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Studies on Hadamard matrices with ``2-transitive'' automorphism groups

1984; Mathematical Society of Japan; Volume: 36; Issue: 1 Linguagem: Inglês

10.2969/jmsj/03610063

1881-1167

Noboru Ιτο, Hiroshi Kimura,

Advanced Topics in Algebra

An Hadamard matrix $H$ of order $n$ is a $\{-1,1\}$ -matrix of degree $n$ such that $HH^{t}=H^{t}H=nI$ , where $t$ denotes the transposition.It is known that $n$ equals one, two or a multiple of four.In this paper we assume that $n$ is greater than eight.For the basic fact on Hadamard matrices see [1] or [7].Let $P$ be the set of $2n$ points 1, 2, $\cdots$ , $n,$ $1^{*},$ $2^{*},$ $\cdots$ , $n^{*}$ .Then we define an n-subset $\alpha_{i}$ of $P$ as follows: $\alpha_{i}$ contains $j$ or $j^{*}$ according as the $(i, j)$ -entry of $H$ equals +1 or $-1(1\leqq i, j\leqq n)$ .Let $\alpha_{i}^{*}=P-\alpha_{i}$ .We call $\alpha_{i}$ and $\alpha_{i}^{*}$ blocks $(1\leqq i\leqq n)$ .Let $B$ be the set of $2n$ blocks $\alpha_{1},$ $\alpha_{2},$ $\cdots$ , $\alpha_{n},$ $\alpha_{1}^{*},$ $\alpha_{2}^{*},$ $\cdots$ , $\alpha_{n}^{*}$ .Then $M(H)=(P, B)$ is called the matrix design of $H$ .By definition each point belongs to exactly $n$ blocks.By the orthogonality of columns of $H$ each point pair not of the shape $\{a, a^{*}\}$ belongs to exactly $n/2$ blocks, and each point trio not containing a point pair of the shape $\{a, a^{*}\}$ belongs to exactly $n/4$ blocks.$\{a, a^{*}\}$ does not belong to any block.Similarly by the orthogonality of rows of $H$ each block pair not of the shape $\{\alpha, \alpha^{*}\}$ intersects in exactly $n/2$ points, and each block trio not containing a block pair of the shape $\{\alpha, \alpha^{*}\}$ intersects in exactly $n/4$ points.We assume that $a^{**}=a$ .Then $\alpha^{**}=\alpha$ .Let $\mathfrak{G}$ be the group of all permuta- tions $\sigma$ on $P$ such that $\sigma$ leaves $B$ as a whole.Then we call $\mathfrak{G}$ the automor- phism group of $M(H)$ .Obviously $\mathfrak{G}$ is isomorphic to the automorphism group of $H$ .Since $\zeta=\prod_{a=1}^{n}(a, a^{*})=\prod_{i=1}^{n}(\alpha_{i}, \alpha_{i}^{*})$ belongs to the center of $\mathfrak{G},$ $\mathfrak{G}$ is imprimitive on $P$ .For the basic facts on permutation groups see [9] or [10].Now let $\overline{P}$ and $\overline{B}$ be the set of point pairs $\overline{a}=\{a, a^{*}\}$ and block pairs $\overline{\alpha}=\{\alpha, \alpha^{*}\}$ , where $a\in P$ and $\alpha\in B$ , respectively.Then $\mathfrak{G}$ may be considered as permutation groups on $\overline{P}$ and on $\overline{B}$ .We notice that $\zeta$ is trivial on $\overline{P}$ and on $\overline{B}$ , and that there is no apparent incidence relation between $P$ and $\overline{B}$ .In this paper we assume that $\mathfrak{G}$ on $\overline{P}$ is doubly transitive and that $\mathfrak{G}$ on $\overline{P}$ contains a regular normal subgroup $\mathfrak{N}$ on $\overline{P}$ .Then $\mathfrak{N}$ on $\overline{P}$ is an elementary Abelian 2-group of order $n$ , and so $n$ $*)$