Research Of Finite Groups With The Number Of The Same Order Elements

Let G be a finite group andÏ€_e(G) the set of its element orders. Denote by m_i(G):=|{g∈G|o(g)=i}|the size of elements of order i in G, and nse(G) := {m_i|i∈π_e(G)}the set of sizes of the elements with the same order.Let Mt(G) := {g∈G|g^t=1}. Finite groups G₁ and G₂ are of the same order type if and only if |M_d(G₁)|=|M_d(G₂)|,d=1,2,….It is an important subject to study the influence on the structure of finite groups by their quantitative relationship. Many scholars have obtained a lot of important results. For example, the famous " Sylow's theorem", "Lagrange's theorem", "solvability of groups of odd order", "Burnside's theorem" etc.In 1987, professor Shi put forward characterization of finite simple groups by their quantitative relationship: charactering finite simple groups only use the set of its element ordersÏ€_e(G)and and its own order |G|(see[41],[42],[43], [44],[45],[46], [47],[48],[49], [50],[51],[52],[53], [54],[55], [56]).Some group theoriests characterize finite simple groups by the orders of their solvable subgroups (see [1],[36],[59]).In 1987, Professor J.G. Thompson posed the following problem in his letter to professor Shi:Thompson's Problem Suppose groups G₁ and G₂ are of the same order type. Suppose G₁ is solvable, is it true that G₂ is necessarily solvable?Some authors who work on group theorem study the topic of influence on the number of maximal order on a finite group to study the Thompson's Problem nondirectly. They obtain a lot of exciting results(see [8],[11],[18],[24], [25],[26],[27],[28],[30], [37], [61]). This problem was opened in 1990 by Professor Wujie Shi(see [47]). Unfortunately, no one can solve it completely, even give a counterexample up to now, which implies the difficulty of Thompson's Problem. It is easy to see that if G₁ and G₂ are of the same order type, then nse(G₁) = nse(G₂), |M(G₁)| = |M(G₂)| and |G₂| = |G₂|.We haven't found any one who characterize finite group G in termes of the set nse(G) so far. In this paper, we consider to work out the structure of a finite group G by set nse(G) or |M(G)|, and get a series of results. It consists of the three following chapters:In Chapter 1, we introduce some symbols and basic concepts that we usually use in the paper.In Chapter 2, we characterize finite groups by the set nse(G). We divide this chapter into three parts.1. We characterize some finite simple groups by nse(G) and |G|, and get the following theorem:Let G be a finite group and M a finite simple group, where M is one of finite simple K₃-groups, or K₄-groups, sporadic simple groups or L₂(q), where q is a prime or q=2^m and 2^m+ 1 or 2^m-1 is a prime, then G≌M if and only if nse(G) = nse(M) and |G| = |M|.2. We classify the finite group G with the property that the elements of nse(G) are consecutive integers, and get the following theorems:Let Gbea finite group, if the elements of nse(G) are consecutive integers, that is, nse(G) ={1,2,…,n}, then n≤3 and one of the following holds:â… . if n = l, thenG≤C₂.â…¡. if n = 2, then G≌C₃,C₄or C₆.â…¢. if n = 3, then G≌S₃.3. We study the finite group G with nse(G) = {1,15, 20, 24}, and get the following theorem:Let G be a finite group, then G≌A₅ if and only if nse(G) = {1,15,20,24}.In Chapter 3, we classify the finite groups with |M(G)| = 24, and get the following theorem: Let G be a finite group and k the maxiamal element order of G. If |M(G)| = 24, then G is one of the following groups :1. If k = 4, then the following claimes hold:1.1. G = N(?) C₃ is a Probenius group with kernel N and complement C₃, and N≌G₄, G₅ or G₆.1.2.G=C₂⁴(?)S₃and P₂=D₈×C₂×C₂.2. If k = 5, then the following claimes hold:2.1. G≌C₅×C₅.2.2. G is a Probenius group with the kernel P₅(?)G and complementH and P₅ = C₅×G₅, where|H||24.2.3. G≌A₅.3.IF k= 6, 8, 9, 12, 16, 18 or 24, then |G||2^α.3^β, whereα≤7 andβ≤4.4. If k=10, then the following claimes hold:4.1. P₅=C₅×C₅(?)G,C_G(P₅)=P₅×C₂and|G/C_G(P₅)||2².4.2. A₅×C₂or SL₂(5).4.3. A₅×C₂(?)Gand|G| = 240.4.4. G/Z(G)≌S₅and |Z(G)| = 2.5. If k = 20, then G/C_G(x)≤C₄, G_G(x) =G₅×HandH≌Q₈orS₄.6. If Jk = 28, then |G||2⁴·3·7,P₇(?)GandC_G(P₇)=C₂₈×C₂7. If k = 30, then |G||2⁴·3·5,P₅(?)GandC_G(P₅)=C₃₀×C₂8. If k = 36, then (?)G,G/C_G()(?)Aut(C₃₆)and C_G()≌C₃₆×C₂or C₄.9. If k satisfiesφ(k)= 24, then C_G(< x >) =< x >(?)G and G/C_G(< x >)(?)Aut(C_k), where o(x) = k.