Dear Prof. Cameron, Thanks for your answer. Sorry to my question's condition should be |H^c \cap Hg|=|G|/4, where H^c the complement of H in G, and all g. So sorry!
Yours, Rulin Shen ---原始邮件--- 发件人: "Peter Cameron "<pj...@st-andrews.ac.uk> 发送时间: 2017年2月2日 19:06:42 收件人: "Rulin Shen"<shenru...@hotmail.com>; 主题: Re: [GAP Forum] A question of GAP Dear Rulin Shen, I believe that the situation you describe is impossible. Suppose that |G|=n and |H|=m. Let us count in two ways the pairs (g,x) with g<>1 and x in H\cap Hg. There are n-1 choices of g and then n/4 choices of x. On the other hand, there are m choices of x in H, and then (m-1) choices of g such that x in Hg (this requires x=hg for some h in H, so g=xh^-1; and we are not allowed to choose h=x here but any other choice is fine. So m(m-1)=(n-1)n/4. Now it is easy to see that this equation has no solution. (n is even, and the numbers n/2(n/2-1) and (n/2+1)n/2 lie on either side of n(n-1)/4.) Yours, Peter Cameron. On 01/02/17 05:17, Rulin Shen wrote: > Dear forum, > > > How to find that the finite group G has a subset H such that |H\cap > Hg|=|G|/4 for all elements g in G different from 1? > > > Thanks in advance! > > Best wishes > > Rulin Shen > > _______________________________________________ > Forum mailing list > Forum@mail.gap-system.org > http://mail.gap-system.org/mailman/listinfo/forum _______________________________________________ Forum mailing list Forum@mail.gap-system.org http://mail.gap-system.org/mailman/listinfo/forum
