# Finite Groups of Orderp2qr in which the Number of Elements of Maximal Order Is p3q∗

Authors: Qingliang Zhang; Liang Xu
Journal: Journal of Mathematics
 (2022)

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## Abstract

LetG be a finite group. We know that the order of G and the number of elements of maximal order in G are closely related to the structure of G. This topic involves Thompson’s conjecture. In this paper, we classify the finite groups of order p2qr in which the number of elements of maximal order is p3q, where p<q<r are different primes.


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## Body

## 1. Introduction

All groups considered in the present paper are finite. Letn be an integer. We denote by πn the set of all prime divisors of n. Let G be a finite group. Then πG is denoted by πG. And we denote by kG and mG the maximal order of elements in G and the number of elements of order kG in G, respectively. We write H char G if H is a characteristic subgroup of G. G=N×Q stands for the split extension of a normal subgroup N of G by a complement Q. By M≲G we denote M is isomorphic to a subgroup of G. And we denote by Zn a cyclic group of order n. All unexplained notations are standard and can be found in [1].As is well known, for a finite groupG, G and mG have an important influence on the structure of G. The authors in [2–4] proved that finite groups with mG=lp are soluble, where l=2, 4, or 18. In [5] it was proved that finite groups with mG=2p2 are soluble. The authors in [6, 7] gave a classification of the finite groups with mG=30 and mG=24. These studies are closely related to the following conjecture.


### 1.1. Thompson’s Conjecture

LetG be a finite group. For a positive integer d, define Gd=x∈G‖x|=d. If H is a soluble group, Gd=Hd for d=1,2,…, then G is soluble.In this paper we classify the finite groups of orderp2qr in which the number of elements of maximal order is p3q, where p<q<r are different primes. Our result is:


### 1.2. Main Theorem

Suppose thatG is a finite group satisfying G=p2qr and mG=p3q, where p<q<r are different primes. Then one of the following statements holds:(1)
G≅M<imesZr, where M is a group of order 4q. Moreover, M/Z2≲ Aut Zr, M/Z2<imesZr is a Frobenius group and r−1=8q.(2)
G≅Z4r⋊Zq. Moreover, Zq≲ Aut Z4r and r−1=4q.(3)
G≅H<imesZqr, where H is a group of order 4, q=3 and r=13 or q=5 and r=11. Moreover, H is isomorphic to a subgroup of Aut Zqr. Hence H≲Z2×Z12 or H≲Z4×Z10.(4)
G≅Z84.(5)
G≅Z2×Z2×Z15.(6)
G≅M<imesZqr, where M is a group of order 4, q=3 and r=13 or q=5 and r=11. Moreover, M/Z2 is isomorphic to a subgroup of Aut Zqr. Hence M/Z2≲Z2×Z12 or M/Z2≲Z4×Z10.(7)
G is a Frobenius group and G≅Z4q<imesZr. Moreover, r−1=8q.(8)
G≅A5.


## 1.1. Thompson’s Conjecture

LetG be a finite group. For a positive integer d, define Gd=x∈G‖x|=d. If H is a soluble group, Gd=Hd for d=1,2,…, then G is soluble.In this paper we classify the finite groups of orderp2qr in which the number of elements of maximal order is p3q, where p<q<r are different primes. Our result is:


## 1.2. Main Theorem

Suppose thatG is a finite group satisfying G=p2qr and mG=p3q, where p<q<r are different primes. Then one of the following statements holds:(1)
G≅M<imesZr, where M is a group of order 4q. Moreover, M/Z2≲ Aut Zr, M/Z2<imesZr is a Frobenius group and r−1=8q.(2)
G≅Z4r⋊Zq. Moreover, Zq≲ Aut Z4r and r−1=4q.(3)
G≅H<imesZqr, where H is a group of order 4, q=3 and r=13 or q=5 and r=11. Moreover, H is isomorphic to a subgroup of Aut Zqr. Hence H≲Z2×Z12 or H≲Z4×Z10.(4)
G≅Z84.(5)
G≅Z2×Z2×Z15.(6)
G≅M<imesZqr, where M is a group of order 4, q=3 and r=13 or q=5 and r=11. Moreover, M/Z2 is isomorphic to a subgroup of Aut Zqr. Hence M/Z2≲Z2×Z12 or M/Z2≲Z4×Z10.(7)
G is a Frobenius group and G≅Z4q<imesZr. Moreover, r−1=8q.(8)
G≅A5.


