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 a subgroup of $\mathit{G}\mathbf{\times }\mathit{G}\mathbf{\times }\mathit{G}$.

Short answer

It is proved that $D$is a subgroup of $G\times G\times G$.

Step-by-step solution

 Step 1: Definition of Subgroup

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,$a{b}^{-1}\in H$.

Define identity elements of G×G×G

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

We have to prove that$D$ is a subgroup of$G\times G\times G$ .

We have$D\subset G$ and$(e_G,e_G,e_G)\in D$ be the identity of$G\times G\times G$ .

To prove D is a subgroup of  G×G×G

Let$(a,a,a),(b,b,b)\in D$.

Then,$a{b}^{-1}\in G$.

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

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

Therefore,$(a{b}^{-1},a{b}^{-1},a{b}^{-1})\in D$ from the definition of Subgroup.

Hence,role="math" localid="1652777896490" $D$ is a subgroup of$G\times G\times G$ .