Prove that left multiplication defines an action of $G$ on the coset space $\frac{G}{H}$ and that the kernel of the associated homomorphism from $G$ to $\operatorname{Aut}\left(\frac{G}{H}\right)$ is the largest normal subgroup contained in $H$.
Let $\sigma: G \times \frac{G}{H} \rightarrow \frac{G}{H}$ denote the group action and $\tilde \sigma: G \rightarrow$ Aut$\frac{G}{H}$ denote the associated homomorphism. I want to show that ker$(\tilde \sigma)$ is a normal subgroup of $G$ and that for any normal subgroup $N \subseteq H$ that is normal to $G$, $N \subseteq$ ker$(\tilde \sigma)$.
I can show the second part but the first part I am completely stuck. So far I have...
Let $g \in$ ker$\tilde \sigma$ and $kH \in \frac{G}{H}$. Then $\sigma_g(kH)=g \cdot kH=kH$. So in particular, $\sigma_g(H) = gH = H$ So $g \in H$. Since $H$ is a subgroup of $G$, $g^{-1} \in H$ so $gHg^{-1} = H$. So $H$ is normal in $G$. Which is not what I am trying to prove at all...