MPEMathematical Problems in Engineering1563-51471024-123XHindawi10.1155/2020/65856476585647Research ArticleFinite Groups Which Have 20 Elements of Maximal OrderHanZhangjia1XieLongjiang1https://orcid.org/0000-0001-5712-3122GuoPengfei23DaiYing1School of Applied MathematicsChengdu University of Information TechnologyChengdu 610225Chinacuit.edu.cn2School of Mathematics and StatisticsHainan Normal UniversityHaikou 571158Chinahainnu.edu.cn3Key Laboratory of Data Science and Intelligence Education (Hainan Normal University)Ministry of EducationHaikou 571158Chinameb.gov.tr202015920202020270420201506202015920202020Copyright © 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.

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

National Natural Science Foundation of China11661031Natural Science Foundation of Hainan Province119MS039
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 AB denotes the center product of groups A and B. For positive integers m and n, Cnm=a1×a2××am, where ai1im 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,dba1=ab,dcd1=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

Set Gr=xGx=r with r1. 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 fixed MG and gave some meaningful results (for example, , etc.). In particular, Chen and Shi  classified groups with MG=30. Jiang and Shao  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 that G is satisfied with MG=20 and k=kG as mentioned above. Then,

If k=4, then GC23C4,D8×C4,Q16×C2,D8Q8,G1,G2, or G3

If k=6, then

G=S5, the symmetric group of degree 5

G=2α3β with α2,5 and β1,2

If k=10, then G=C5×C5C2 and CGC2=10

If k=12, then

2235G2435

G2732

If k=25, then G=C25K is a Frobenius group, where KC20

If k33,44,50,66, then GCk or G is metacyclic and G=a,bak=1,bl=at,bab1=ar, where l20.

Corollary 1.

If G satisfying MG=20 is A5-free, then G is solvable.

2. PreliminariesLemma 1. [12, Lemma <xref ref-type="statement" rid="lem1">1</xref>].

Assume G 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 Lemma 1, we easily follow the result as given by computations.

Lemma 2.

If MG=20, then possible values of n, k=kG, and ϕk are shown as follows:

Lemma 3. [7, Lemma 4].

If a nonabelian 2-group G has the exponent 4, then G contains at least GCGx/2 elements with order 4 for xG\ZG with x=2.

Lemma 4.

If a 2-group G has the exponent 4 and MG=20, then GC23C4,D8×C4,Q16×C2,D8Q8,G1,G2, or G3.

Proof.

Let G=2t. If G is abelian, then 2t120 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 xZG with x=2, we claim G64. If G128, then there is K<G satisfying KC2×C2×C2×C2×C4, because xZG. Clearly, n4K is 32, and it is impossible. Hence, G64. If G is nonabelian with yG\ZG and y=2, we have G64 by Lemma 3. By , there is no group of order 64 which is satisfied with the assumption of our lemma. If G=32, then GC23C4,D8×C4,Q16×C2,D8Q8,G1,G2, or G3 by .

Lemma 5.

If a 2-group G has the exponent 8 and G64, then 8 divides n8G.

Proof.

Obviously, G is noncyclic. Setting H=AB for two different maximal subgroups A and B in G, then HG and G/HC2×C2. So ABC=G and ABC=H, where C is a 3-maximal subgroup in G satisfying HC. Since ϕ8=4 divides n8H, n8G=n8A\H+n8B\H+n8C\H+n8Hn8A+n8B+n8C (mod 8).

We show the result by induction on G.

If G=64, then n8G is divisible by 8 by applying a result in . If G>64, then 8 divides n8A, n8B, and n8C by induction, and so 8 divides n8G.

3. Proof of TheoremProof.

Since MG=20, we have that k2,5 and 11 by [3, Lemma 6] and k22 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=NQ is a Frobenius group, where NC3t,QC2 or NC22t,QC3. Suppose QC2. 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, k3.

Case 2.

k=4. Now G is a 2-group. Using Lemma 4, GE8C4,D8×C4,Q16×C2,D8Q8,G1,G2, or G3. Thus, (1) holds.

Case 3.

k=6. By a result of [3, Lemma 8], G=2α3β5γ, where α,β0 and 0γ1. 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 1HKG satisfying HG; G/K are π1-groups, K/H is simple, and G/K must divide OutK/H using [4, Theorem 2.1]. So K/HA5, 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/HU42. Hence, K/H may be A5 or A6. Now, we have G/K2. Thus, G/HA5,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 GS5 since S5 has 20 elements of order 6. Thus, (2.1) follows.

Assume that G/HA5,A6, or S5 and H1. Let xG with x=6. Then, 5NGx. 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 2u13v1 elements of order 6. Note that HG 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, 2u1=5s and 3v1=5t for s and t, 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 first H=2u. Then, G=2u+235 or 2u+3325 or 2u+335 since G/HA5 or A6 or S5. Since CGx has at most 20 elements with order 6, CGx=2r3 or 2r32. Hence, S2 is 2r, 2r+1, or 2r+2, where S2Syl2G. As CGx has no element with order 4, we have r4. Otherwise, CGx contains more than 20 elements with order 6, and it is impossible. So H24, ZH is 2t (1t4). Furthermore, 5H1 and 5ZH1. Hence, H and ZH are 16 and G=2635 or 27325 or 2735 and CGx=243 or 2432 or 243. Thus, CGx has at least 30 elements with order 6, and it contradicts the hypothesis.

Suppose next H=3v. Then, G=223v+15 or 233v+25 or 233v+15 since G/HA5 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 GA6 or S5. Thus, we can let CGx=23r or 43r. Since G:NGx=5 or 10, we know that S3 is 3r and 3r+1, where S3Syl3G. It means that vr. By hypothesis, r3 and so v3. Thus, H|1 (mod 5), a contradiction.