## 2. Preliminaries

We need the following lemmas to show the main theorem.Lemma 1. (see [8]).
LetG be a finite group. Then the number of elements whose orders are multiples of n is either zero, or a multiple of the greatest divisor of G that is prime to n.Lemma 2. (see [2]).
LetG be a finite group. We denote by Ai1≤i≤s a complete representative system of conjugate classes of cyclic subgroups of order kG, respectively. Then we have the following:(1)
mG=φkG∑ni, where φkG is the Euler function, ni=G:NGAi and 1≤i≤s.(2)
G=G:NGAiNGAi:CGAiCGAi, where 1≤i≤s.(3)
|NGAi:CGAi‖φkG, where 1≤i≤s.(4)
πCGAi=πAi, where 1≤i≤s.Lemma 3. (see [9]).
LetG be a soluble group of order mn, where m is prime to n. Then the number of subgroups of G of order m may be expressed as a product of factors, each of which (i) is congruent to 1 modulo some prime factor of m and (ii) is a power of a prime and divides the order of some chief factor of G.Lemma 4. (see [10]).
LetG be a finite simple group. If πG=3, then we call G a simple K3-group. If G is a simple K3-group, then G is isomorphic to one of the following groups: A5, A6, L27, L28, L217, L33, U33 or U42.


