×

zbMATH — the first resource for mathematics

Some results on Hadamard groups. (English) Zbl 0864.05021
Kim, A. C. (ed.) et al., Groups - Korea ’94. Proceedings of the international conference, Pusan, Korea, August 18-25, 1994. Berlin: Walter de Gruyter. 149-155 (1995).
Let \(G\) be a group of order \(2n\) such that \(G\) contains a central involution \(e^*\). Let \(D\) be a transversal of \(G\) with respect to subgroup \(N\) generated by \(e^*\). Then there exist exactly \(2^n\) transversals. For every transversal \(D\) we have that \(D=De\), where \(e\) denotes the identity element of \(G\), \(D\cap De^*=\varnothing\) and \(G=D\cup De^*\). If for some transversal \(D\) we have that \(|D\cap Da|=n/2\) for every element \(a\) of \(G\) other than \(e\) and \(e^*\), where \(|X|\) denotes the number of elements in a finite set \(X\), namely if \(D\) is a \((n,n/2,N)\) relative difference set, then we call \(D\) and \(G\) an Hadamard subset and an Hadamard group respectively. Hadamard groups are introduced so that we may scrutinize the structure of automorphism groups of Hadamard matrices.
Let \(D\) be any transversal. Then it holds obviously that \[ |D\cap Da|=|D\cap Da^{-1}|\quad\text{for any element \(a\) of }G\tag{1} \] and that \[ |D\cap Da|+|De^*\cap Da|=n\quad\text{for any element \(a\) of }G.\tag{2} \] From (1) and (2) follows the following lemma.
Lemma 1. Let \(D\) be any transversal. If \(a\) is an element of order 4 such that \(a^2=e^*\), then we have that \(|D\cap Da|=n/2\).
Lemma 2. Let \(H\) be a non-trivial subgroup of \(G\). If a transversal \(D\) is a union of left cosets of \(H\), then \(D\) is not an Hadamard subset.
Proposition 1. Let \(C_m\) and \(Q\) denote the cyclic group of order \(m\) and the quaternion group of order 8 respectively. Let \(G\) be an Hadamard group with the property that every transversal is an Hadamard subset. Then \(G\) is isomorphic to either \(C_4\) or \(Q\).
Proposition 2. Let \(G\) be an Hadamard group of order \(8n\) and \(S\) a Sylow 2-subgroup of \(G\). Assume that \(S=C_{2^{m+1}}\times C_2\) and that \(a\) and \(b\) are generators for \(C_{2^{m+1}}\) and \(C_2\) respectively. If \(e^*=a^{2^m}\), then \(m=1\) or 2.
(Remarks on the strong Hadamard conjecture and a conjecture of Ryser follow).
For the entire collection see [Zbl 0857.00028].

MSC:
05B20 Combinatorial aspects of matrices (incidence, Hadamard, etc.)
20D60 Arithmetic and combinatorial problems involving abstract finite groups
PDF BibTeX XML Cite