Question

Let$\mathit{f}\mathbf{:}{\mathit{Z}}_{\mathbf{6}}\mathbf{\to }{\mathit{Z}}_{\mathbf{2}}\mathbf{\times }{\mathit{Z}}_{\mathbf{3}}$ be the bijection given by $\mathbf{0}\mathbf{\to }\left(0,0\right)\mathbf{,}\mathbf{}\mathbf{1}\mathbf{\to }\left(1,1\right)\mathbf{,}\mathbf{}\mathbf{2}\mathbf{\to }\left(0,2\right)\mathbf{,}\mathbf{}\mathbf{3}\mathbf{\to }\left(1,0\right)\mathbf{,}\mathbf{}\mathbf{4}\mathbf{\to }\left(0,1\right)\mathbf{,}\mathbf{}\mathbf{5}\mathbf{\to }\left(1,2\right)$

Use the addition and multiplication tables of ${\mathit{Z}}_{\mathbf{6}}$ and ${\mathit{Z}}_{\mathbf{2}}\mathbf{\times }{\mathit{Z}}_{\mathbf{3}}$ to show that $\mathit{f}$is an isomorphism.

Short answer

It is proved that $f$is an isomorphism

Step-by-step solution

Property of Rings

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

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

Addition Table 

The given mapping is:$f:Z_6\to Z_2\times Z_3$

Now, let there be elements as: $a,b\in Z_6$

Then, for the addition table, we have:

+	0	1	2	3	4	5
0	0	1	2	3	4	5
1	1	2	3	4	5	0
2	2	3	4	5	0	1
3	3	4	5	0	1	2
4	4	5	0	1	2	3
5	5	0	1	2	3	4

+	(0,0)	(1,1)	(0,2)	(1,0)	(0,1)	(1,2)
(0,0)	(0,0)	(1,1)	(0,2)	(1,0)	(0,1)	(1,2)
(1,1)	(1,1)	(0,2)	(1,0)	(0,1)	(1,2)	(0,0)
(0,2)	(0,2)	(1,0)	(0,1)	(1,2)	(0,0)	(1,1)
(1,0)	(1,0)	(0,1)	(1,2)	(0,0)	(1,1)	(0,2)
(0,1)	(0,1)	(1,2)	(0,0)	(1,1)	(0,2)	(1,0)
(1,2)	(1,2)	(0,0)	(1,1)	(0,2)	(1,0)	(0,1)

Clearly we have$f\left(a+b\right)=f\left(a\right)+\left(b\right)$

Multiplication Table

Similarly, let there be elements as$a,b\to Z_2\times Z_3$:

Then, for the multiplication table, we have

$·$	0	1	2	3	4	5
0	0	0	0	0	0	0
1	0	1	2	3	4	5
2	0	2	4	0	2	4
3	0	3	0	3	0	3
4	0	4	2	0	4	2
5	0	5	4	3	2	1

+	(0,0)	(1,1)	(0,2)	(1,0)	(0,1)	(1,2)
(0,0)	(0,0)	(0,0)	(0,0)	(0,0)	(0,0)	(0,0)
(1,1)	(0,0)	(1,1)	(0,2)	(1,0)	(0,1)	(1,2)
(0,2)	(0,0)	(0,2)	(0,1)	(0,0)	(0,2)	(0,1)
(1,0)	(0,0)	(1,0)	(0,0)	(1,0)	(0,0)	(1,0)
(0,1)	(0,0)	(0,1)	(0,2)	(0,0)	(0,1)	(0,2)
(1,2)	(0,0)	(1,2)	(0,1)	(1,0)	(0,2)	(1,1)

Clearly, we have:$f\left(ab\right)=f\left(a\right)f\left(b\right)$

Hence, fis an isomorphism