Question

Let ${\mathit{i}}_{\mathbf{1}}\mathbf{,}{\mathit{i}}_{\mathbf{2}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\mathit{i}}_{\mathbf{n}}$be a permutation of integers $\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}\mathit{n}$. Prove that ${\mathit{G}}_{\mathbf{i}}\mathbf{\times }{\mathit{G}}_{{\mathbf{i}}_{\mathbf{2}}}\mathbf{\times }\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\times }{\mathit{G}}_{{\mathbf{i}}_{\mathbf{n}}}$is isomorphic to ${\mathit{G}}_{\mathbf{1}}\mathbf{\times }{\mathit{G}}_{\mathbf{2}}\mathbf{\times }\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\times }{\mathit{G}}_{\mathbf{n}}$.[Exercise 4 is the case $\mathit{n}\mathbf{=}\mathbf{2}$.]

Short answer

Answer:

It is proved that, $G_{\mathrm{i}}\times G_{{\mathrm{i}}_2}\times ...\times G_{{\mathrm{i}}_{\mathrm{n}}}$is isomorphic to $G_1\times G_2\times ...\times G_{\mathrm{n}}$.

Step-by-step solution

To define mappings φ and φ'

Definition of group homomorphism:

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

A permutation localid="1656491434926" $i_1,i_2,...,i_n$of $1,2,...,n$is a bijective map $\sigma :\left\{1,2,...,n\right\}\to \left\{1,2,..,n\right\}$defined by $\sigma \left(j\right)=i_j$.

Let $i_1,i_2,...,i_n$be the permutation of integers $1,2,...,n$.

We have to show that $G_{i_1}\times G_{i_2}\times ...\times {G_i}_n$is isomorphic to $G_1\times G_2\times ...G_n$.

Define a map $\phi :G_1\times G_2\times ...\times G_n\to G_{i_1}\times G_{i_2}\times ...\times G_{i_n}$by $\phi \left(\left(x_1,x_2,...,x_n\right)\right)=\left(x_{\sigma \left(1\right)},x_{\sigma \left(2\right)},...,x_{\sigma \left(n\right)}\right)$.

Also, define localid="1656493384080" $\phi \text{'}:G_{i_1}\times G_{i_2}\times ...\times G_{i_n}\to G_1\times G_2\times ...\times G_n$by $\phi \text{'}\left(\left(x_1,x_2,...,x_n\right)\right)=\left(x_{{\sigma }^{-1}\left(1\right)},x_{{\sigma }^{-1}\left(2\right)},...,x_{{\sigma }^{-1}\left(n\right)}\right)$.

To show φ and φ' are homomorphism

Let $\left(x_1,x_2,...,x_n\right)\left(y_1,y_2,...,y_n\right)\in G_1\times G_2\times ...\times G_n$.

Find $\phi \left(\left(x_1,x_2,...,x_n\right)\left(y_1,y_2,...,y_n\right)\right)$as:

role="math" localid="1656494557396" $$\begin{gathered} \phi \left(\left(x_1,x_2,...,x_n\right)\left(y_1,y_2,...,y_n\right)\right)=\phi \left(x_1{y}_1,x_2{y}_2,..,x_n{y}_n\right) \\ =\left(x_{\sigma \left(1\right)}{y}_{\sigma \left(1\right)},x_{\sigma \left(2\right)}{y}_{\sigma \left(2\right)},...,x_{\sigma \left(n\right)}{y}_{\sigma \left(n\right)}\right) \\ =\left(x_{\sigma \left(1\right)},x_{\sigma \left(2\right)},...,x_{\sigma \left(n\right)}\right)\left(y_{\sigma \left(1\right)},y_{\sigma \left(2\right),..,}{y}_{\sigma \left(n\right)}\right) \\ =\phi \left(\left(x_1,x_2,...,x_n\right)\right)\phi \left(\left(x_1,x_2,...,x_n\right)\right) \end{gathered}$$

Therefore, $\phi \left(\left(x_1,x_2,...,x_n\right)\left(y_1,y_2,...,y_n\right)\right)=\phi \left(\left(x_1,x_2,...,x_n\right)\right)\phi \left(\left(x_1,x_2,...,x_n\right)\right)$

Hence, $\phi$is a homomorphism.

Also, let $\left(x_1,x_2,...,x_n\right)\left(y_1,y_2,...,y_n\right)\in G_{i_1}\times G_{i_2}\times ...\times G_{i_n}$.

