| Let G be a finite group.Denote by e(G) the set of all orders of elements in G, k is the maximal number in e(G), n is the number of cyclic groups with order k in G, l is a natural number,denote by Ml(G) the set of elements with order k in G, M(G) = Mk{G). denote by t{G) the number of prime graph component in G. H x K is a semidirect product of H and K. Let p be a prime, pl|||G| means pl| |G| and pl+1 |G|. In this paper, we prove the following two theorems:Theoreml.l Let | M(G) |= 2m, (m, 30) = 1, then the finite group G are solvable.Theorem 2.1 Let Zk be the cyclic subgroup of order k in G. If | M{G) |= 2pq (5 < q < p, p and q are distinct primes), we can suppose |G| = 2u3v(2q + 1)v(2p + 1)vl, (1,6(2q+1)(2p+1))=1, Then1 when n=l, G is one of the following groups(1) G = Zk i.e. G is cyclic groups of order k , where k = 2 (2q + 1)2 , e = 0,1, p = 2q + 1 and 2q + 1 is a prime , or k = 2 (2pq + 1), = 0,1, 2pq + 1 is a prime ;(2) G =< a > < b >, ak = , bg = a', b-1ab = ar, where rg = 1(modk), t(r -1) = 0(modk), g|2pq, k = 2 (2q+1)2 , = 0,1, p = 2q + 1 is prime , or k = 2 (2pq + 1), = 0,1, 2pq + 1 is a prime ;2, If n = p( but when it = 2(2q +1),let t(G) > 2), then G is one of the following groups(1) k = 2q + 1, t(G) = 1 , and G has 2pq elements of maximal order if and only if p = (2q+1)v-1 + ... + (2q + 1) + 1 is a prime, v is an odd number;(2) k = 2q + 1; G is a Frobenius group , G = H K, |H| = (2q + l)u, |K|| 2q, p = (2q + 1)v-1+ ... + (2q+ 1) + 1 is a prime,v > 3, v is an odd number. Especially, when 2 e(K), H is an elementary abelian group ;(3) k = 2(2q+1), G is a Frobenius group. G = H K, |H| = 2u(2q+1)v, p = 2u + 1 and 2u + l, v= 1, or p = (2g+ 1)v-1 + ... + (2q + 1) + 1, u= 1;(4) k = 2(2q + 1), G is a 2-Frobenius groupwith normal series H K G, s.t. |H| = (2q + 1)v2u-1, and |G/K| = 2.3> If n = q then the maximal order k of elements in G is equal to 2(2p + 1). and when t(G) > 2, G is one of the following groups(1) G is a Frobenius group , G = H x K, \H\ = 2u(2p +1), p = 2U - 1, tt(G) = {2,2p+l}(2) G is 2-Frobenius group with normal series H< K < G, s.t. \H\ = 2u~1(2p+1)", and \G/K\ = 2.4 ?If n = pq, k ^ 4(but when k = 6 let i(G) > 2), then G is one of the following groups(1) k = 3, G is Frobenius group , G = H x K, \H\ = 2, \K) = 3", and K is an elementary abelian group;(2) k = 3, G is 3-group, |G| = 3v, and group G has 2pq elements of maximal orderif and only if 2pq = 3v - 1;(3) k = 6, G is Frobenius group, G = H K, |K| = 5, |H| = 2u3v pq = (2u-1 + ... + 2 + 1)(3v-1 + ... + 3 + 1);(4) k = 6,G is 2-Frobenius group with normal series H K G, s.t. |H| = 2u-13v,|G/K| = 2.Question:1 (1) If n = p, k = 2(2q + 1), G is 2-Frobenius group, |H| = (2q + 1)v2u-1 and |G/K| = 2 , What is the structure of group G ?(2) If n = p, k = 2{2q + 1) and t(G) = 1 , What is the structure of group G ?2 (1) If n = q, k = 2(2p + 1), G is 2-Frobenius group, \H\ = (2p + 1)v2u-1 and |G/K| = 2, What is the structure of group G ?(2) when n = q, k ?2(2p + 1) and t(G) - 1, What is the structure of group G ?3 If n = pq(1) If k = 6, G is 2-Frobenius group , |H| = 2u-13v, |G/K| = 2 and |K/H| = 5, What is the structure of group G ?(2) If k = 6 and t(G) = 1 , What is the structure of group G ?(3) If k = 4 , What is the structure of group G ?...
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