In [2], we also obtained a complete classification of groups G in which every noncyclic proper subgroup is nonnormal; all such groups G satisfy 1 (G).By |(G)| we denote the order of (G). Note that we cannot ensure that 1 (G) for any solvable group G with |(G)| = selleckbio n �� 1. For example, let GD2pn be a dihedral group of order 2pn, where n �� 1 and p is an odd prime. Then (D2pn) = p, p2,��, pn, so 1 (D2pn). For the nonsolvable group of the smallest order PSL(2,5), it is easy to see that (PSL(2,5)) = 5, 6, 10, and so |(PSL(2,5))| = 3.For the influence of |(G)| on the solvability of groups, we have the following result, the proof of which is given in Section 3.Theorem 1 �� Let G be a group.If |(G)|��2, then G is solvable.G is a nonsolvable group with |(G)| = 3 if and only if GPSL(2,5) or PSL(2,13) or SL(2,5) or SL(2,13).

The following two corollaries are direct consequences of Theorem 1.Corollary 2 �� Let G be a group with |(G)|��3. Then G is nonsolvable if and only if (G) = 5,6, 10 or 14,78,91. Corollary 3 �� Let G be a group and (G) the set of the numbers of conjugates of nontrivial subgroups of G.If |(G)|��2, then G is solvable.G is a nonsolvable group with |(G)| = 3 if and only if GPSL(2,13). Let G be a group and *(G) the set of the numbers of conjugates of nonnormal noncyclic proper subgroups of G. Obviously *(G)(G).Arguing as in the proof of Theorem 1, we can obtain the following result.Theorem 4 �� Let G be a group. If |*(G)|��2, then G is solvable. Remark 5 ��If we assume that G is a nonsolvable group with |*(G)| = 3, we cannot get that ��(G) = Z(G).

For example, let GPSL(2,5) �� p, where p �� 7 is a prime. It is easy to see that |*(G)| = 3. But ��(G) = 1 and Z(G) = p.Let G be a group and (G) the set of the numbers of conjugates of nonabelian proper subgroups of G. Obviously (G)(G). Arguing as in the proof of Theorem 1, we can also obtain the following result.Theorem 6 �� Let G be a group. If |(G)|��2, then G is solvable. 2. PreliminariesIn this section, we collect some essential lemmas needed in the sequel.Lemma 7 (see [3]) �� Let G be a group. If all nonnormal maximal subgroups of G have the same order, then G is solvable. Lemma 8 (see [4]) �� Let G be a nonsolvable group having exactly two classes of nonnormal maximal subgroups of the same order; then G/S(G)PSL(2,7), Cilengitide where S(G) is the largest solvable normal subgroup of G.