.Case 7.
k=25. Let S5∈Syl5G. Then, S5≥25. If S5>25, then S5 may be noncyclic, and its number whose elements with order 25 exceeds 20, a contradiction. Thus, S5=25 and S5⊴G. Since G/CGS5≲AutC25≅C20 , we have (5) holds.Case 8.
k∈33,44,50,66. Let x∈G with x=k. Then CGx=x⊴G, G/x≲AutCk≅C20, which implies that G≅Ck or G is metacyclic. Hence, G≅Ck or G=a,bak=1,bl=at,bab−1=ar, where l20. 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 groupsG in which its number whose elements of maximal order of G is 20. The main result enriches the research of finite group theory and provides theoretical guidance for the development of various disciplines.



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