Question

Prove that the function$\mathit{f}\mathbf{:}\mathit{K}\mathbf{\to }\mathit{N}\mathit{K}\mathbf{/}\mathit{N}$ given by$\mathit{f}\mathbf{\left(}\mathbf{k}\mathbf{\right)}\mathbf{=}\mathit{N}\mathit{k}$ is a surjective homomorphism with kernel $\mathit{K}\mathbf{\cap }\mathit{N}$.

Short answer

It is proved that the function$f:K\to NK/N$ given by$f\left(k\right)=Nk$ is a surjective homomorphism with kernel $K\cap N$.

Step-by-step solution

Step by Step Solution Step 1: Proving f is a homomorphism

Consider that $f:K\to NK/N$given by $f\left(k\right)=Nk$.

For any arbitrary elements $k_1,k_2\in K$

$\begin{array}{c}f(k_1,k_2)=N{k}_1{k}_2\\ =N{k}_{1}N{k}_2\\ =f\left(k_1\right)f\left(k_2\right)\end{array}$

Therefore, f is a homomorphism.

Proving f is surjective

For any arbitrary $a\in NK/N,$there exists $n\in N,k\in K$such that $a=nk$.

$\begin{array}{c}Na=NnK\\ =Nk\\ =f\left(k\right)\end{array}$

Therefore, f is surjective.

Finding ker f

We can find kernel of f as:

$\ker f=\text{}\{k\in K:\text{\hspace{0.17em}\hspace{0.17em}}f\left(k\right)=N\}$

$\ker f=\text{}\{k\in K:\text{\hspace{0.17em}\hspace{0.17em}}Nk=N\}$

From this, we can see that, $Nk=N$.

Therefore,$k\in N$implies$k\in \ker f$if $k\in N$.

Thus, $\ker f=N\cap K$.

Hence, from steps 1, 2, and 3, it is proved that the function$f:K\to NK/N$ given by$f\left(k\right)=Nk$ is a surjective homomorphism with kernel $K\cap N$.