Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $G$ is a group and $H$ be a subgroup of it. Then $G$ can act on the following set $$\Omega= \{Hg|g\in G\}$$ by $\forall Hg\in\Omega$ and $x\in G$; $(Hg)^x=Hgx$ (I don't know if I can call this action right regular representation of $G$?). It can easily be found that the kernel of this action is: $$N=\{x\in G|(Hg)^x=Hg\}=\bigcap_{g\in G}g^{-1}Hg$$

Clearly, if $H=\{1\}$ then $N=\{1\}$, so the action is faithful. Now, I am thinking about the condition(s) that we can consider for $G$ until above action has non-trivial kernel. For example, if the group be cyclic, abelian or our subgroup is normal in $G$ then $N=H$. Of course I assume $H\neq\{1\}$. Does this problem make any sense? Thanks.

share|improve this question
    
Strictly speaking, you are probably looking for conditions on $G$ and $H.$ –  Tim Duff Jun 22 '12 at 20:32
    
@TimDuff: I am looking for finite groups than subgroups of them. –  B. S. Jun 22 '12 at 20:34
1  
The word "regular" is only used when $H=1$. I think this is just the "natural action on the cosets of H". –  Jack Schmidt Jun 22 '12 at 20:45
add comment

1 Answer

up vote 4 down vote accepted

The subgroup $$\cap_{g\in G}g^{-1}Hg$$ is known as the core of $H$ in $G$. It is the largest normal subgroup of $G$ that is contained in $H$.

Therefore, the kernel of the action is trivial if and only if $H$ is corefree in $G$: it does not contain any nontrivial normal subgroup of $G$.

share|improve this answer
    
May I ask you note me which subgroups of a given group of corefree? May I ask you to elaborate your answer? Thanks. –  B. S. Jun 22 '12 at 20:40
    
@BabakSorouh: The subgroups of a given group $G$ that are corefree are the ones that don't contain any nontrivial normal subgroup of $G$. There is nothing to elaborate. –  Arturo Magidin Jun 22 '12 at 20:40
1  
@BabakSorouh: I looked for core-free subgroups fairly hard in large groups. It is very difficult to say too much about them. If the subgroup is maximal and core-free, then you can say a fair amount, but not all groups have such maximal subgroups, and even second maximal (maximal in maximal) are much harder to describe. –  Jack Schmidt Jun 22 '12 at 20:44
    
Sorry for many asking but, what about a nilpotent group? –  B. S. Jun 22 '12 at 20:47
    
@BabakSorouh: they are the worst. I studied nilpotent groups with no non-identity core-free subgroups for a while (until I realized JG Thompson already proved everything I did, but 40 years earlier). There are lots of them. The quaternion group of order 8 is one, but there are many more. –  Jack Schmidt Jun 22 '12 at 20:49
show 1 more comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.