Question

If $\mathit{k}\mathbf{|}\mathit{n}$, prove that ${\mathbb{Z}}_n/⟨k⟩\cong {\mathbb{Z}}_k$. [Exercise 11 may be helpful.]

Short answer

It is proved that, ${\mathbb{Z}}_n/K\cong {\mathbb{Z}}_k$.

Step-by-step solution

First Isomorphism Theorem

Let $f:G\to H$be a surjective homomorphism of groups with kernel K. Then the quotient group $G/K$ is isomorphic to H.

 f is surjective 

Let n,k be positive integer such that$k|n$ .

Define a map $f:{\mathbb{Z}}_n\to {\mathbb{Z}}_k$by$f\left({\left[a\right]}_n\right)={\left[a\right]}_k$ .

Let $x\in {\mathbb{Z}}_k$ and choose$n\le a\le k$ such that $x={\left[a\right]}_k$.

Then, there exists an element ${\left[a\right]}_n$ in${\mathbb{Z}}_n$ such that$f\left({\left[a\right]}_n\right)={\left[a\right]}_k=x$.

Thus, f is surjective.

f  is a homomorphism

Let .${\left[a\right]}_n,{\left[b\right]}_n\in {\mathbb{Z}}_n$

Then, prove f is a homomorphism as follows:

$\begin{array}{c}f({\left[a\right]}_n+{\left[b\right]}_n)=f\left({[a+b]}_n\right)\\ ={[a+b]}_k\\ ={\left[a\right]}_k+{\left[b\right]}_k\\ =f\left({\left[a\right]}_n\right)+f\left({\left[b\right]}_n\right)\end{array}$

Therefore, f is a homomorphism.

Kernel  f

The kernel f will be the set $Ker\left(f\right)=\{{\left[a\right]}_n\in {\mathbb{Z}}_n:f\left({\left[a\right]}_n\right)=0\text{\hspace{0.17em}in\hspace{0.17em}}{\mathbb{Z}}_k\}$that can be written as$Ker\left(f\right)=\{{\left[a\right]}_n\in {\mathbb{Z}}_n:a\equiv 0\mathrm{mod}k\}$.

Thus, $\ker f$ consists of all elements $\{0,1,2,\dots ,n-1\}$which are divisible by k that is, $\ker f=\{0,k,2k,\dots ,\frac{n}{k}k\}$.

Hence, $\ker f=⟨k⟩$.

Here, f is a surjective homomorphism with the kernel $⟨k⟩$ thenbyFirst Isomorphism Theorem,${\mathbb{Z}}_n/K\cong {\mathbb{Z}}_k$ .

Therefore, if $k|n$ then ${\mathbb{Z}}_n/K\cong {\mathbb{Z}}_k$.