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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.

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Strictly speaking, you are probably looking for conditions on $G$ and $H.$ – user17794 Jun 22 '12 at 20:32
@TimDuff: I am looking for finite groups than subgroups of them. – Babak S. Jun 22 '12 at 20:34
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
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 $G$.

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May I ask you note me which subgroups of a given group of corefree? May I ask you to elaborate your answer? Thanks. – Babak 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
@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? – Babak 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

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