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.