Let $(\mathbb{Q},+)$ be additive group of rational numbers. If $G_1, G_2$ are two nonzero subgroup of $(\mathbb{Q},+)$, prove that $G_1\cap G_2\neq\{0\}$.
My approach :
Take $p/q\in G_1$ and $a/b \in G_2$. Then $p/q+p/q+..=p \in G_1 $ ($q$ times addition). Similarly $a\in G_2$. Now we can always choose $p$ and $a$ as positive because in group their inverse are of opposite parity.
Now set $l=lcm(p,a)$. Since $l=\frac{pa}{d}$ where $d=\gcd(p,a)$ one finds $l\in G_1$.
Reason : Just $p+...p=l$ where (addition is done $\frac{a}{d}$ times).
similarly $l\in G_2$.
So $l\in G_1\cap G_2 $. Hence $G_1\cap G_2\neq \{0\}$. Proved
I would like to know if my solution is correct. This was supposed to be a tough problem. So it feels like i have committed some mistake in proof. Thank you.
seems ok! both groups are Z-modules so $ bp\frac{a}{b}=pa \in G_1$ and similarly $aq\frac{p}{q} = ap \in G_2$