# Finite Groups with the Same Order and the Same Number of Elements of Maximal Order as the Projective Special Linear GroupL2q

Authors: Qingliang Zhang
Journal: Journal of Mathematics
 (2022)

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## Abstract

Assume thatG is a finite group. It is widely known that G and the number of elements of maximal order in G have something to do with the structure of G. This subject is related to Thompson’s conjecture. In the present paper, we give a complete classification of the groups with the same order and the same number of elements of maximal order as L2q, where q=7,17.


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## Body

## 1. Introduction

The groups discussed in our paper are all finite. For some integerl, πl stands for the set of all prime divisors of l. For a group G, πG denotes πG. M char G implies that M is characteristic in G. πeG stands for xx∈G. Let snG=x∈Gx=n∈πeG. We denote by μG the largest element in πeG and define υG=x∈Gx=μG. We use nμG to represent the number of cyclic subgroups of order μG in G. Assume p∈πG. Then OpG means the maximal normal p-subgroup of G, and SylpG denotes the set of Sylow p-subgroups of G. G=K⋊L represents G which is a semidirect product of L with K. And Cn stands for a cyclic group of order n. Symbols that are not introduced are standard; readers may refer to [1].As everyone knows,G and υG have a certain impact on the structure of group G. In [2–4], it was shown that groups G with υG=kp are solvable, where k=4, 18, or 2 and p is a prime. In [5], it was shown that groups G with υG=2p2 are solvable. The authors in [6, 7] classified the groups G with υG=30 and 24. In [8, 9], it was proved that the Mathieu groups and some projective special linear groups can be characterized by a group order and set of numbers of elements with the same order. All above studies have a close relationship with the following conjecture.Thompson’s conjecture. For a finite groupH and some positive integer l, define Hl=x∈Hx=l. Suppose that G is a solvable group and Hl=Gl for l=1,2,⋯. Then H is also solvable.The problem we will consider is related to this conjecture too. For finite groupsH and G, suppose that Hl=Gl. Then we can see that H=G and υH=υG. Therefore, we want to think over the impact of G and υG on the structure of G. Our research method is somewhat similar to that in [10]. In the present paper, we classify the groups G satisfying G=L2q and υG=υL2q, where q=7,17. Our results are:Theorem 1.
Assume thatG is a group. If G=L27 and υG=υL27, then one of the following conclusions holds:
(i)
G≅M⋉C7. Furthermore, CMC7≅A4 and M≅A4×C2 or S4(ii)
G≅A4×C2×C7(iii)
G≅SL2F3×C7(iv)
G≅C2×C12×C7(v)
G≅C168(vi)
G≅C3⋉C7⋉P is a 2-Frobenius group. Furthermore, P is elementary abelian,P=8, P⊴G,G/P≅C3⋉C7, and πeG=1,2,3,6,7(vii)
G≅L27Theorem 2.
Assume thatG is a group. If G=L217 and υG=υL217, then one of the following conclusions holds:
(i)
G≅L⋉C153. Furthermore, L=16 and CLC153≅C2×C2(ii)
G≅L217Corollary 3.
Thompson’s conjecture holds for finite groups (i-vi) of Theorem1 and (i) of Theorem 2.


## 2. Preliminaries

Here we present some lemmas which will be useful in the sequel.Lemma 4 (see [11]).
Suppose thatG is a finite group and n is a positive integer. 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 5 (see [2]).
For a finite groupG, Bk1≤k≤t represent a complete representative system of conjugate classes of cyclic subgroups of order μG. Then all the following hold:
(1)
υG=φμGnμG=φμG∑nk, where φμG is Euler function, nk=G:NGBi, and 1≤k≤t(2)
G=CGBkNGBk:CGBkG:NGBk, where 1≤k≤t(3)
NGBk:CGBkφμG, where 1≤k≤t(4)
πCGBk=πBk, where 1≤k≤tLemma 6 (see [12]).
LetH be a finite group.
(1)
IfπeH=1,2,3, then H is a Frobenius group and H=K⋊M, where K≅C3t and M≅C2 or K≅C22t and M≅C3, and C3t stands for the direct product of t copies of C3(2)
IfπeH=1,2,3,4, then H=K⋊M. Furthermore, one of the following conclusions holds:(i)
expK=4, class of N is less than or equal to 2 and M≅C3(ii)
K=C22t and M≅S3, where C22t stands for the direct product of 2t copies of C2(iii)
K=C32t and M≅C4 or Q8 and H is a Frobenius group, where Q8 is the generalized quaternion groupLemma 7 (see [13]).
LetG be a finite group. Then
(1)
G=L27 and μG=μL27 if and only if G≅L27 or G≅C3⋉C7⋉P is a 2-Frobenius group; moreover, P is elementary abelian, P=8, P⊴G,G/P≅C3⋉C7, and πeG=1,2,3,6,7(2)
G=L217 and μG=μL217 if and only if G≅L217
We callG a simple K3-group if G is simple and πG=3. The next lemma classifies the simple K3-groups.Lemma 8 (see [14]).
Suppose thatG is a simple K3-group. Then G is isomorphic to one of the following groups: A5, A6, L27, L28, L217, L33, U33, and U42.Lemma 9 (see Frobenius [15]).
LetG be a finite group and n be a positive integer dividing G. Then nfn, where fn=g∈Ggn=1.Lemma 10 (see [16]).
LetG be a solvable group satisfying G=mn, and let m,n=1. 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 11.
No groupH of order 2α·32 satisfies μH=6 and υH=18, where 2≤α≤4.Proof.
Suppose thatα=3. If O3H=3, then Syl3H=4 by the theorem of Sylow. It follows that s3H=26. From Lemma 4 we obtain 8s3H+s6H, namely, 844, which is a contradiction. If O3H=9, then we can also get a contradiction similarly. If O3H=1, then 22≤s3H≤32, which is impossible since 8s3H+s6H and 31+s3H by Lemmas 4 and 9. Therefore, no group H of order 23·32 satisfies μH=6 and υH=18.
Similar to the above, we can show that no groupH of order 22·32 satisfies μH=6 and υH=18.
Suppose thatα=4. Then by Lemmas 4 and 9 we can get that s3H=62 by simple calculation. Hence, Syl3H=16 by the theorem of Sylow. It follows that NHP3=CHP3, where P3∈Syl3H. Therefore, H is 3-nilpotent by Burnside’s theorem, and so the Sylow 2-subgroup of H is normal in H. Note that πeH⊆1,2,3,4,6, thus 1+s2H+s3H+s4H+s6H=H. Namely, 96=144, which is impossible. Hence, no group H of order 24·32 satisfies μH=6 and υH=18.


