Let $G$ be a finite group, $D=\{(g,g):g\in G\}$ is a subgroup of the direct product $G\times G$. Show that $G$ is simple if and only if $D$ is a maximal subgroup of $G\times G$.
I tried to prove by contradiction. First assume that $G$ is simple and suppose there exist a subgroup $H$ of $G\times G$ such that $D<H<G\times G$. But i couldn't use the simplicity of $G$ to get a contradiction.
On
One can give an alternative solution using the following fact:
Let $L$ be a group, acting transitively on a set $A$. Let $a \in A$. Then $L$ acts primitively on $A$ iff the stabiliser $L_a$ is a maximal subgroup.
Here $L = G \times G$ acts on $A = G$ with $(g, h)$ sending $x \mapsto g^{-1} x h$. Clearly $D = L_1$ is the stabiliser of $1$.
We will show that the block systems for the action of $L$ and $A$ are of the form $$ \mathfrak{N} = \{ N^{g} : g \in G\}, $$ for $N$ a normal subgroup of $G$. This implies that $L$ acts primitively on $A$ if and only if $G$ is simple.
Suppose $N$ is a normal subgroup of $G$. Then $N$ is a block. In fact, if $$g^{-1} N h \cap N = g^{-1} h N \cap N$$ is non-empty, then $g^{-1} N h = g^{-1} h N = N$, as the (left) cosets of $N$ are a partition of $G$.
Conversely, consider a block system $\mathfrak{N}$, and let $N \in \mathfrak{N}$ be the block containing 1. We claim that $N$ is a normal subgroup of $G$. If $g, h \in N$, then $h = g^{-1} g h \in g^{-1} N h \cap N$, so $g^{-1} N h = N$ and $g^{-1} h = g^{-1} 1 h \in N$. This shows that $N$ is a subgroup of $G$. Also, if $x \in G$, then $x^{-1} N x$ contains 1, so that it equals $N$. Thus $N$ is normal in $G$.
If $G$ is simple, and $D < M \le G \times G$, let $(a,b) \in M \setminus D$, so that $a \ne b$. Then $(1, 1) \ne (a,a)^{-1} (a, b) = (1, a^{-1} b) = (1, c) \in M$.
Conjugating $(1, c)$ by $(g, g)$, we see that $M$ contains all $(1, c^g)$, for $g \in G$. As $G$ is simple, the subgroup of $G$ generated by the conjugacy class $\{c^g : g \in G \}$ (which is a non-trivial normal subgroup of $G$) is $G$. Thus $M$ contains $\{1\} \times G$, and as $M$ contains $D$, $M$ also contains $G \times \{1\}$, so that $M = G \times G$.
We have proved that if $G$ is simple, then $D$ is maximal in $G \times G$.
For the converse, suppose $N$ is non-trivial normal subgroup of $G$.
Then $D$ normalizes $N \times \{1\} = \{(n, 1):n\in N\}$, so $ND$ is a subgroup of $G \times G$ properly containing $D$.
If $D$ is maximal in $G \times G$, then $ND = G \times G$. In particular for each $g \in G$ one has $(1, g^{-1}) \in ND$, so that $(1, g^{-1}) = (n, 1) (g^{-1}, g^{-1})$ for some $n \in N$, so $g = n \in N$, and $N = G$, so that $G$ is simple.