On the Set of the Numbers of Conjugates of Noncyclic Proper Subgroups of Finite Groups

Let G be a finite group and 𝒩𝒞(G) the set of the numbers of conjugates of noncyclic proper subgroups of G. We prove that (1) if |𝒩𝒞(G)| ≤ 2, then G is solvable, and (2) G is a nonsolvable group with |𝒩𝒞(G)| = 3 if and only if G≅PSL(2,5) or PSL(2,13) or SL(2,5) or SL(2,13).


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 [1], 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 ≅ (2,5) or (2,5). 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 ≅ (2,5) or (2, 5) (see [2]). Let be a group and NC( ) the set of the numbers of conjugates of noncyclic proper subgroups of . It is clear that a group with NC( ) = 0 is either a cyclic group or a minimal noncyclic group, and a group with NC( ) = {1} is a group in which every noncyclic proper subgroup is normal. In [2], we also obtained a complete classification of groups in which every noncyclic proper subgroup is nonnormal; all such groups satisfy 1 ∉ NC( ).
The following two corollaries are direct consequences of Theorem 1.    Arguing as in the proof of Theorem 1, we can obtain the following result.
Let be a group and NA( ) the set of the numbers of conjugates of nonabelian proper subgroups of . Obviously NA( ) ⊆ NC( ). Arguing as in the proof of Theorem 1, we can also obtain the following result. Theorem 6. Let be a group. If |NA( )| ≤ 2, then is solvable.

Preliminaries
In this section, we collect some essential lemmas needed in the sequel.
Lemma 7 (see [3]). Let be a group. If all nonnormal maximal subgroups of have the same order, then is solvable.

Proof of Theorem 1
The proof of Theorem 1 follows from the following two lemmas. (1) Suppose that 1 ∈ MS( ). Since is nonsolvable, must have nonnormal maximal subgroups. Let be any nonnormal maximal subgroup of ; one has | : ( )| = | : |. Since |MS( )| ≤ 2, 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.
Thus, our assumption is not true, so is solvable.
Proof. The sufficiency part is evident, and we only need to prove the necessity part.
Since |MS( )| ≤ 3, 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.