{"title":"有限群的分解问题","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":"{\"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}","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}
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.