Question

Let $\mathit{H}\mathbf{,}\mathit{K}\mathbf{,}\mathit{M}\mathbf{,}\mathit{N}$be groups such that $\mathit{K}\mathbf{\cong }\mathit{M}\mathbf{\times }\mathit{N}$. Prove that $\mathit{H}\mathbf{\times }\mathit{K}\mathbf{\cong }\mathit{H}\mathbf{\times }\mathit{M}\mathbf{\times }\mathit{N}$.

Short answer

Answer:

It is proved that, $H\times K\cong H\times M\times N$.

Step-by-step solution

Definition of Group Homomorphism, Kernel of a Function, and Group Isomorphism

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 $f\left(a*b\right)=f\left(a\right)*\text{'}f\left(b\right),\forall a,b\in G$.

Definition of kernel of a function:

Let $f:G\to H$be a homomorphism of groups. Then, the kernel of $f$is defined by the set localid="1656474663536" $\left\{a\in G:f\left(a\right)=e_H\right\}$, where $e_H$is an identity element.

It is the mapping from elements in $G$onto an identity element in $H$by the homomorphism $f$.

Definition of Group Isomorphism:

A mapping i$\psi :G\to H$is called a group isomorphism if it is a homomorphism and there exists a homomorphism $\psi \text{'}:H\to G$such that $\psi \psi \text{'}=l{d}_H$and $\psi \text{'}\psi =l{d}_G$.

Defining projection map

Let $H,K,M,N$be groups.

We have to show that $H\times K\cong H\times M\times N$.

Let $\psi :K\to M\times N$and $\psi \text{'}:M\times N\to K$be the isomorphisms.

Let ${\gamma }_1:M\times N\to M$be the projection map.

Then by problem 8, it is a surjective map on the first co-ordinate such that , ${\gamma }_1\left(m,n\right)=m$, for $\left(m,n\right)\in M,N$.

Also, let ${\gamma }_2:M\times N\to$such that ${\gamma }_2\left(m,n\right)=n$, for $\left(m,n\right)\in M\times N$.

Now, we define a map $\phi :H\times K\to H\times M\times N$by $\phi \left(h,k\right)=\left(h,{\gamma }_1\left(\psi \left(k\right)\right),{\gamma }_2\left(\psi \left(k\right)\right)\right)$.

To show φ is a homomorphism

let $\left(h_1,k_1\right),\left(h_2,k_2\right)\in H\times K$

Find $\phi \left(\left(h_1,k_1\right),\left(h_2,k_2\right)\right)$as:

$$\begin{gathered} \phi \left(\left(h_1,k_1\right),\left(h_2,k_2\right)\right)=\phi \left(\left(h_1{h}_2,k_1{k}_2\right)\right) \\ =\left(h_1{h}_2{\gamma }_1\left(\psi \left(k_1{k}_2\right)\right),{\gamma }_2\left(\psi \left(k_1{k}_2\right)\right)\right) \\ =\left(h_1{h}_2{\gamma }_1\left(\psi \left(k_1\right)\right),{\gamma }_1\left(\psi \left(k_2\right)\right),{\gamma }_2\left(\psi (k_1\right)\right),{\gamma }_2\left(\psi \left(k_2\right)\right)) \\ =\left(h_1,{\gamma }_1\left(\psi \left(k_1\right)\right),{\gamma }_2\left(\psi \left(k_1\right)\right)\right)\left(h_2,{\gamma }_1\left(\psi \left(k_2\right)\right),{\gamma }_2\left(\psi \left(k_2\right)\right)\right) \end{gathered}$$

Further simplify the equation as follows: $\phi \left(\left(h_1,k_1\right),\left(h_2,k_2\right)\right)=\phi \left(\left(h_1,k_1\right)\right)\phi \left(\left(h_2,k_2\right)\right)$

Therefore,$\phi \left(\left(h_1,k_1\right),\left(h_2,k_2\right)\right)=\phi \left(\left(h_1,k_1\right)\right)\phi \left(\left(h_2,k_2\right)\right)$.

Hence, $\phi$is a homomorphism.

To show H×K≅H×M×N

Let $\phi \text{'}:H\times M\times N\to H\times K$be the mapping defined by $\phi \left(h,m,n\right)=\left(h,\psi \text{'}\left(m,n\right)\right)$.

Now, prove that $\phi \text{'}$is a homomorphism.

Let $\left(h_1,m_1,n_1\right)\left(h_2,m_2,n_2\right)\in H\times M\times N$.

Find $\phi \text{'}\left(\left(h_1,m_1,n_1\right)\left(h_2,m_2,n_2\right)\right)$as:

$$\begin{gathered} \phi \text{'}\left(\left(h_1,m_1,n_1\right)\left(h_2,m_2,n_2\right)\right)=\phi \text{'}\left(\left(h_1{h}_2,m_1{m}_2,n_1{n}_2\right)\right) \\ =\left(h_1{h}_2\psi \text{'}\left(m_1{m}_2,n_1{n}_2\right)\right) \\ =\left(h_1{h}_2,\psi \text{'}\left(m_1,n_1\right)\psi \text{'}\left(m_2,n_2\right)\right) \\ =\left(h_1,\psi \text{'}\left(m_1,n_1\right)\right)\left(h_2,\psi \text{'}\left(m_2,n_2\right)\right) \end{gathered}$$

Further simplify the equation as follows. $\phi \text{'}\left(\left(h_1,m_1,n_1\right)\left(h_2,m_2,n_2\right)\right)=\phi \text{'}\left(\left(h_1,m_1,n_1\right)\right)\phi \text{'}\left(\left(h_2,m_2,n_2\right)\right)$

Therefore, $\phi \text{'}\left(\left(h_1,m_1,n_1\right)\left(h_2,m_2,n_2\right)\right)=\phi \text{'}\left(\left(h_1,m_1,n_1\right)\right)\phi \text{'}\left(\left(h_2,m_2,n_2\right)\right)$.

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

Now, let $\left(h,k\right)\in H\times K$then find $\phi \text{'}\left(\phi \left(h,k\right)\right)$as:

$$\begin{gathered} \phi \text{'}\left(\phi \left(h,k\right)\right)=\phi \text{'}\left(\left(h,{\gamma }_1\left(\psi \left(k\right)\right),{\gamma }_2\left(\psi \left(k\right)\right)\right)\right) \\ =\left(h,\psi \text{'}\left({\gamma }_1\left(\psi \left(k\right)\right),{\gamma }_1\left(\psi \left(k\right)\right)\right)\right) \\ =\left(h,k\right) \end{gathered}$$

Therefore, $\phi \text{'}\left(\phi \left(h,k\right)\right)=h,k$.

Hence, $\phi \text{'}\phi =l{d}_{H\times K}$ ……… (i)

Let $\left(h,m,n\right)\in H\times M\times N$then find $\phi \phi \text{'}\left(h,m,n\right)$as:

$$\begin{gathered} \phi \phi \text{'}\left(h,m,n\right)=\phi \left(h,\psi \text{'}\left(m,n\right)\right) \\ =\phi \left(h,\psi \left(\psi \text{'}\left(m,n\right)\right)\right) \\ =\left(h,m,n\right) \end{gathered}$$

Therefore, $\phi \phi \text{'}\left(h,m,n\right)=\left(h,m,n\right)$.

Hence, $\phi \phi \text{'}=l{d}_{H\times M\times N}$ ……… (ii)

Thus, from (i) and (ii) $H\times K\cong H\times M\times N$.