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 $0\leq r<n$ 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 $\langle g^{r}\rangle=G$. 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.