Factoring nonabelian finite groups into two subsets

R. Bildanov, V. Goryachenko, A. Vasil’ev
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引用次数: 8

Abstract

A group $G$ is said to be factorized into subsets $A_1, A_2, \ldots, A_s\subseteq G$ if every element $g$ in $G$ can be uniquely represented as $g=g_1g_2\ldots g_s$, where $g_i\in A_i$, $i=1,2,\ldots,s$. We consider the following conjecture: for every finite group $G$ and every factorization $n=ab$ of its order, there is a factorization $G=AB$ with $|A|=a$ and $|B|=b$. We show that a minimal counterexample to this conjecture must be a nonabelian simple group and prove the conjecture for every finite group the nonabelian composition factors of which have orders less than $10\,000$.
将非abel有限群分解为两个子集
如果$G$中的每个元素$G$可以唯一地表示为$G =g_1g_2\ldots g_s$,其中$g_i\在A_i$中,$i=1,2,\ldots,s$,则称群$G$被分解为子集$A_1, A_2, \ldots, A_s $。我们考虑以下猜想:对于每一个有限群$G$和它阶的每一个分解$n=ab$,存在一个分解$G= ab$且$| a |=a$和$|B|= B$。我们证明了这个猜想的最小反例必须是一个非abel简单群,并证明了对于每一个非abel组成因子的数量级小于$10\ 000$的有限群的猜想。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
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