Uniqueness of a subgroup of a given order

Question

Let $G$ be a cyclic group with order $n$. Prove that for every divisor $d$ of $n$ there is a unique subgroup with order $d$.

For the existence, let $x$ be a generator of $G$. It is easy to check that the order of $x^{n/d}$ is $d$.

I'm struggling with uniqueness. Let $H$ be a subgroup of $G$ with order $d$. I want to prove that $H=<x^{n/d}>$.

If $H$ is not trivial, there is some $r>0$ such that $<x^r>= H$. The order of $x^r$ in $G$ is both $d$ and $\frac{n}{\gcd(n,r)}$, hence $d=\frac{n}{\gcd(n,r)}$

So ${\gcd(n,r)} = \frac{n}{d}$ and it remains to prove that $\gcd(n,r)=r$, but why should $r$ divide $n$ ?

Answer 1

Hint.

Suppose that you have two subgroups $H_1$ and $H_2$ both having order $d$. Then $H_1 \cap H_2$ is a subgroup of $G$ which is cyclic. Hence $H_1 \cap H_2$ is cyclic. Try then to derive a contradiction is you suppose $|H_1 \cap H_2| < d$

Answer 2

The link posted here Subgroups of a cyclic group and their order. actually addresses my concerns.

According to theorem 3.6 one has $<x^r>=<x^{\gcd(n,r)}>$. Therefore, it may be assumed WLOG that $r$ divides $n$, which answers my question.