Question

Let $\mathit{G}$be a group and let$\mathit{D}\mathbf{=}\mathbf{\left\{}\mathbf{(}\mathbf{a}\mathbf{,}\mathbf{a}\mathbf{,}\mathbf{a}\mathbf{,}\mathbf{\text{}}\mathbf{)}\mathbf{|}\mathbf{a}\mathbf{\in }\mathbf{G}\mathbf{\right\}}$ .

Prove that$\mathit{D}$ is normal in $\mathit{G}\mathbf{\times }\mathit{G}\mathbf{\times }\mathit{G}$if and only if$\mathit{G}$ is abelian.

Short answer

It is proved that,$D$ is normal in$G\times G\times G$ if and only if$G$ is abelian.

Step-by-step solution

Step-by-Step Solution Step 1: Definition of Subgroup, Normal Subgroup, and Abelian Group

Definition of Subgroup:

Let$(G,\ast )$be a group under binary operation, say $\ast$. A non-empty subset$H$of$G$ is said to be a subgroup of$G$, if $(H,\ast )$is itself a group.

In other words, if $e_G\in H$and $a,b\in H$ then $a{b}^{-1}\in H$.

Definition of Normal Subgroup:

A subgroup $N$of a group role="math" localid="1652778704658" $G$is called a normal subgroup of $G$if.

$Na=aN,\forall a\in G$

Definition of Abelian group:

A group$G$ is called abelian if$ab=ba$ , $\forall a,b\in G$.

To prove G is abelian

Let $G$be a group and $D=\left\{(a,a,a)|a\in G\right\}$.

Suppose $e_G\in G$is an identity element in $G$.

We have to prove that$D$ is normal in$G\times G\times G$if and only if$G$is abelian.

Let us prove the first part, i.e., to prove$G$ is abelian.

Let role="math" localid="1652778906686" $D$be a normal subgroup of role="math" localid="1652778920424" $G\times G\times G$.

Let $(g,g,g)\in D$.

Then,$(a,e_G,e_G)\in G\times G\times G$and$D$is normal in$G\times G\times G$.

Find $(a,e_G,e_G)(g,g,g){(a,e_G,e_G)}^{-1}$as:

$\begin{array}{c}(a,e_G,e_G)(g,g,g){(a,e_G,e_G)}^{-1}=(ag{a}^{-1},g,g)\\ \in D\end{array}$

Further, it can be written as:

$\begin{array}{c}ag{a}^{-1}=g\\ ag=ga\end{array}$

Therefore, $G$is abelian by thedefinition of Abelian group.

To prove is D normal in G×G×G

Assume that $G$is an abelian group.

Let $D$ be a subgroup of $G\times G\times G$.

Then, by Example 7,$G_1\times \dots \times G_n$ is abelian if and only if every$G$ is abelian.

It follows that,$G\times G\times G$ is abelian.

Therefore, every subgroup is normal.

Thus,$D$ is normal in $G\times G\times G$, since$D$ is a subgroup of$G\times G\times G$ .