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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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    The word "regular" is only used when $H=1$. I think this is just the "natural action on the cosets of H".2012-06-22

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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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    Thanks for your help.2012-06-22