Exercise 4.13

Proof. Suppose $G=⟨g⟩$, with $\mathrm{Card}G=n$, so that the order of $g$ is $n$.

Let $x$ be another generator of $G$, then $x=g^k$, and $g=x^l,k,l\in \mathbb{Z}$, so $g=g^{\mathit{kl}},g^{\mathit{kl}-1}=e:n\mid \mathit{kl}-1$, then $\mathit{kl}-1=\mathit{qn},q\in \mathbb{Z}$, so $n\wedge k=1$.

Conversely, if $u\wedge k=1$, there exist $u,v\in \mathbb{Z}$ such that $\mathit{un}+\mathit{vk}=1$, so $g=g^{\mathit{un}+\mathit{vk}}={(g^n)}^u{(g^k)}^v=x^v\in ⟨x⟩$, so $G\subset ⟨x⟩$, $G=⟨x⟩$, i.e. $x$ is a generator of $G$.

Conclusion: if $g$ is a generator of $G$, all the other generators are the elements $g^k$, where $k\wedge n=1$, $n=|G|$. □