Prove that if $G$ is a finite group and $H$ is a proper normal subgroup of largest order, then $G/H$ is simple. We define the group action of $G$ on the set of left cosets of $H$ in $G$ by left multiplication. Let $\Phi$ be the associated permutation representation.
Then by Generalised Cayley's Theorem, $G/ \text{Ker}(\Phi)$ is isomorphic to a subgroup of symmetric group of order equal to index of $H$ in $G$.Would this mean that $H=\text{Ker}(\Phi)$ ($\text{Ker}(\Phi)$ is the largest subgroup of $G$ contained in $H$)?
And if so, does this help prove this?

A normal subgroup of $G/H$ is of the form $K/H$ where $K \le G$ is a normal subgroup containing $H$. By maximality of $H$ you have $K=H$. Hence $G/H$ has no proper normal subgroups.
