# $G$ 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.

• You can find the solution in Rotman , Introduction in the theory of groups pg 80 .It's a corollary to Sylow theorems.It is just the existence.. – alexb Dec 14 '15 at 15:02
• Use the Sylow theorems to find a nice way to write down $G$, then use a well-known corollary of the Chinese remainder theorem to show that $G$ is cyclic. – Mathematician 42 Dec 14 '15 at 15:05
• Use the abelian group structure theorem to reduce the problem to the case where $|G|$ is a prime power. – rogerl Dec 14 '15 at 15:05
• Is there any other way besides using Sylow theorems? Since we haven't learn it yet there should be a more elementary way to prove it. – mathchai Dec 14 '15 at 15:19

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.
• I don't understand the way you define $H$. Could you give more explanation to it? – mathchai Dec 15 '15 at 15:59