# 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?

## 3 Answers

Recall that the normal subgroups of $G/H$ are in bijection with the normal subgroups of $G$ which contain $H$. This is the famous fourth isomorphism theorem, also called correspondence theorem.

Suppose that $G/H$ isn't simple: then there there is a proper normal subgroup $M$ of $G/H$, which corresponds to a proper normal subgroup $M'$ of $G$ which contains $H$, but this contradicts the hypothesis.

Suppose $K$ is a nontrivial normal subgroup of $G/H$. What can you say about its inverse image in $G$?

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.