#### Abstract

Let be a finite group and the set of the numbers of conjugates of noncyclic proper subgroups of . We prove that (1) if , then is solvable, and (2) is a nonsolvable group with if and only if or or or .

#### 1. Introduction

In this paper, all groups are assumed to be finite. It seems interesting to investigate the influence of some arithmetic properties of noncyclic proper subgroups on the solvability of groups. In , Li and Zhao proved that any group having at most three conjugacy classes of noncyclic proper subgroups is solvable, and a group having exactly four conjugacy classes of noncyclic proper subgroups is nonsolvable if and only if or . As a generalization of the above result, we showed that any group having at most three conjugacy classes of nonnormal noncyclic proper subgroups is solvable, and a group having exactly four conjugacy classes of nonnormal noncyclic proper subgroups is nonsolvable if and only if or (see ).

Let be a group and the set of the numbers of conjugates of noncyclic proper subgroups of . It is clear that a group with is either a cyclic group or a minimal noncyclic group, and a group with is a group in which every noncyclic proper subgroup is normal. In , we also obtained a complete classification of groups in which every noncyclic proper subgroup is nonnormal; all such groups satisfy .

By we denote the order of . Note that we cannot ensure that for any solvable group with . For example, let be a dihedral group of order , where and is an odd prime. Then , so . For the nonsolvable group of the smallest order , it is easy to see that , and so .

For the influence of on the solvability of groups, we have the following result, the proof of which is given in Section 3.

Theorem 1. Let be a group.(1)If , then is solvable.(2) is a nonsolvable group with if and only if or or or .

The following two corollaries are direct consequences of Theorem 1.

Corollary 2. Let be a group with . Then is nonsolvable if and only if or .

Corollary 3. Let be a group and the set of the numbers of conjugates of nontrivial subgroups of .(1)If , then is solvable.(2) is a nonsolvable group with if and only if .

Let be a group and the set of the numbers of conjugates of nonnormal noncyclic proper subgroups of . Obviously .

Arguing as in the proof of Theorem 1, we can obtain the following result.

Theorem 4. Let be a group. If , then is solvable.

Remark 5. If we assume that is a nonsolvable group with , we cannot get that . For example, let , where is a prime. It is easy to see that . But and .
Let be a group and the set of the numbers of conjugates of nonabelian proper subgroups of . Obviously . Arguing as in the proof of Theorem 1, we can also obtain the following result.

Theorem 6. Let be a group. If , then is solvable.

#### 2. Preliminaries

In this section, we collect some essential lemmas needed in the sequel.

Lemma 7 (see ). Let be a group. If all nonnormal maximal subgroups of have the same order, then is solvable.

Lemma 8 (see ). Let be a nonsolvable group having exactly two classes of nonnormal maximal subgroups of the same order; then , where is the largest solvable normal subgroup of .

Lemma 9 (see [5, 6]). Let be a group having exactly classes of maximal subgroups of the same order, where ; then one of the following statements holds:(1)suppose that is a group with , and then is a -group for some prime ;(2)suppose that is a nonsolvable group with , and then , where , and is a semidirect product of the normal subgroup and the subgroup ;(3)suppose that is a nonsolvable group with , and then ; , ; ; ; , and is a prime; .

#### 3. Proof of Theorem 1

The proof of Theorem 1 follows from the following two lemmas.

Lemma 10. Let be a group. If , then is solvable.

Proof. Assume that is nonsolvable. Then by [7, Exercise 10.5.7], all maximal subgroups of are noncyclic. Let be the set of the numbers of conjugates of maximal subgroups of . It follows that . Then .(1)Suppose that . Since is nonsolvable, must have nonnormal maximal subgroups. Let be any nonnormal maximal subgroup of ; one has . Since , we know that has at most one class of nonnormal maximal subgroups of the same order. It follows that is solvable by Lemma 7, a contradiction.(2)Suppose that . It follows that all maximal subgroups of are nonnormal. By the hypothesis, has at most two classes of maximal subgroups of the same order. Since is nonsolvable and has no normal maximal subgroups, one has by Lemma 9 (1) and (2), where . It is easy to see that and . It follows that , a contradiction.
Thus, our assumption is not true, so is solvable.

Lemma 11. A group is a nonsolvable group with if and only if or or or .

Proof. The sufficiency part is evident, and we only need to prove the necessity part.
By the hypothesis, . We claim that
Otherwise, assume that . Then has at most two classes of nonnormal maximal subgroups of the same order. Since is nonsolvable, one has by Lemmas 7 and 8. It is easy to see that and . It follows that , a contradiction. Thus, .
Since , we have that has at most three classes of maximal subgroups of the same order.
By Lemma 9 (1), cannot have exactly one class of maximal subgroups of the same order.
If has exactly two classes of maximal subgroups of the same order, according to Lemma 9 (2), one has since has no normal maximal subgroups, where . Since , it follows that , a contradiction.
Thus, has exactly three classes of maximal subgroups of the same order. By Lemma 9 (3), might be isomorphic to or or or or or , and is a prime or . If is an isomorphism to or , or or or or , and is an odd prime or . It is easy to see that by [8, 9], which implies that , a contradiction. Thus, or .
Note that and . It follows that , so is cyclic. We claim that
Otherwise, assume that . Let be a maximal subgroup of such that . Then . It is obvious that . Moreover, , since is cyclic. It follows that . Therefore, . Let , and . By N/C-theorem, . That is, . Note that is abelian since is cyclic. Moreover, is a nonabelian simple group and . Here . Therefore, one has or . If , it follows that is abelian, a contradiction. If , then . It follows that and then ; this contradicts that all maximal subgroups of are nonnormal. Thus, our assumption is not true, so .
It follows that or .
If , then or .
Next, suppose that . Let be any prime divisor of . We claim that . Otherwise, assume that . Let be a subgroup of such that . That is, . Then or . Since and Schur multipliers of both and are , we have that or . Note that and . It follows that , a contradiction. Thus, , so is a cyclic -group. If , let be a subgroup of such that . Then or . We have that or . Let be a subgroup of such that . Then or . Since Schur multipliers of both and are trivial, we have that or ; this contradicts that all maximal subgroups of are nonnormal. Thus, . It follows that or .

Lemmas 10 and 11 combined together give Theorem 1.

#### Acknowledgments

The authors are grateful to the referees who gave valuable comments and suggestions. Jiangtao Shi was supported by NSFC (Grant nos. 11201401 and 11361075). Cui Zhang was supported by H.C. Ørsted Postdoctoral Fellowship at DTU (Technical University of Denmark).