[GAP Forum] 回复： A question of GAP
Rulin Shen
shenrulin at hotmail.com
Thu Feb 2 11:45:56 GMT 2017
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 "<pjc20 at st-andrews.ac.uk>
发送时间: 2017年2月2日 19:06:42
收件人: "Rulin Shen"<shenrulin at 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
>
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