Fastest way to list the subgroup which has rank $5$ in $\Bbb Z_{30}$?

Question

What is the probably fastest way to list the subgroup of rank $5$ in $\Bbb Z_{30}$? We know that $\Bbb Z_{30}$ is cyclic, and $5\mid 30$, hence $\Bbb Z_{30}$ has exactly one subgroup $H$ such that $|H|=5$. However, there're several thoughts to be used to list the all elements of $H$. A method I come up in mind is that: since $\langle \overline{1}\rangle=\Bbb Z_{30}$, then $|\langle \overline{1}\rangle^i|=\frac{30}{\gcd(i,30)}=5$. We want $\gcd(i,30)=6$. And we know that $i=6$ is a solution. So $H=\{\overline{0},~\overline{6},\cdots\}$ and so on. However, is there a more elegant way or the probably fastest way? I feel my solution is a bit ugly.

Answer 1

$H$ being a subgroup of a cyclic group has to be cyclic, and let $a$ be the generator of $H$, then $5a=0$, so the obvious choice for $a$ is $30/5=6$. and $$H=\{0,6,12,18,24\}$$