Criterion for being a simple group

Question

In this work it's written that

A group $G$ is simple if and only if the diagonal subgroup of $G \times G$ is a maximal subgroup.

How can I prove it?

Answer 1

($\Rightarrow$): Suppose $G$ is simple, and $H\le G\times G$ is a subgroup containing the diagonal subgroup and some $(g,h)$ with $g\ne h$. Let $N\le G$ be the subgroup of elements $n$ such that $(e,n)\in H$, which is nontrivial since $(e,g^{-1}h)\in H$. Furthermore $N$ is normal, since if $(e,n)\in H$ then $(g,g)(e,n)(g^{-1},g^{-1})=(e,gng^{-1})\in H$. Thus by simplicity $N=G$, so $(e,g)\in H$ for all $g\in G$. By an identical argument, $(g,e)\in h$ for all $g\in G$, so clearly $H=G\times G$. Thus the diagonal subgroup is maximal.

($\Leftarrow$): Suppose $G$ is not simple, so we have some nontrivial proper normal subgroup $N$. Let $H=\{(g,ng):g\in G, n\in N\}$, which is a subgroup since if $g,g'\in G$ and $n,n'\in N$ we have $$(g,ng)(g',n'g')=(gg',ngn'g')=(gg',ngn'g^{-1}gg')\in H$$ since $gg'\in G$ and $ngn'g^{-1}\in N$. This contains the diagonal subgroup but is still proper, so the diagonal subgroup is not maximal.