If $G$ is a cyclic group and $g \in G$ then $g$ is a generator of $G$ if $\langle g \rangle =G$

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 $\langle g^z \rangle =G$ Then $g \in G= \langle g^z \rangle$ 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 $|g|=|\langle g \rangle |=|G|$ 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 $\varphi(n)$ generators, where $\varphi$ 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 $|g|=|\langle g \rangle |=|G|=n$ 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 $0 \le r<n$ such that $z=qn+r$ Thus, $g^z=g^{qn+r}=g^{qn}g^r=(g^n)^qg^r=(e_G)^qg^r=e_Gg^r=g^r$ Since $g^r$ is a generator of $G$ it must be the case that $\langle g^r \rangle =G$ Thus, $\displaystyle n=|G|=|\langle g^r \rangle|=|g^r|=\frac{|g|}{\gcd(r,|g|)}=\frac {n}{\gcd(r,n)}$ Therefore, $\gcd(r,n)=1$ and the result follows.