Question

Let S be the subring $\left\{0,2,4,6,8\right\}$ of ${\mathit{Z}}_{\mathbf{10}}$and let ${\mathit{Z}}_{\mathbf{5}}\mathbf{=}\left\{\overline{0},\overline{1},\overline{2},\overline{3},\overline{4}\right\}$(notation as in Example 1). Show that the following bijection from ${\mathit{Z}}_{\mathbf{5}}$ to $\mathit{S}$is not an isomorphism:

$\overline{\mathbf{0}}\mathbf{\to }\mathbf{0}\mathbf{}\mathbf{}\mathbf{}\overline{\mathbf{1}}\mathbf{\to }\mathbf{2}\mathbf{}\mathbf{}\overline{\mathbf{2}}\mathbf{\to }\mathbf{4}\mathbf{}\mathbf{}\overline{\mathbf{3}}\mathbf{\to }\mathbf{6}\mathbf{}\mathbf{}\mathbf{}\overline{\mathbf{4}}\mathbf{\to }\mathbf{8}$

Short answer

It is proved that $S$is not an isomorphism

Step-by-step solution

Property of Rings

If any ringRhas elements such that, $a,b\in R$, then, addition and multiplication of the function of its elements is respectively given by:

$$\begin{gathered} f\left(a+b\right)=f\left(a\right)+f\left(b\right) \\ f\left(ab\right)=f\left(a\right)f\left(b\right) \end{gathered}$$

Isomorphism

The given subring can be expressed in the form of bijection as:

$f:Z_6\to S=\left\{0.2,6,4,8\right\}$such that:

$\frac{f}{2}\mapsto 4$and$\frac{f}{4}\mapsto 8$

Now, we get:

$$\begin{gathered} f\left(\overline{22}\right)=f\left(\overline{4}\right)=8 \\ f\left(\overline{2}\right)f\left(\overline{2}\right)=4.4=6 \end{gathered}$$

Clearly, we see:

$f\left(\overline{22}\right)\ne f\left(\overline{2}\right)f\left(\overline{2}\right)$

Hence proved, $f$ is not an isomorphism.