Question

Show that$S=\left\{0,4,8,12,16,20,24\right\}$ is a subring of ${\mathrm{\mathbb{Z}}}_{28}$. Then, prove that the map$f$:${\mathrm{\mathbb{Z}}}_7\to S$ is given by$f\left({\left[x\right]}_7\right)={\left[8x\right]}_{28}$ is an isomorphism.

Short answer

It is proved that S is a subring of ring ${\mathrm{\mathbb{Z}}}_{28}$. The map $f$:${\mathrm{\mathbb{Z}}}_7\to S$ is an isomorphism.

Step-by-step solution

Show that S is a subring

Consider $r=4$and$s=7$, then

$\begin{array}{rcl}S& =& \left\{0,4,8,12,16,20,24\right\}\\ & =& \left\{0,r,2r,....,\left(s-\right)r\right\}\end{array}$

This implies that$\overline{)4}1.7+1\left(1\le 1<4\right)$, that is, there exists integer $k$such that $\overline{)r}ks+1$and $1\le k<r$.

Therefore, S is a subringof ring ${\mathrm{\mathbb{Z}}}_{28}$.

Show surjection and injection 

It is observed that:

$\begin{array}{rcl}{\left[x\right]}_7& =& {\left[y\right]}_7\\ x& \equiv & y\left(mod7\right)\\ & \Rightarrow & \overline{)7}x-y\end{array}$

Therefore, we can say that:

$$\begin{gathered} 28\overline{)\left(4x-y\right)\Rightarrow 28\overline{)8\left(x-y\right)}} \\ \Rightarrow {\left[8x\right]}_{28}={\left[8x\right]}_{28} \end{gathered}$$

We can say that$f$:${\mathrm{\mathbb{Z}}}_7\to S$ is well defined.

Verify that:

$\begin{array}{rcl}f\left({\left[0\right]}_7\right)& =& {\left[8·0\right]}_{28}\\ & =& {\left[0\right]}_{28}\end{array}$

$\begin{array}{rcl}f\left({\left[1\right]}_7\right)& =& {\left[8·1\right]}_{28}\\ & =& {\left[8\right]}_{28}\\ f\left({\left[2\right]}_7\right)& =& {\left[8·2\right]}_{28}\\ & =& {\left[16\right]}_{28}\\ & & \end{array}$

$\begin{array}{rcl}f\left({\left[3\right]}_7\right)& =& {\left[8·3\right]}_{28}\\ & =& {\left[24\right]}_{28}\end{array}$

$\begin{array}{rcl}f\left({\left[4\right]}_7\right)& =& {\left[8·4\right]}_{28}\\ & =& {\left[32\right]}_{28}\\ & =& {\left[4\right]}_{28}\end{array}$

role="math" localid="1648458520101" $\begin{array}{rcl}f\left({\left[5\right]}_7\right)& =& {\left[8.5\right]}_{28}\\ & =& {\left[40\right]}_{28}\\ & =& {\left[12\right]}_{28}\end{array}$

$\begin{array}{rcl}f\left({\left[4\right]}_7\right)& =& {\left[8·6\right]}_{28}\\ & =& {\left[48\right]}_{28}\\ & =& {\left[20\right]}_{28}\end{array}$

This implies that function is surjection and injection.

Show that function is isomorphism

Consider two arbitrary integers, $x$and $y$, this implies that:

$\begin{array}{rcl}f\left({\left[x+y\right]}_7\right)& =& {\left[x+y\right]}_{28}\\ & =& {\left[x\right]}_{28}+{\left[y\right]}_{28}\\ & =& f\left({\left[x\right]}_7\right)+\left({\left[y\right]}_7\right)\end{array}$

$\begin{array}{rcl}f\left({\left[xy\right]}_7\right)& =& {\left[8xy\right]}_{28}\\ & =& {\left[64xy\right]}_{28}\\ & =& {\left[8x·8y\right]}_{28}\\ & =& {\left[8x\right]}_{28}{\left[8y\right]}_{28}\\ & =& f\left({\left[x\right]}_7\right)f\left({\left[y\right]}_7\right)\end{array}$

Hence, we can say that$f$ is homomorphism. Therefore,$f$ is an isomorphism.