Prove that $G$ is an abelian group if $\{(g, g):g\in G\}$ is a normal subgroup.

Question

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?

Answer 1

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.

Answer 2

I think you've solved your own question correctly. Kudos. Here's a slightly different, perhaps more direct, approach:

$$D\lhd G\times G\iff\forall\,g\in G\;,\;\;(a,b)(g,g)(a,b)^{-1}\in D\iff$$

$$(aga^{-1}\,,\,bgb^{-1})\in D$$

and this means $\;aga^{-1}=bgb^{-1}\;$ for all $\;a,b,g\in\Bbb G\;$ . Take now $\;g=b\;$ , and the above says that for all $\;a,b\in G\;$ we have

$$aba^{-1}=bbb^{-1}=b\implies ab=ba\;\;\;\;\;\;\;\;\square$$

Answer 3

Since D is normal in GxG So using the property of cosets and for (g,g) in GxG We have => D(g,g)=(g,g)D . Let (d,d) in D ve arbitray where d is in G . =>(d,d)(g,g)=(g,g)(d,d) => (dg,dg)=(gd,gd) . Now comparing corresponding entries we have dg=gd and g,d both are in G . Hence G is abelian