Let G be solvable with G=2α3β5γ, where α,β1 and γ0. If γ1, 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 2u13v1 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 HKG 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, NG with N=3. Choosing yG 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=2u3v. 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=2r3s. 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, s2. In addition, it follows from CGx has no element with order 4 and arbitrary 2-element z satisfying z,x=1 that r3 by MG=20.

If G:NGx=8 or 9, then it has yG 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 1G:NGx4. Furthermore, we obtain that G is a divisor of 2CGx or 4CGx. Hence, G2532. Thus, (2.2) follows.

Case 4.

k=8. If G=32, then n8G=20 is divisible by 8 by , a contradiction. If G is a 2-group with G64, then there is no group which satisfies our assumption using Lemma 5. If G is a 2,5-group, then G23α+35. Choosing xG with x=8, we get G:NGx=1,2,4, or 5, as G has precisely 5 cyclic subgroups with order 8. If yG 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 yG\C with y=8, then CGyC has 4 or 8 elements of order 8 and so CGyC has 28 or 24 elements of order 8. Assume that CGyC has 4 elements with order 8, and let zCGyC with z=8. Then, z is not in x, z is not in y, and CGz is conjugate to C. As CCGy is abelian, CGyCCGz. Hence, CCGyCGz contains 16×38=40 elements of order 8, and it contradicts the hypothesis. Assume that CGyC has 8 elements with order 8, and let zCGyC be any element with z=8. Then, CCGyCGz has 16×316=32 elements with order 8. Let tG\CCGyCGz be any element with t=8. Then, we get that CCGyCGzCGt contains 16×48×6+8×48=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 CGtC contains no element of order 8 for any tG\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 CC8 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, GC5C8, which is impossible since k=8.

Case 5.

k=10. By [3, Lemma 8], we can set G=2α5β, where α,β>0. As G has 5 cyclic subgroups with order 10, we have G:NGx is 1,2,4, or 5 if xG with x=8.

If G:NGx=2 or 4, then G has y with y=8 satisfying G:NGy=1 or 2. Let CGx=2u5v. Then, u2 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.

If G:NGx=5, then all cyclic subgroups with order 10 are conjugate in G. Let CGx=2u5v. Then, u2 and v=1. We always have that NGx/CGx divides 4 and CGx is a 2,5-group. Hence, G2452. Let S5Syl5G. If S5G, then G:NGS5=16 using Sylow’s Theorem. Thus, u=2, G=2452, and NGx/CGx=4. Clearly, O5G=5. If O2G=1, then CGO5GO5G, a contradiction. Hence, O2G1. 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 S5G and G/CGS54. If 2CGS5, then G contains at least 24 elements with order 10, and it is impossible. Hence, CGS5=S5 and S24. If G=2252, 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=252. If G=x×yz, 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 xG with x=12. Then, CGx=2u3v. By [8, Lemma 2.5], CGx has at least 2u1 elements with order 4. In addition, its 3-elements in CGx are of order 3. So 22u1+23v1420 using [8, Lemma 2.5] and the hypothesis. Thus, v=2 and 2u3 or v=1 and 2u4. Using Lemma 2, G has 5 cyclic subgroups with order 12, and so 1G:NGx5.

If G:NGx=5, we let C=CGx,S2Syl2C, and S3Syl3C. Clearly, xZC. Assume 32C. Then, x<C. If C is abelian, then C=2232 and C has precisely 16 elements with order 12. If we choose y in GC with y=12, then CGy has 16 elements with order 12. And CCGy 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 S3NCS3 and x3NCS3, we know that S3 is normal in C. Then, CCFC=S3×O2C since C is solvable. Let yS2\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 yCGx and y=12. Then, CCGy has no element with order 12. If not, let zCCGy with z|=12. Then, CGyCGz and CCGz 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 S2C4×C2C2. Now S2 has 8 elements with order 4. Let yGC with y|=12. Then, CCGy has at least 4 elements with order 12. If CCGy has exactly 4 elements with order 12, then CCGy must contain 28 elements of order 12, and it contradicts the hypothesis. So CCGy has at least 8 elements with order 12. Thus, CCGy is 24, and CCGyC12×C2. Let zCCGy with z=12. Then, z is neither in x nor in y and CGz is conjugate to C. As CCGy is abelian, CCGyCGz. Hence, CCGyCGz contains 32 elements with order 12, and it also contradicts the hypothesis. Thus, u=2 and C=x. Therefore, 2235nG and G2435. It is noted that G is solvable since the simple groups A5 and A6 have no element with order 12. Thus, (4.1) holds.

If G:NGx=3 or 4, then there exists yG with y=12 satisfying G:NGx=1 or 2 as G has 5 cyclic subgroups with order 12 and by using Lemma 2. Noting 5G, if there is aG with a=12 satisfying G:NGa=1, then a is normal in G, and hence both a3 and a4 are normal in G. Let S3Syl3G. Then, S3CGa. If S3=3, then S3¯G, and hence, n3G=2. Thus, 20 = MG = n12G = n3Gn4CGS3 = 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 P3G, 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=n3Gn4CGS3=8n4CGS3, which is impossible. Hence, there must exist an element a of G with a=12 satisfying G:NGa=2. Therefore, G2732. Thus, (4.2) holds.

Case 7.

k=25. Let S5Syl5G. Then, S525. If S5>25, then S5 may be noncyclic, and its number whose elements with order 25 exceeds 20, a contradiction. Thus, S5=25 and S5G. Since G/CGS5AutC25C20 , we have (5) holds.

Case 8.

k33,44,50,66. Let xG with x=k. Then CGx=xG, G/xAutCkC20, which implies that GCk or G is metacyclic. Hence, GCk or G=a,bak=1,bl=at,bab1=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 groups G 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.

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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