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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