TITLE: A finite group which has a unique subgroup of order $d$ for each $d\mid n$.
QUESTION [14 upvotes]: Problem
Suppose G is a finite group of order $n$ which has a unique subgroup of order $d$ for each $d\mid n$. Prove that $G$ must be a cyclic group.
My idea:   I try to prove it by induction. Let $p|n$ be a prime. Then by condition, there exists a unique subgroup $H$ of order $n/p$. Since $|gHg^{-1}|=|H|$, we must have $gHg^{-1}=H$ by the uniqueness part of the condition. So, H is a normal subgroup.
Now, $|G/H|=p$ and thus $G/H=\langle x\rangle$ where $x^p \in H$.  
However, I cannot continue.
My another idea is that first consider the case when $|G|$ is a power of some prime $p$. But, it still doesn't work.

REPLY [2 votes]: Keith Conrad (citing Trevor Hyde for part of the proof) gives a proof in Appendix A of this article.
Here is an outline of the proof:
Proposition. Let $G$ be a finite group. Suppose that for each positive divisor $d$ of $|G|$, there is at most one subgroup of order $d$. Then $G$ is cyclic.
Proof.
First suppose that $|G|=p^m$, where $p$ is a prime.
Let $g\in G$ be an element with a maximum order.
Let $h\in G$ be any element.
Let $|g|=p^k$ and $|h|=p^\ell$.
Then $p^k\geq p^\ell$, so $p^\ell\mid p^k$.
Since $\langle g\rangle$ is cyclic, it has a (unique) subgroup of order $p^\ell$.
But $\langle h\rangle$ has order $p^\ell$ as well, so our assumption implies that they are the same set. Hence, $\langle h\rangle\subseteq\langle g\rangle$, and in particular $h\in\langle g\rangle$. Since $h$ was arbitrary, we then have $G\subseteq\langle g\rangle$, i.e., $G=\langle g\rangle$.
Next, suppose that $|G|=p_1^{m_1}\cdots p_r^{m_r}$, where $p_1,\ldots,p_r$ are distinct primes.
Let $H_i\in\operatorname{Syl}_{p_i}(G)$.
Our assumption implies that $n_{p_i}=1$ for each $i=1,\ldots,r$, so
$$G\simeq H_1\times\cdots\times H_r.$$
On the other hand, by the first part of the proof, $H_i\simeq\mathbb{Z}_{p_i^{m_i}}$ for each $i=1,\ldots,r$. By the Chinese remainder theorem, $G$ is cyclic.
Corollary. Let $G$ be a finite group. TFAE:

$G$ is cyclic.
For each positive divisor $d$ of $|G|$, there is at most one subgroup of order $d$.
For each positive divisor $d$ of $|G|$, there exists a unique subgroup of order $d$.