Question

Let $\mathit{i}$be an integer with $\mathbf{1}\mathbf{\le }\mathit{i}\mathbf{\le }\mathit{n}$. Let ${\overline{\mathbf{G}}}_{\mathbf{i}}$be the subset of ${\mathit{G}}_{\mathbf{1}}\mathbf{\times }{\mathit{G}}_{\mathbf{2}}\mathbf{\times }\mathbf{\dots }\mathbf{\times }{\mathit{G}}_{\mathbf{n}}$consisting of those elements whose $\mathit{i}$th coordinate is any element ${\mathit{G}}_{\mathbf{i}}$of and whose other coordinates are each the identity element, that is,

${\overline{\mathbf{G}}}_{\mathbf{i}}\mathbf{=}\mathbf{\left\{}\mathbf{(}{\mathbf{e}}_{\mathbf{1}}\mathbf{,}{\mathbf{e}}_{\mathbf{2}}\mathbf{,}\mathbf{\dots }\mathbf{,}{\mathbf{e}}_{\mathbf{i}\mathbf{-}\mathbf{1}}\mathbf{,}{\mathbf{e}}_{\mathbf{i}}\mathbf{,}{\mathbf{e}}_{\mathbf{i}\mathbf{+}\mathbf{1}}\mathbf{,}\mathbf{\dots }\mathbf{,}{\mathbf{e}}_{\mathbf{n}}\mathbf{)}\mathbf{|}{\mathbf{e}}_{\mathbf{i}}\mathbf{\in }{\mathbf{G}}_{\mathbf{i}}\mathbf{\right\}}$

Prove that,

${\overline{\mathbf{G}}}_{\mathbf{i}}\mathbf{\cong }{\mathit{G}}_{\mathbf{i}}$

Short answer

It is proved that,${\overline{G}}_i\cong G_i$

Step-by-step solution

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

Definition of Group Homomorphism:

Let $(G,\ast )$and $(G\text{'},\ast \text{'})$be any two groups. A function $f:G\to G\text{'}$is said to be a group homomorphism if $f(a\ast b)=f\left(a\right)\ast \text{'}f\left(b\right),\forall a,b\in G$.

Definition of Normal Subgroup:

A subgroup $N$of a group $G$is called a normal subgroup of $G$ if .

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

To define mappings

Let ${\overline{G}}_i$be the subset of $G_1\times G_2\times \dots \times G_n$.

Then, ${\overline{G}}_i=\left\{e_1,e_2,\dots ,g_i,\dots ,e_n|g_i\in G_i\right\}$

We have to prove that ${\overline{G}}_i\cong G_i$

Now, we define a mapping $\psi :{\overline{G}}_i\to G_i$by $\psi (e_1,e_2,\dots ,g_i,\dots ,e_n)=g_i$and ${\psi }^{\text{'}}:G_i\to {\overline{G}}_i$by ${\psi }^{\text{'}}\left(g{}_i\right)=(e_1,e_2,\dots ,g_i,\dots ,e_n)$.

Then,$\psi {\psi }^{\text{'}}=I{d}_{G_i}$and ${\psi }^{\text{'}}\psi =I{d}_{{\overline{G}}_i}$

We now prove that both are homomorphisms.

To show ψ is a homomorphism

Consider mapping, $\psi :{\overline{G}}_i\to G_i$

Let $(e_1,e_2,\dots ,g_i,\dots e_n)(e_1,e_2,\dots ,{g^{\text{'}}}_i,\dots e_n)\in {\overline{G}}_i$

Then,

$\psi \left((e_1,e_2,\dots ,g_i,\dots e_n)(e_1,e_2,\dots ,{g^{\text{'}}}_i,\dots e_n)\right)=\psi \left((e_1,e_2,\dots ,g_i,\dots e_n)\right)\psi \left((e_1,e_2,\dots ,{g^{\text{'}}}_i,\dots e_n)\right)$

Hence,$\psi$ is a homomorphism.

To show ψ' is a homomorphism and G¯i≅Gi

Now, consider a mapping, ${\psi }^{\text{'}}:G_i\to {\overline{G}}_i$

Let $g_i,{g_i}^{\text{'}}\in G_i$.

Find ${\psi }^{\text{'}}\left(g_i{g_i}^{\text{'}}\right)$as:

$$\begin{gathered} {\psi }^{\text{'}}\left(g_i{g_i}^{\text{'}}\right)=(e_1,e_2,\dots ,g_i{g_i}^{\text{'}},\dots ,e_n) \\ =(e_1,e_2,\dots ,g_i,\dots ,e_n)(e_1,e_2,\dots ,{g_i}^{\text{'}},\dots ,e_n) \\ ={\psi }^{\text{'}}\left(g_i\right){\psi }^{\text{'}}\left({g_i}^{\text{'}}\right) \end{gathered}$$

Therefore,${\psi }^{\text{'}}\left(g_i{g_i}^{\text{'}}\right)={\psi }^{\text{'}}\left(g_i\right){\psi }^{\text{'}}\left({g_i}^{\text{'}}\right)$.

Hence,${\psi }^{\text{'}}$ is a homomorphism.

Hence, ${\overline{G}}_i\cong G_i$.