Exercise 9.1.7

Proof. Let $\zeta$ be a primitive $n$th root of unity, so $o(\zeta )=n$ (where we write $o(x)$ the order of an element $x$ in a group $G$). We have proved in Exercise 3 that for all $i\in \mathbb{Z}$,

In particular, if $i$ and $n$ are relatively prime ( $n\wedge i=1$), then $o({\zeta }^i)=n$, so ${\zeta }^i$ is a primitive $n$th root of unity.

If $\xi$ is any primitive $n$th root of unity, as $\zeta$ is a generator of ${𝕌}_n$, $\xi ={\zeta }^i,0\le i<n$. As ${\zeta }^i$ is a primitive $n$th root of unity, $o({\zeta }^i)=n=\frac{n}{n\wedge i}$, so $n\wedge i=1$. (b) Let $i\in \mathbb{Z}$ relatively prime to $n$. Consider

$\phi$ is a group homomorphism.

If $\lambda \in \ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )$, then $\lambda ={\zeta }^k,k\in \mathbb{Z}$, and $1={\lambda }^i={\zeta }^{\mathit{ki}}$, thus $n\mid \mathit{ki}$. Since $n\wedge i=1$, $n\mid k$, hence $\lambda ={\zeta }^k=1$, so $\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )=\{1\}$.

The group homomorphism $\phi$ is injective, so the images of the distinct ${\gamma }_1,\cdots ,{\gamma }_r\in {𝕌}_n$ are distinct.

Conclusion: if $i\wedge n=1$, $\zeta \mapsto {\zeta }^i$ is a bijection from the set of primitive $n$th roots of unity on itself. □