Group Theory – Prove G is Abelian if {(g, g): g in G} is a Normal Subgroup

group-theorynormal-subgroups

Let $G$ be a group and let $D=\{(g, g):g\in G\}$. If $D$ is a normal subgroup of $G\times G$, prove that $G$ is an abelian group.

My attempt:

$D$ is a normal subgroup of $G\times G$.

$\implies(a, b)D=D(a, b)\ \forall(a, b)\in G\times G$

So for a given $(a, b)\in G$ and $(g, g)\in D$, $\exists(g', g')\in D$ such that

$(a, b)(g, g)=(g', g')(a, b)$

$(ag, bg)=(g'a, g'b)$

$ag=g'a$ and $bg=g'b$

I feel like I've used all of the information given but don't know how to conclude that $G$ is abelian. Any suggestions?

I think I've got it!

Let $g=a$.

Then, $aa=g'a\implies g'=a$

So, $bg=g'b\implies ba=ab$

So, $G$ is abelian.