## 3. Proof of the Results

Proof of Theorem 1.
By [17] we get that G=23·3·7 and υG=48. From Lemma 4, we obtain πG⊆πυG⋃πμG, and so 7∈πμG. Suppose that G is solvable, and we discuss four cases in the following.Case 1.
Suppose thatπμG=3,7. Then μG=21. Assume that a∈G is an arbitrary element satisfying a=21 and assume that a=B. It is evident that a3 belongs to ZCGB, which implies that P7≤ZCGB for some P7∈Syl7G and thus P7 char CGB. From CGB⊴NGB, we get that P7⊴NGB and thus NGB≤NGP7. It follows that G:NGP7 divides G:NGB. Since G:NGB=υG/φμG=4 by Lemma 5, we have G:NGP7=1 by the theorem of Sylow, and so P7⊴G. It is obvious that B≤CGB≤CGP7; thus, s21CGP7=υG. Since P7≤ZCGP7, by the theorem of Schur-Zassenhaus, we obtain that CGP7=H×P7. Furthermore, H=2α·3, s3H=8, and πeH=1,2,3, where 1≤α≤3. Since G/CGP7AutP7, we have 2≤α≤3.
Assume thatα=2. Then H≅A4 by (1) of Lemma 6. Since H char CGP7⊴G, we obtain H⊴G. Since G=M⋉P7 by the theorem of Schur-Zassenhaus and H⊴M, we have M≅A4×C2 or S4 by [18]. Note that P7≅C7 and CMP7=H; thus, (i) holds.
Assume thatα=3. Then H=24, which is a contradiction since such group H does not exist by (1) of Lemma 6.Case 2.
Suppose thatπμG=2,7. Then μG=14, 28, or 56.
Assume thatμG=56. Then from Lemma 5, we obtain that nμG=υG/φμG=48/φ56=2, which is impossible obviously.
Suppose thatμG=28. Assume that a∈G is an arbitrary element satisfying a=28, and assume that a=B. It is evident that a4 belongs to ZCGB, which implies that P7≤ZCGB for some P7∈Syl7G. Then similar to Case 1, we can get P7⊴G and s28CGP7=υG. Since P7≤ZCGP7, by the theorem of Schur-Zassenhaus, we obtain CGP7=H×P7, where H is a group satisfying H2α·3β, 2≤α≤3, and 0≤β≤1. Furthermore, μH=4 and s4H=8. It is not difficult to find that such group H does not exist through simple calculation.
Assume thatμG=14. Then we can also get a contradiction similarly.Case 3.
Suppose thatπμG=2,3,7. Then μG=42, 84, or 168.
Assume thatμG=42. Assume that a∈G is an arbitrary element satisfying a=42, and assume that a=B. It is evident that a6 belongs to ZCGB, which implies that P7≤ZCGB for some P7∈Syl7G. Then similar to Case 2, we can get P7⊴G and s42CGP7=υG. Since P7≤ZCGP7, by the theorem of Schur-Zassenhaus, we obtain CGP7=H×P7. Furthermore, H=2α·3, μH=6, and υH=8, where 1≤α≤3. It follows that H≅A4×C2 or SL2F3 by [18]. Therefore, G≅A4×C2×C7 or G≅SL2F3×C7. Thus, (ii) and (iii) hold.
Assume thatμG=84. Then similar to the above proof of μG=42, we can get that G≅C2×C12×C7. Thus, (iv) holds.
Assume thatμG=168. Then φμG=48=υG, and so G≅C168. Thus, (v) holds.Case 4.
Suppose thatπμG=7. Then μG=7. By (1) of Lemma 7, we obtain that G≅C3⋉C7⋉P is a 2-Frobenius group. Furthermore, P is elementary abelian, P=8, P⊴G, G/P≅C3⋉C7, and πeG=1,2,3,6,7. Hence, (vi) holds.
Assume thatG is nonsolvable. Then G≅L27 by Lemma 8, and so (vii) holds.Proof of Theorem 2.
By [17] we get that G=24·32·17 and υG=288. From Lemma 4, πG⊆πυG⋃, and it follows that 17∈πμG. Now we discuss the following four cases.Case 1.
Suppose thatπμG=2,17. Then μG=2α·17, where 1≤α≤2 since φμGυG by Lemma 5.
Assume thatα=1. Assume that a∈G is an arbitrary element satisfying a=34, and assume that a=B. It is clear that a2 belongs to ZCGB, which implies that P17≤ZCGB for some P17∈Syl17G. Then similar to the above, we have G:NGP17 divides G:NGB. Since G:NGB72 by Lemma 5, we obtain G:NGP17=1 or 18 by the theorem of Sylow.
