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

Abstract

Assume that is a finite group. It is widely known that and the number of elements of maximal order in have something to do with the structure of . 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 , where .

1. Introduction

The groups discussed in our paper are all finite. For some integer , stands for the set of all prime divisors of . For a group , denotes . char implies that is characteristic in . stands for . Let . We denote by the largest element in and define . We use to represent the number of cyclic subgroups of order in . Assume . Then means the maximal normal -subgroup of , and denotes the set of Sylow -subgroups of . represents which is a semidirect product of with . And stands for a cyclic group of order . Symbols that are not introduced are standard; readers may refer to [1].

As everyone knows, and have a certain impact on the structure of group . In [2–4], it was shown that groups with are solvable, where , 18, or 2 and is a prime. In [5], it was shown that groups with are solvable. The authors in [6, 7] classified the groups with 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 group and some positive integer , define . Suppose that is a solvable group and for . Then is also solvable.

The problem we will consider is related to this conjecture too. For finite groups and , suppose that . Then we can see that and . Therefore, we want to think over the impact of and on the structure of . Our research method is somewhat similar to that in [10]. In the present paper, we classify the groups satisfying and , where . Our results are:

Theorem 1. Assume that is a group. If and , then one of the following conclusions holds: (i). Furthermore, and or (ii)(iii)(iv)(v)(vi) is a 2-Frobenius group. Furthermore, is elementary abelian,, , and (vii)

Theorem 2. Assume that is a group. If and , then one of the following conclusions holds: (i). Furthermore, and (ii)

Corollary 3. Thompson’s conjecture holds for finite groups (i-vi) of Theorem 1 and (i) of Theorem 2.

2. Preliminaries

Here we present some lemmas which will be useful in the sequel.

Lemma 4 (see [11]). Suppose that is a finite group and is a positive integer. Then the number of elements whose orders are multiples of is either zero or a multiple of the greatest divisor of that is prime to .

Lemma 5 (see [2]). For a finite group , represent a complete representative system of conjugate classes of cyclic subgroups of order . Then all the following hold: (1), where is Euler function, , and (2), where (3), where (4), where

Lemma 6 (see [12]). Let be a finite group. (1)If , then is a Frobenius group and , where and or and , and stands for the direct product of copies of (2)If , then . Furthermore, one of the following conclusions holds:(i)exp, class of is less than or equal to 2 and (ii) and , where stands for the direct product of copies of (iii) and or and is a Frobenius group, where is the generalized quaternion group

Lemma 7 (see [13]). Let be a finite group. Then (1) and if and only if or is a 2-Frobenius group; moreover, is elementary abelian, , , and (2) and if and only if We call a simple -group if is simple and . The next lemma classifies the simple -groups.

Lemma 8 (see [14]). Suppose that is a simple -group. Then is isomorphic to one of the following groups: , , , , , , , and .

Lemma 9 (see Frobenius [15]). Let be a finite group and be a positive integer dividing . Then , where .

Lemma 10 (see [16]). Let be a solvable group satisfying , and let . Then the number of subgroups of of order may be expressed as a product of factors, each of which (i) is congruent to 1 modulo some prime factor of and (ii) is a power of a prime and divides the order of some chief factor of .

Lemma 11. No group of order satisfies and , where .

Proof. Suppose that . If , then by the theorem of Sylow. It follows that . From Lemma 4 we obtain , namely, , which is a contradiction. If , then we can also get a contradiction similarly. If , then , which is impossible since and by Lemmas 4 and 9. Therefore, no group of order satisfies and .
Similar to the above, we can show that no group of order satisfies and .
Suppose that . Then by Lemmas 4 and 9 we can get that by simple calculation. Hence, by the theorem of Sylow. It follows that , where . Therefore, is 3-nilpotent by Burnside’s theorem, and so the Sylow 2-subgroup of is normal in . Note that , thus . Namely, , which is impossible. Hence, no group of order satisfies and .

3. Proof of the Results

Proof of Theorem 1. By [17] we get that and . From Lemma 4, we obtain , and so . Suppose that is solvable, and we discuss four cases in the following.

Case 1. Suppose that . Then . Assume that is an arbitrary element satisfying and assume that . It is evident that belongs to , which implies that for some and thus char . From , we get that and thus . It follows that divides . Since by Lemma 5, we have by the theorem of Sylow, and so . It is obvious that ; thus, . Since , by the theorem of Schur-Zassenhaus, we obtain that . Furthermore, , , and , where . Since , we have .
Assume that . Then by (1) of Lemma 6. Since char , we obtain . Since by the theorem of Schur-Zassenhaus and , we have or by [18]. Note that and ; thus, (i) holds.
Assume that . Then , which is a contradiction since such group does not exist by (1) of Lemma 6.

