Question

Question: Prove that the additive group${\mathit{Z}}_{\mathbf{6}}$is isomorphic to the multiplicative group of non zero elements in ${\mathit{Z}}_{\mathbf{7}}$.

Short answer

It has been proved that additive group $Z_6$is isomorphic to the multiplicative group of non zero elements in ${\mathrm{\mathbb{Z}}}_7$.

Step-by-step solution

Step-By-Step SolutionStep 1: Shoe that the multiplicative group of non -zero elements in ℤ7is cyclic

Group of non-zero elements in ${\mathrm{\mathbb{Z}}}_7$is denoted by ${\mathrm{\mathbb{Z}}}_7^{*}$.

${\mathrm{\mathbb{Z}}}_7^{*}=\{1,2,3,4,5,6\}$

Clearly, by computing the order of 3, it can be seen that is a cyclic group with generator 3.

Use theorem 7.19

Theorem 7.19 states that, “If G is a cyclic group of finite order n, then G is isomorphic to the additive group $Z_n$”

Now, ${\mathrm{\mathbb{Z}}}_7^{*}$is a cyclic group with order 6.

Hence, it is isomorphic to additive group${\mathrm{\mathbb{Z}}}_6$

Conclusion

Hence, additive group ${\mathrm{\mathbb{Z}}}_6$is isomorphic to the multiplicative group of non-zero elements in ${\mathrm{\mathbb{Z}}}_7$.