Question

Prove that$\mathbf{\text{Aut}}{\mathit{\mathbb{Z}}}_{\mathbf{n}}\mathbf{\cong }{\mathit{U}}_{\mathbf{n}}$ .

Short answer

It is proved that$\text{Aut}{\mathbb{Z}}_n\cong U_n$

Step-by-step solution

 Step 1: Isomorphic Groups

Two groups $\mathit{G}$and$\mathit{H}$are isomorphic if there exists an isomorphism$\mathit{f}$from$\mathit{G}$onto$\mathit{H}$.

Prove that Autℤn≅Un

According to exercise 7.3.44, the generators of ${\mathrm{\mathbb{Z}}}_n$are elements in $U_n$. Exercise 26 tells that any automorphism of ${\mathrm{\mathbb{Z}}}_n$send 1 to some element in $U_n$, and exercise 25 tells that both options indeed induce an automorphism of $\mathrm{\mathbb{Z}}$.

Thus, homomorphism from cyclic groups is completely evaluated by where the generators are sent. Therefore, it can be concluded that an isomorphism between $U_n$and role="math" localid="1651581116530" $\text{Aut}{\mathbb{Z}}_n$given by the following map.

$\phi :U_n\to \text{Aut}{\mathbb{Z}}_n;\phi \left({\left[a\right]}_n\right)\left({\left[b\right]}_n\right)={\left[ba\right]}_n$

Therefore, the map $\phi :U_n\to \text{Aut}{\mathbb{Z}}_n;\phi \left({\left[a\right]}_n\right)\left({\left[b\right]}_n\right)={\left[ba\right]}_n$is an

isomorphism between $U_n$and $\text{Aut}{\mathbb{Z}}_n$

Hence, it is proved that$\text{Aut}{\mathbb{Z}}_n\cong U_n$