Find $\phi \left(\left(x_1,x_2,...,x_n\right)\left(y_1,y_2,...,y_n\right)\right)$as:

$$\begin{gathered} \phi \text{'}\left(\left(x_1,x_2,...,x_n\right)\left(y_1,y_2,...,y_n\right)\right)=\phi \text{'}\left(x_1{y}_1,x_2{y}_2,..,x_n{y}_n\right) \\ =\left(x_{{\sigma }^{-1}\left(1\right)}{y}_{{\sigma }^{-1}\left(1\right)},x_{{\sigma }^{-1}\left(2\right)}{y}_{{\sigma }^{-1}\left(2\right)},...,x_{{\sigma }^{-1}\left(n\right)}{y}_{{\sigma }^{-1}\left(n\right)}\right) \\ =\left(x_{{\sigma }^{-1}\left(1\right)},x_{{\sigma }^{-1}\left(2\right)},...,x_{{\sigma }^{-1}\left(n\right)}\right)\left(y_{{\sigma }^{-1}\left(1\right)},y_{{\sigma }^{-1}\left(2\right),..,}{y}_{{\sigma }^{-1}\left(n\right)}\right) \\ =\phi \text{'}\left(\left(x_1,x_2,...,x_n\right)\right)\phi \text{'}\left(\left(y_1,y_2,...,y_n\right)\right) \end{gathered}$$

Therefore, $\phi \text{'}\left(\left(x_1,x_2,...,x_n\right)\left(y_1,y_2,...,y_n\right)\right)$$=\phi \text{'}\left(\left(x_1,x_2,...,x_n\right)\right)\phi \text{'}\left(\left(y_1,y_2,...,y_n\right)\right)$

Hence, $\phi$is a homomorphism.

To prove isomorphism andGi1×Gi2×...×Gin≅G1×G2×...×Gn

Now, we prove isomorphism.

Let $\left(x_1,x_2,...,x_n\right)\in G_1\times G_2\times ...\times G_n.$

Find $\phi \text{'}\phi \left(\left(x_1,x_2,...,x_n\right)\right)$as:

$$\begin{gathered} \phi \text{'}\phi \left(\left(x_1,x_2,...,x_n\right)\right)=\phi \text{'}\left(x_{\sigma \left(1\right)},x_{\sigma \left(2\right)},...,x_{\sigma \left(n\right)}\right) \\ =\left({x_{{\sigma }^{-1}}}_{\sigma \left(1\right)},{x_{{\sigma }^{-1}}}_{\sigma \left(2\right)},...,{x_{{\sigma }^{-1}}}_{\sigma \left(n\right)}\right) \\ =\left(x_1,x_2,...,x_n\right) \end{gathered}$$

Therefore, $\phi \text{'}\phi \left(\left(x_1,x_2,...,x_n\right)\right)$$=\left(x_1,x_2,...,x_n\right)$

Hence, $\phi \text{'}\phi =l{d}_{G_1\times G_2\times ...\times G_n}$.

Let $\left(x_1,x_2,...,x_n\right)\in G_{i_1}\times G_{i_2}\times ...\times G_{i_n}$.

Find$\phi \text{'}\phi \left(\left(x_1,x_2,...,x_n\right)\right)$as:

$$\begin{gathered} \phi \phi \text{'}\left(\left(x_1,x_2,...,x_n\right)\right)=\phi \left(x_{{\sigma }^{-1}\left(1\right)},x_{{\sigma }^{-1}\left(2\right)},...,x_{{\sigma }^{-1}\left(n\right)}\right) \\ =\left({x_{\sigma {\sigma }^{-1}}}_{\left(1\right)},{x_{\sigma {\sigma }^{-1}}}_{\left(2\right)},...,{x_{\sigma {\sigma }^{-1}}}_{\left(n\right)}\right) \\ =\left(x_1,x_2,...,x_n\right) \end{gathered}$$

Therefore, $\phi \phi \text{'}\left(\left(x_1,x_2,...,x_n\right)\right)$$=\left(x_1,x_2,...,x_n\right)$.

Hence, $\phi \phi \text{'}=l{d}_{{G_i}_1\times G_{i_2}\times ...\times G_{i_n}}$.

Thus, ${G_i}_1\times G_{i_2}\times ...\times G_{i_n}$is isomorphic to $G_1\times G_2\times ...\times G_n$.

i.e., ${G_i}_1\times G_{i_2}\times ...\times G_{i_n}\cong G_1\times G_2\times ...\times G_n$.