Subgroup of a cyclic finite group

Question

Let $G$ be a cyclic group of order $n$ and let $m$ be a positive integer dividing $n$. Show that $G$ contains one and only subgroup of order $m$.

I have started the proof by stating the smallest positive integer is $m$ such that $a^m$ exists in $H$.

Answer 1

Hints:

(a) A subgroup of a cyclic group of order $n$, generated by $a$ is generated by some $a^k\enspace(0\le k<n)$.

(b) The order of $a^k\,$ is equal to $\,\dfrac n{\gcd(n,k)}$.