Any finitely generated subgroup of $(\mathbb{Q},+)$ is cyclic. 
$\fbox{1}$ 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.
 A: Hint 1. Given any two elements $a,b\in\mathbb{Q}$, can you find an element $r\in\mathbb{Q}$ such that $a=mr$ and $b=nr$ for some integers $m$ and $n$? If so, then $\langle a,b\rangle\subseteq \langle r\rangle$; what do we know about subgroups of cyclic groups.
Hint 2. Is there a finitely generated subgroup of $\mathbb{Q}\times\mathbb{Q}$ that is not cyclic?
A: 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.
A: Hint 1. Suppose $\frac{m_1}{n_1},\frac{m_2}{n_2},\ldots,\frac{m_k}{n_k}$ are the generators of $G\leq\Bbb Q$. We can use these generators to construct an element $\frac mn$ which already generates $G$.
Hint 2. Suppose $f:\Bbb Q\to\Bbb Q\times\Bbb Q$ is an isomorphism. How are $f(1)$ and $f(\frac mn)$ related?
A: If you are stuck, then it often helps to look at special cases. Here it turns out that if you can prove that subgroups generated by two elements are cyclic, then this will quickly imply the full result. To be even more concrete, let's consider the subgroup generated by $\frac12$ and $\frac13$.
How is this going to work? The subgroup consists of elements of the form, for $n, m \in \mathbb Z$,
\[
\frac n2 + \frac m3 = \frac{3n + 2m}6.
\]
The numerator here should set off some bells, and I think you can now prove that the subgroup is generated by $\frac16$. Can you make this work for two general rational numbers?
One way to do the second part is to look at the finitely generated subgroup $\mathbb Z \times \mathbb Z$ of $\mathbb Q \times \mathbb Q$.