Case 2. Suppose that . Then , 28, or 56.
Assume that . Then from Lemma 5, we obtain that , which is impossible obviously.
Suppose that . Assume that is an arbitrary element satisfying , and assume that . It is evident that belongs to , which implies that for some . Then similar to Case 1, we can get and . Since , by the theorem of Schur-Zassenhaus, we obtain , where is a group satisfying , , and . Furthermore, and . It is not difficult to find that such group does not exist through simple calculation.
Assume that . Then we can also get a contradiction similarly.

Case 3. Suppose that . Then , 84, or 168.
Assume that . Assume that is an arbitrary element satisfying , and assume that . It is evident that belongs to , which implies that for some . Then similar to Case 2, we can get and . Since , by the theorem of Schur-Zassenhaus, we obtain . Furthermore, , , and , where . It follows that or by [18]. Therefore, or . Thus, (ii) and (iii) hold.
Assume that . Then similar to the above proof of , we can get that . Thus, (iv) holds.
Assume that . Then , and so . Thus, (v) holds.

Case 4. Suppose that . Then . By (1) of Lemma 7, we obtain that is a 2-Frobenius group. Furthermore, is elementary abelian, , , , and . Hence, (vi) holds.
Assume that is nonsolvable. Then by Lemma 8, and so (vii) holds.

Proof of Theorem 2. By [17] we get that and . From Lemma 4, , and it follows that . Now we discuss the following four cases.

Case 1. Suppose that . Then , where since by Lemma 5.
Assume that . Assume that is an arbitrary element satisfying , and assume that . It is clear that belongs to , which implies that for some . Then similar to the above, we have divides . Since by Lemma 5, we obtain or 18 by the theorem of Sylow.
Assume that . Then . Since , we get that . Furthermore, . Since , by the theorem of Schur-Zassenhaus, we obtain , where is a group satisfying . Hence, , which is impossible obviously.
Assume that . Then by Lemma 10, it follows that (mod 17) since is solvable by Lemmas 7 and 8, which is a contradiction.
Assume that . Assume that is an arbitrary element satisfying and assume that . It is clear that belongs to , which implies that for some and so char . From , we have and thus . It follows that divides . Since by Lemma 5, we obtain or 18 by the theorem of Sylow. If , then similarly to the proof of Case 1 of Theorem 1 and by (2) of Lemma 6, we can get a contradiction. If , then similarly to the above proof of , we can get a contradiction.

Case 2. Suppose that . Then or .
Assume that . Then we can get a contradiction similarly to Case 1 of Theorem 1.
Assume that . Assume that is an arbitrary element satisfying and assume that . It is clear that belongs to , which implies that for some . Then similar to the above, we have . From , we obtain . Since and , we get that by the theorem of Schur-Zassenhaus, where and . Thus, . Therefore, , which is contradict to the theorem of Sylow.

Case 3. Assume that . Then or since by Lemma 5.
Assume that . Assume that be an arbitrary element satisfying and assume that . It is evident that belongs to , which implies that for some . Then similar to the above, we can get that . Since and , we have , where . Note that ; thus, by the theorem of Schur-Zassenhaus, we obtain that , where is a group of order satisfying and . Obviously, . Now we get a contradiction since such group does not exist by Lemma 11.
Assume that . Assume that is an arbitrary element satisfying , and assume that . Since belongs to , we obtain that there exists a cyclic subgroup of order of satisfying . So char . From we obtain . So . Consequently, divides .
From Lemma 5 we obtain that . Hence, . Since is solvable by Lemmas 7 and 8, we have by Lemma 10. Consequently, and hence by the theorem of Schur-Zassenhaus, where and . Since , we get that . Consequently, . Note that ; thus, (i) holds.

Case 4. Suppose that . Then . Therefore by (2) of Lemma 7. Hence, (ii) holds.

Proof of Corollary 3. It is evident by Theorems 1 and 2.
Now the proofs of our results are complete.

Data Availability

Our paper is theoretical research. Relevant data are all in the paper.

Conflicts of Interest

The author declares that they have no conflicts of interest.

Acknowledgments

This work is supported by the National Natural Science Foundation of China (11401324). The author would like to thank the referee with deep gratitude for pointing out some questions in the previous version of the paper. His/her valuable suggestions help us improve the quality of the paper.