Assume thatG:NGP17=1. Then P17⊴G. Since B≤CGB≤CGP17, we get that s34CGP17=υG. Furthermore, CGP1724·17. Since P17≤ZCGP17, by the theorem of Schur-Zassenhaus, we obtain CGP17=H×P17, where H is a group satisfying H16. Hence, s2H=288/16=18, which is impossible obviously.
Assume thatG:NGP17=18=2·32. Then by Lemma 10, it follows that 2≡1(mod 17) since G is solvable by Lemmas 7 and 8, which is a contradiction.
Assume thatα=2. Assume that a∈G is an arbitrary element satisfying a=68 and assume that a=B. It is clear that a4 belongs to ZCGB, which implies that P17≤ZCGB for some P17∈Syl17G and so P17 char CGB. From CGB⊴NGB, we have P17⊴NGB and thus NGB≤NGP17. It follows that G:NGP17 divides G:NGB. Since G:NGB36 by Lemma 5, we obtain G:NGP17=1 or 18 by the theorem of Sylow. If G:NGP17=1, then similarly to the proof of Case 1 of Theorem 1 and by (2) of Lemma 6, we can get a contradiction. If G:NGP17=18, then similarly to the above proof of α=1, we can get a contradiction.Case 2.
Suppose thatπμG=3,17. Then μG=3·17 or 32·17.
Assume thatμG=3·17. Then we can get a contradiction similarly to Case 1 of Theorem 1.
Assume thatμG=32·17. Assume that a∈G is an arbitrary element satisfying a=32·17 and assume that a=B. It is clear that a9 belongs to ZCGB, which implies that P17≤ZCGB for some P17∈Syl17G. Then similar to the above, we have P17⊴G. From B≤CGB≤CGP17, we obtain s32·17CGP17=υG. Since G/CGP17AutP17 and P17≤ZCGP17, we get that CGP17=H×P17 by the theorem of Schur-Zassenhaus, where H≤G and H=2α·321≤α≤4. Thus, s9H=18. Therefore, Syl3H=3, which is contradict to the theorem of Sylow.Case 3.
Assume thatπμG=2,3,17. Then μG=2·3·17 or 2·32·17 since φμGυG=288 by Lemma 5.
Assume thatμG=2·3·17. Assume that a∈G be an arbitrary element satisfying a=2·3·17 and assume that a=B. It is evident that a6 belongs to ZCGB, which implies that P17≤ZCGB for some P17∈Syl17G. Then similar to the above, we can get that P17⊴G. Since B≤CGB≤CGP17 and G/CGP17AutP17, we have CGP17=2α·32·17, where 1≤α≤4. Note that P17≤ZCGP17; thus, by the theorem of Schur-Zassenhaus, we obtain that CGP17=H×P17, where H is a group of order 2α·32 satisfying μH=6 and υH=18. Obviously, α≠1. Now we get a contradiction since such group H does not exist by Lemma 11.
Assume thatμG=2·32·17. Assume that a∈G is an arbitrary element satisfying a=2·32·17, and assume that a=B. Since a2 belongs to ZCGB, we obtain that there exists a cyclic subgroup M of order 32·17 of G satisfying M≤ZCGB. So M char CGB. From CGB⊴NGB we obtain M⊴NGB. So NGB≤NGM. Consequently, G:NGM divides G:NGB.
From Lemma5 we obtain that nμG=υG/φμG=3. Hence, G:NGB≤3. Since G is solvable by Lemmas 7 and 8, we have G:NGM=1 by Lemma 10. Consequently, M⊴G and hence G=L⋉M by the theorem of Schur-Zassenhaus, where L≤G and L=16. Since B≤CGB≤CGM, we get that s306CGM=υG. Consequently, CLM≅C2×C2. Note that M≅C153; thus, (i) holds.Case 4.
Suppose thatπμG=17. Then μG=17. Therefore G≅L217 by (2) of Lemma 7. Hence, (ii) holds.Proof of Corollary 3.
It is evident by Theorems1 and 2.
Now the proofs of our results are complete.



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