generator

If $G$ is a cyclic group and $g\in G$, then $g$ is a generator of $G$ if $⟨g⟩=G$.

All infinite cyclic groups have exactly $2$ generators. To see this, let $G$ be an infinite cyclic group and $g$ be a generator of $G$. Let $z\in \mathbb{Z}$ such that $g^z$ is a generator of $G$. Then $⟨g^z⟩=G$. Then $g\in G=⟨g^z⟩$. Thus, there exists $n\in \mathbb{Z}$ with $g={(g^z)}^n=g^{nz}$. Therefore, $g^{nz-1}=e_G$. Since $G$ is infinite and $|g|=|⟨g⟩|=|G|$ must be infinity, $nz-1=0$. Since $nz=1$ and $n$ and $z$ are integers, either $n=z=1$ or $n=z=-1$. It follows that the only generators of $G$ are $g$ and $g^{-1}$.

A finite cyclic group of order $n$ has exactly $\phi (n)$ generators, where $\phi$ is the Euler totient function. To see this, let $G$ be a finite cyclic group of order $n$ and $g$ be a generator of $G$. Then $|g|=|⟨g⟩|=|G|=n$. Let $z\in \mathbb{Z}$ such that $g^z$ is a generator of $G$. By the division algorithm, there exist $q,r\in \mathbb{Z}$ with such that $z=qn+r$. Thus, $g^z=g^{qn+r}=g^{qn}{g}^r={(g^n)}^q{g}^r={(e_G)}^q{g}^r=e_G{g}^r=g^r$. Since $g^r$ is a generator of $G$, it must be the case that $⟨g^r⟩=G$. Thus, $n=|G|=|⟨g^r⟩|=|g^r|={\displaystyle \frac{|g|}{\gcd (r,|g|)}}={\displaystyle \frac{n}{\gcd (r,n)}}$. Therefore, $\gcd (r,n)=1$, and the result follows.