Question

Suppose that $\mathit{k}\mathbf{,}\mathit{n}\mathbf{,}\mathit{r}$ are positive integers such that$\mathit{k}\mathbf{|}\mathit{n}$ .Show that the function $\mathit{f}\mathbf{:}{\mathbf{\mathbb{Z}}}_{\mathbf{n}}\mathbf{\to }{\mathbf{\mathbb{Z}}}_{\mathbf{k}}$given by$\mathit{f}\mathbf{\left(}{\mathbf{\left[}\mathbf{a}\mathbf{\right]}}_{\mathbf{n}}\mathbf{\right)}\mathbf{=}{\mathbf{\left[}\mathbf{r}\mathbf{a}\mathbf{\right]}}_{\mathbf{k}}$ is well defined (meaning that if ${\mathbf{\left[}\mathbf{a}\mathbf{\right]}}_{\mathbf{n}}\mathbf{=}{\mathbf{\left[}\mathbf{b}\mathbf{\right]}}_{\mathbf{n}}$, then${\mathbf{\left[}\mathbf{r}\mathbf{a}\mathbf{\right]}}_{\mathbf{k}}\mathbf{=}{\mathbf{\left[}\mathbf{r}\mathbf{b}\mathbf{\right]}}_{\mathbf{k}}$ ).

Short answer

It is proved that,$f$ is well defined.

Step-by-step solution

To define elements of ℤn

Definition of Group Homomorphism

Let $(G,\ast )$and $(G\text{'},\ast \text{'})$be any two groups. A function $f:G\to G\text{'}$is said to be a group homomorphism if $f(a\ast b)=f\left(a\right)\ast \text{'}f\left(b\right),\forall a,b\in G$.

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 $\{a\in G:f\left(a\right)=e_H\}$, where $e_H$is an identity element.

It is the mapping from elements in $G$onto an identity element in $H$by the homomorphismrole="math" localid="1654594222682" $f$.

Let $k,n,r$be the positive integers such that $k|n$.

We have to show that the function $f$is well defined.

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

Then, we have to show that ${\left[ra\right]}_k={\left[rb\right]}_k$.

To show f is well defined

Now, we have${\left[a\right]}_n={\left[b\right]}_n$ , $n|b-a$$\Rightarrow n|r(b-a)$.

Since $k|n,k|r(b-a)$, simplify as:

$$\begin{gathered} {\left[ra\right]}_k={\left[rb\right]}_k \\ f\left({\left[a\right]}_k\right)=f\left({\left[b\right]}_k\right) \end{gathered}$$

Hence, role="math" localid="1654594734325" $f$is well defined.