A note on factorizations of finite groups

G. Bergman
{"title":"A note on factorizations of finite groups","authors":"G. Bergman","doi":"10.30504/jims.2020.108338","DOIUrl":null,"url":null,"abstract":"In Question 19.35 of the Kourovka Notebook, M. H. Hooshmand asks whether, given a finite group $G$ and a factorization $\\mathrm{card}(G)= n_1\\ldots n_k$, one can always find subsets $A_1,\\ldots,A_k$ of $G$ with $\\mathrm{card}(A_i)=n_i$ such that $G=A_1\\ldots A_k;$ equivalently, such that the group multiplication map $A_1\\times\\ldots\\times A_k\\to G$ is a bijection. \nWe show that for $G$ the alternating group on 4 elements, $k=3$, and $(n_1,n_2,n_3) = (2,3,2)$, the answer is negative. We then generalize some of the tools used in our proof, and note an open question.","PeriodicalId":8427,"journal":{"name":"arXiv: Group Theory","volume":null,"pages":null},"PeriodicalIF":0.0000,"publicationDate":"2020-03-28","publicationTypes":"Journal Article","fieldsOfStudy":null,"isOpenAccess":false,"openAccessPdf":"","citationCount":"6","resultStr":null,"platform":"Semanticscholar","paperid":null,"PeriodicalName":"arXiv: Group Theory","FirstCategoryId":"1085","ListUrlMain":"https://doi.org/10.30504/jims.2020.108338","RegionNum":0,"RegionCategory":null,"ArticlePicture":[],"TitleCN":null,"AbstractTextCN":null,"PMCID":null,"EPubDate":"","PubModel":"","JCR":"","JCRName":"","Score":null,"Total":0}
引用次数: 6

Abstract

In Question 19.35 of the Kourovka Notebook, M. H. Hooshmand asks whether, given a finite group $G$ and a factorization $\mathrm{card}(G)= n_1\ldots n_k$, one can always find subsets $A_1,\ldots,A_k$ of $G$ with $\mathrm{card}(A_i)=n_i$ such that $G=A_1\ldots A_k;$ equivalently, such that the group multiplication map $A_1\times\ldots\times A_k\to G$ is a bijection. We show that for $G$ the alternating group on 4 elements, $k=3$, and $(n_1,n_2,n_3) = (2,3,2)$, the answer is negative. We then generalize some of the tools used in our proof, and note an open question.
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有限群的分解问题
在Kourovka Notebook的问题19.35中,M. H. Hooshmand问,给定一个有限群$G$和一个因式分解$\ mathm {card}(G)= n_1\ldots n_k$,是否总能找到$G$的子集$A_1,\ldots,A_k$,使得$G=A_1\ldots A_k$;$等价地,使得$A_1\ ldots\乘以A_k\到G$的群乘法映射$A_1\ ldots\乘以A_k$是双射。我们证明了对于4个元素的交替群$G$, $k=3$和$(n_1,n_2,n_3) =(2,3,2)$,答案是负的。然后,我们推广了证明中使用的一些工具,并注意到一个开放的问题。
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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