Question

Question: Let $\mathit{G}\mathbf{,}\mathit{H}\mathbf{,}{\mathit{G}}_{\mathbf{1}}\mathbf{,}{\mathit{H}}_{\mathbf{1}}$be groups such that$\mathit{G}\mathbf{\cong }{\mathit{G}}_{\mathbf{1}}$ and $\mathit{H}\mathbf{\cong }{\mathit{H}}_{\mathbf{1}}$. Prove that$\mathit{G}\mathbf{\times }\mathit{H}\mathbf{\cong }{\mathit{G}}_{\mathbf{1}}\mathbf{\times }{\mathit{H}}_{\mathbf{1}}$

Short answer

It has been proved that$G\times H\cong G_1\times H_1$

Step-by-step solution

Step-By-Step SolutionStep 1: Define two isomorphic maps

Let, $f:G\to G_1$and $g:H\to H_1$be two isomorphic maps.

Define a mapping$\phi :G\times G_1\to H\times H_1$ such that$\phi \left(x,y\right)=\left(f\left(x\right),g\left(y\right)\right)$

Show that is φbijection

Since f, g are isomorphism therefore, they both must be surjective

Now, $\phi \left(x,y\right)=\left(f\left(x\right),g\left(y\right)\right)$,therefore, it must also be a surjective.

Also, let $\phi \left(x,y\right)=\phi \left(z,w\right)$

role="math" localid="1651260934291" $$\begin{gathered} \left(f\left(x\right),g\left(y\right)\right)=\left(f\left(z\right),g\left(w\right)\right) \\ f\left(x\right)=f\left(z\right),g\left(y\right)=g\left(w\right) \end{gathered}$$

Now, since are injective therefore,

$$\begin{gathered} dx=z,y=w \\ \left(x,y\right)=\left(z,w\right) \end{gathered}$$

Thus $\phi$is also injective

Hence, $\phi$being injective and surjective is a bijection.

Show that φis homomorphism

Consider, $\phi \left(\left(x,y\right)*\left(w,z\right)\right)$

$$\begin{gathered} \phi \left(\left(x,y\right)*\left(w,z\right)\right)=\phi \left(\left(x*w,y*z\right)\right) \\ =\left(f\left(x*w\right),g\left(y*z\right)\right) \end{gathered}$$

Since f,g are homomorphism therefore,

$$\begin{gathered} \left(f\left(x*w\right),g\left(y*z\right)\right)=\left(f\left(x\right)*f\left(w\right),g\left(y\right)*g\left(z\right)\right) \\ =\left(f\left(x\right),g\left(y\right)\right)*\left(f\left(w\right),g\left(z\right)\right) \\ =\phi \left(x,y\right)*\phi \left(w,z\right) \end{gathered}$$

Hence $\phi$is a homomorphism.

Conclusion

Since $\phi$is bijective and homomorphic, therefore, $\phi$is an isomorphism

hence ,$G\times H\cong G_1\times H_1$