Question

Conclude that $\mathit{K}\mathbf{/}\mathbf{(}\mathbf{N}\mathbf{\cap }\mathbf{K}\mathbf{)}\mathbf{\cong }\mathit{N}\mathit{K}\mathbf{/}\mathit{N}$.

Short answer

It is proved that,$K/(N\cap K)\cong NK/N$

Step-by-step solution

Step by Step Solution Step 1: First Isomorphism Theorem

Let $\mathit{f}\mathbf{:}\mathit{G}\mathbf{\to }\mathit{H}$ be a surjective homomorphism of a group with kernel K. Then, quotient group $G/K$is isomorphic to H.

Proving f is a homomorphism

Consider that $f\left(k\right)=Nk$

For any arbitrary $k_1,k_2\in K$, find $f\left(k_1{k}_2\right)$as:

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

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$a=nk$.

Find $Na$as:

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

Therefore, f is surjective.

Finding ker f

We know that kernel of f can be written 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, $Nk=N$.

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

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

Since f is surjective homomorphism with $\ker f=N\cap K$, byFirst Isomorphism Theorem $K/(N\cap K)\cong NK/N$.