$G$ is a finite abelian group. For every prime $p$ that divides $|G|$, there is a unique subgroup of order $p$. $G$ is a finite abelian group. Assume that for every prime $p$ that divides $|G|$, there is a unique subgroup of order $p$. I'd like to prove that $G$ is cyclic. I'm thinking about the approach of induction but not able to develop a complete proof yet.

Set $G = \{g_{1}, g_{2}, ..., g_{n}\}$, with $|g_{i}| =d_{i}$.
Consider the group $H = \prod_{i=1}^{n}\mathbb{Z}\backslash d_{i}\mathbb{Z}$.
Since $G$ is abelian, the map $f: H \rightarrow G: f(g_{1}, g_{2}, ..., g_{n}) = \prod_{i=1}^{n}g_{i}^{k_{i}}$ is a well-defined surjective group homomorphism so that by one of the isomorphism theorems, $\prod_{i}^{n}d_{i} = |H| = |\ker f| |G| \implies p|d_{j}$ for some $j$. The element $g = g_{i}^{\frac{d_{j}}{p}}$ has order $p$ just as it was promised.
