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Whether there is a subgroup $G \leq (\mathbb{Q},+)$ so that $G\simeq (\mathbb{Z},+) \times (\mathbb{Z},+)$

I think I know how to prove that $(\mathbb{Q},+) \neq (\mathbb{Z},+) $ or $(\mathbb{Q},+) \neq (\mathbb{Z},+) \times (\mathbb{Z},+)$ but I have no idea how to start when the question is about a subgroup not the whole group

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    Any two nonzero elements of $\mathbb Q$ have a common multiple. – 2017-02-19

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Hint: in $\mathbb{Z}\times\mathbb{Z}$ there are two linearly independent elements $x$ and $y$, in the sense that $$ ax+by=0 \implies a=b=0 $$ (for integer $a$ and $b$).

Can you find two such elements in $\mathbb{Q}$?

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    So $f$ - isomorphism, then $0 = f(ax+by) = af(x)+bf(y)$ so if thera are such elements in $\mathbb{Z} \times \mathbb{Z}$ then there have to be such elements in $\mathbb{Q}$ ? $(1,0) and (0,1)$ are example in $\mathbb{Z} \times \mathbb{Z}$ but i still do not see why $\mathbb{Q}$ for every $x,y$ there exist $a,b \neq 0$ that the equality holds? – 2017-02-19
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    @dyrAnd If $f\colon \mathbb{Z}\times\mathbb{Z}\to\mathbb{Q}$ is an injective homomorphism with image $G$, then $x=f(1,0)$ and $y=f(0,1)$ are linearly independent in $\mathbb{Q}$. Can they be? Write $x=\frac{m}{n}$ and $y=\frac{p}{q}$. Then $pnx-qmy=0$ and… – 2017-02-19
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Hint: try to prove that every finitely generated subgroup of $\mathbb{Q}$ is actually cyclic.