3. Proof of Theorem 1The proof of Theorem 1 follows from the following two lemmas.Lemma 10 �� Let G be a group. If |(G)|��2, then G is solvable. Proof �� Assume that G is nonsolvable. Then by [7, Exercise 10.5.7], all maximal subgroups of G are noncyclic. Let (G) be the set of the numbers of conjugates of maximal subgroups selleck chemicals Pazopanib of G. It follows that (G)(G). Then |(G)|��2.Suppose that 1 (G). Since G is nonsolvable, G must have nonnormal maximal subgroups. Let M be any nonnormal maximal subgroup of G; one has |G : NG(M)| = |G : M|. Since |(G)|��2, we know that G has at most one class of nonnormal maximal subgroups of the same order. It follows that G is solvable by Lemma 7, a contradiction.Suppose that 1 (G). It follows that all maximal subgroups of G are nonnormal.
By the hypothesis, G has at most two classes of maximal subgroups of the same order. Since G is nonsolvable and G has no normal maximal subgroups, one has G/��(G)23iPSL(2,7) by Lemma 9 (1) and (2), where i = 0, 1,��. It is easy to see that (G/��(G))(G) and |(23iPSL(2,7))|>2. It follows that |(G)|>2, a contradiction.Thus, our assumption is not true, so G is solvable.Lemma 11 �� A group 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). Proof ��The sufficiency part is evident, and we only need to prove the necessity part.By the hypothesis, |(G)|��3. We claim that1???(G).(1)Otherwise, assume that 1 (G). Then G has at most two classes of nonnormal maximal subgroups of the same order. Since G is nonsolvable, one has G/S(G)PSL(2,7) by Lemmas 7 and 8.
It is easy to see that (G/S(G))(G) and |(PSL(2,7))|>3. It follows that |(G)|>3, a contradiction. Thus, 1 (G).Since |(G)|��3, we have that G has at most three classes of maximal subgroups of the same order.By Lemma 9 (1), G cannot have exactly one class of maximal subgroups of the same order.If G has exactly two classes of maximal subgroups of the same order, according toLemma 9 (2), one has G/��(G)23iPSL(2,7) since G has no normal maximal subgroups, where i = 0, 1,��. Since |(23iPSL(2,7))|>3, it follows that |(G)|>3, a contradiction.Thus, G has exactly three classes of maximal subgroups of the same order. By Lemma 9 (3), G/S(G) might be isomorphic to A6 or PSL(2, q), q = 11, 13, 23, 59, 61 or PSL(3,3) or U3(3) or PSL(5,2) or PSL(2, 2f), and f is a prime or PSL(2,7) �� PSL(2,7)����PSL(2,7).
If G/S(G) is an isomorphism to A6 or PSL(2, q), q = 11, 23, 59, 61 or PSL(3,3) or U3(3) or PSL(5,2) or PSL(2, 2f), and f is an odd prime or PSL(2,7) �� PSL(2,7) �� ��PSL(2,7). It is easy to see that |(G/S(G))|>3 by [8, 9], which implies that |(G)|>3, a contradiction. Thus, G/S(G)PSL(2,4)PSL(2,5) or PSL(2,13).Note that 1 (G) and |(G)| = |(G)| = 3. It follows that 1 (G), so S(G) is cyclic. We claim that��(G)=S(G).(2)Otherwise, assume Anacetrapib that ��(G) < S(G). Let M be a maximal subgroup of G such that S(G)M.
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