| Budget Amount *help |
¥3,400,000 (Direct Cost: ¥3,400,000)
Fiscal Year 2003: ¥1,100,000 (Direct Cost: ¥1,100,000)
Fiscal Year 2002: ¥1,100,000 (Direct Cost: ¥1,100,000)
Fiscal Year 2001: ¥1,200,000 (Direct Cost: ¥1,200,000)
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| Research Abstract |
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).
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