Let $G$ be a group and let $H \leq G$ be any subgroup of $G$, supposing also $H$ has finite index $n$. Let $G$ act on the left cosets of $H$ by left multiplication; this induces the homomorphism $\varphi: G \to S_n$ given by $\varphi_x(gH)=xgH$.
The link here asks a similar question and the posted answer mentions that under this action, $\mathrm{ker} \varphi$ is the core of $H$, i.e.
$$\mathrm{ker} \varphi = \mathrm{cor}(H) = \displaystyle\bigcap_{g \in G} gHg^{-1}$$
From this it follows that $\mathrm{ker} \varphi \leq H$.
What circumstances are necessary for $H = \mathrm{ker} \varphi$? One circumstances would be normality because then $\displaystyle\bigcap_{g \in G} gHg^{-1} = \displaystyle\bigcap_{g \in G} H = H$. Is this the only circumstance which would force $\mathrm{ker} \varphi = H$?
