Research on difference sets related to finite geometry

如果R是u的右余集表示，并且不在u中的每一个元素都恰好有R个表示xy^<-1> (x,y∈R)，则mu阶的群G的k子集R称为相对于u阶的子群u的（m,u,k， R）差集。由R，我们可以构造一个可整除的设计（P，B），其中P=G,B ={Ra: a∈G}。2003年，我们研究了非可解群，特别是简单群中的(2n,2,2n,n) -差分集。n . Ito引入了左哈达玛截线的概念，即G=R<t>, t^2=1, RR^<(-1)>=nS_1+2nS_22n，其中S_1， S_2∧G。我们注意到R在R≠xR， 1≠∀x∈G的附加条件下也是一个（2n,2,2n,n）差分集。考虑到G的一个合适的置换表示，我们得到了以下结论。定理假定4n阶群G包含一条左哈达玛截线R w.r.t <t>满足R≠xR， 1≠∀x∈G。若G有一子群H使得G=[G,G]H， t∈H且$ G ^<-1>tg∈G-H (G∈G)，则存在一个v × v积分矩阵B使得BB^ t =B^TB=(n/2)I_v，其中v=([G:H](|t^G|-|t^G∩H|))/(2|t^G|)。在上述定理中，如果G是Ω上的传递置换群，则有：主张让(G,Ω)是一个传递置换群度r(> 4)和t G .假设的退化(G, G)传递Ω。如果t修复r-4分| G |广场免费的部分有一个素因子p, p≡3国防部4,然后向左G没有阿达玛横向r w.r.t < t >满足r≠xR, 1≠∀x∈G.Corollary没有离开阿达玛横向满足r≠xR, 1≠∀x∈G在A_5 S_5, A_7, S_7, PSL(2、7)和PGL(2、7)。

A k-subset R of a group G of order mu is called a (m,u,k,r) -difference set relative to a subgroup U of order u if R is a right coset representative of U and every element not in U has exactly r representations xy^<-1> (x,y ∈ R). From R, we can construct a divisible design (P,B), where P=G and B={Ra: a ∈ G}.In 2003, we study (2n,2,2n,n) -difference sets in non-solvable groups, especially in simple groups. N. Ito introduced the notion of a left Hadamard transversal, that is, G=R<t>, t^2=1, RR^<(-1)>=nS_1+2nS_22n, where S_1, S_2 ⊂ G. We note that R is also a (2n,2,2n,n) -difference set under an additional condition that R ≠ xR, 1 ≠ ∀ x ∈ G. Considering a suitable permutation representation of G, we have obtained the following.Theorem Assume that a group G of order 4n contains a left Hadamard transversal R w.r.t <t> satisfying R ≠ xR, 1 ≠ ∀ x ∈ G. If G has a subgroup H such that G=[G,G]H, t ∈ H and $g^<-1>tg ∈ G-H (∃ g ∈ G), then there exists a v × v integral matrix B such that BB^T=B^TB=(n/2)I_v, where v=([G:H](|t^G|-|t^G ∩ H|))/(2|t^G|).In the above theorem, if G is a transitive permutation group on Ω, then we have the following.Proposition Let (G,Ω) be a transitive permutation group of degree r (>4) and t an involution of G. Suppose that [G,G] is transitive on Ω. If t fixes r-4 points and the square free part of |G| has a prime divisor p such that p ≡ 3 mod 4, then G has no left Hadamard transversal R w.r.t <t> satisfying R ≠ xR, 1 ≠ ∀ x ∈ G.Corollary There is no left Hadamard transversal R satisfying R ≠ xR, 1 ≠ ∀ x ∈ G in A_5, S_5, A_7, S_7, PSL (2,7) and PGL (2,7).