How many elements $a$ with $o(a)\mid 5$ in $\mathbb{Z}/360\mathbb{Z}$

Question

I am trying to understand why there are exactly $5$ elements of order dividing $5$ in $\mathbb{Z}/360\mathbb{Z}$.

Answer 1

Hint: $\mathbb{Z}/360\mathbb{Z}$ is a cyclic group. So there is exactly one subgroup of order $5.$ So...?

$\bf{EDIT:}$ This link (kindly provided by Bill Dubuque) says that, if $G$ is a cyclic group of order $n$ and if $d$ is +ve integer dividing $n$, then there is a unique subgroup of $G$ order $d.$ Applying this result in this particular case $(n = 360, d=5),$ we get that $\mathbb{Z}/360\mathbb{Z}$ has a unique subgroup $H$ of order $5.$ Surely for any $a \in H, o(a)|5$ (why?). Now if there is an element of order $5$ in $G$ which is not in $H$ then that element will generate a subgroup of order $5$ which is different from $H.$ But his can't happen because the above mentioned result says that $H$ is unique.