TITLE: Any finitely generated subgroup of $(\mathbb{Q},+)$ is cyclic.
QUESTION [9 upvotes]: $\fbox{1}$ Prove that any finitely generated subgroup of $(\mathbb{Q},+)$ is cyclic.
$\fbox{2}$ Prove that $\mathbb{Q}$ is not isomorphic to $\mathbb{Q}\times \mathbb{Q}$.

Any hints would be appreciated.

REPLY [12 votes]: Let $\frac{m_1}{n_1},...,\frac{m_k}{n_k}$ be the generators of $G\leq\Bbb Q$ and $g\in G$. Then we have $g=a_1\frac{m_1}{n_1}+...+a_k\frac{m_k}{n_k}=\frac{a_1m_1n_2...n_{k}+...+a_km_kn_1...n_{k-1}}{n_1...n_k}$ for some integers $a_1,...,a_k$. Thus $G$ is a subgroup of the cyclic group $<\frac{1}{n_1...n_k}>$. Hence $G$ must be cyclic.