Counting cyclic subgroups by their generators ($\#\{H\le G:H\text{ cyclic},|H|=m\}=\#\{g:|g|=m\}/\phi (m)$)

In a finite group, a cyclic subgroup of order $m$ has exactly $\phi (m)$ generators, where $\phi$ is the Euler totient function. Each element generates exactly one cyclic subgroup, so dividing the number of elements of order $m$ by $\phi (m)$ counts these subgroups. Include the identity subgroup using $m=1$ and $\phi (1)=1$. In a symmetric group, element counts come from cycle type and order of a group element is the least common multiple of cycle lengths.