## 3. Proof of the Theorem

We know thatπG⊆πmG∪πkG by Lemma 1. Then r∈πkG. In the following, we discuss four cases.Case 1.
IfπkG=p,r, then kG=pr or p2r.
Suppose thatkG=pr. Since φkG=p−1r−1|mG=p3q by Lemma 2, we have p=2. Choose an arbitrary element x of order kG in G and let x=A. It is clear that xp∈ZCGA and so G has a Sylow r-subgroup Pr such that Pr≤ZCGA. Therefore Pr char CGA and it follows that Pr⊲¯NGA since CGA⊲¯NGA. Therefore NGA≤NGPr and thus |G:NGPr‖G:NGA. By Lemma 2 we get that |G:NGA‖2q. So |G:NGPr‖2q.
IfPr⋬G, then G:NGPr=G:NGA=2q by Sylow’s theorem. Consequently, we have 2q|n by Lemma 2, where n is the number of cyclic subgroups of order kG in G. Note that n=mG/φpr=8q/r−1, thus r−1=4 and so r=5. It follows that q=3. It is well known that a group of order 60 is isomorphic to A5, if it is not 5-closed. Therefore G is isomorphic to A5, which is a contradiction since A5 has no elements of order 10.
IfPr⊲¯G, then CGPr contains all the elements of order kG in G since A≤CGA≤CGPr. Note that Pr≤ZCGPr, thus CGPr=2αr, where 1≤α≤2. Moreover, CGPr=H×Pr by Schur–Zassenhaus’s theorem, where H is a group of order 2α. If α=2, then H is an elementary abelian group of order 4. Thus 3r−1=8q and it follows that r−1=8, which is a contradiction since r is a prime. If α=1, then r−1=8q. By Schur–Zassenhaus’s theorem we get that G≅M<imesZr, where M is a group of order 4q. It is obvious that M/Z2≲ Aut Zr and M/Z2<imesZr is a Frobenius group. Hence (1) holds.
Suppose thatkG=p2r. Since φkG=pp−1r−1|mG=p3q by Lemma 2, we get that p−1=1 and so p=2. Choose an arbitrary element x of order kG in G and let x=A. It is clear that xp2∈ZCGA and so G has a Sylow r-subgroup Pr such that Pr≤ZCGA. Therefore Pr char CGA and it follows that Pr⊲¯NGA since CGA⊲¯NGA. So NGA≤NGPr. Then |G:NGPr‖G:NGA. Note that |G:NGA|q by Lemma 2, thus G:NGPr=1 by Sylow’s theorem. It follows that Pr⊲¯G and CGPr contains all the elements of order kG in G. Furthermore, CGPr=4qαr, where 0≤α≤1.
Ifα=0, then CGPr≅Z4r. Therefore φ4r=8q and thus r−1=4q. Since CGPr⊲¯G, we obtain G≅Z4r⋊Zq by Schur–Zassenhaus’s theorem. Moreover, Zq≲ Aut Z4r. Hence (2) holds.
Ifα=1, then CGPr=G and so Pr≤ZG. Hence G has elements of order qr, but has no elements of order 2q since kG=4r is the maximal element order of G. Since kG=4r>qr, we get that q=3. Note that the Sylow 2-subgroup P2 of G is cyclic, then G has a normal 2-complement H. Obviously H≅Z3r and so the Sylow 3-subgroup P3 of G is normal in G. Since G has no elements of 6, we get that P2 acts fixed-point-freely on P3. Therefore P2P3−1, namely, 4|2, which is a contradiction.Case 2.
IfπkG=q,r, then kG=qr.
SinceφkG=q−1r−1|mG=p3q by Lemma 2, we get that q−1|p3. Since 2|q−1, we have 2|p3 and so p=2. Choose an arbitrary element x of order kG in G and let x=A. It is clear that ZCGA contains elements of order r, and so G has a Sylow r-subgroup Pr such that Pr≤ZCGA. Therefore Pr char CGA and it follows that Pr⊲¯NGA since CGA⊲¯NGA. So NGA≤NGPr. Then |G:NGPr‖G:NGA. Note that |G:NGA‖4 by Lemma 2, thus we get that |G:NGPr‖4. Since 2<q<r, we get that G:NGPr=1 by Sylow’s theorem. It follows that Pr⊲¯G. Then G is soluble. It follows that G:NGA=1 or 4 by Lemma 3. If G:NGA=4, then 4≡1 (mod q) by Lemma 3. Therefore q=3 and thus n=24/φkG=24/q−1r−1=4, where n is the number of cyclic subgroups of order kG in G. Then r=4, which is a contradiction since r is a prime. If G:NGA=1, then from φqr=q−1r−1=8q we get that q=3 and r=13 or q=5 and r=11. Furthermore, G≅H<imesZqr by Schur–Zassenhaus’s theorem, where H is a group of order 4. It is evident that the conjugate action of H on Zqr is faithful. Therefore H is isomorphic to a subgroup of Aut Zqr. Hence H≲Z2×Z12 or H≲Z4×Z10 and so (3) holds.Case 3.
IfπkG=p,q,r, then kG=p2qr or pqr.
IfkG=p2qr, then φp2qr=pp−1q−1r−1=p3q. Therefore p−1=1 and so p=2. Consequently, q−1/2⋅r−1/2=q. Since r−1/2>1, we have q−1/2=1 and so q=3. It follows that r=7. Hence G≅Z84 and thus (4) holds.
IfkG=pqr, then by φkG=p−1q−1r−1|p3q we know that p−1=1 and so p=2. Choose an arbitrary element x of order kG in G and let x=A. It is clear that ZCGA contains elements of order qr, and so G has a subgroup H of order qr such that H≤ZCGA. Therefore H char CGA and it follows that H⊲¯NGA since CGA⊲¯NGA. So NGA≤NGH. Then |G:NGH‖G:NGA. Note that |G:NGA=1, thus G:NGH=1 and so H⊲¯G. Therefore CGH contains all the elements of order kG in G and so CGH=2αqr, where 1≤α≤2.
Ifα=2, then H≤ZG. So G=K×H by Schur–Zassenhaus’s theorem. Obviously H is a non-cyclic group of order 4. Hence 3q−1r−1=8q and so q=3, r=5. Therefore G≅Z2×Z2×Z15. Hence (5) holds.
Ifα=1, then CGH≅Z2×H. So q−1r−1=8q. It follows that q=3 and r=13 or q=5 and r=11. Furthermore, G≅M<imesZqr by Schur–Zassenhaus’s theorem, where M is a group of order 4. It is evident that the kernel of the conjugate action of M on Zqr is isomorphic to Z2. Therefore M/Z2 is isomorphic to a subgroup of Aut Zqr. Hence M/Z2≲Z2×Z12 or M/Z2≲Z4×Z10. Thus (6) holds.Case 4.
IfπkG=r, then kG=r.
Sincer−1|p3q and 2|r−1, we get that p=2. We know that the number nr of Sylow r-subgroups of G is equal to 1, 2q, or 4q by Sylow’s theorem.
Ifnr=1, then the Sylow r-subgroup Pr is normal in G and r−1=8q. Moreover, G has an r-complement H of order 4q by Schur–Zassenhaus’s theorem. Note that the conjugate action of H on Pr is fixed-point-free, thus G is a Frobenius group with Frobenius kernel Pr and Frobenius complement H. Note that Pr≅Zr and H is a cyclic group, thus G≅Z4q<imesZr. Hence (7) holds.
Ifnr=4q, then 4qr−1=8q, which is impossible.
Ifnr=2q, then G is non-soluble by Lemma 3 and so G≅A5 by Lemma 4. Hence (8) holds.
Now the proof of the theorem is complete.



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*Source: 2294627-2022-03-18.xml*