Consider the additive group $(\mathbb{Q},+)$ and let $p,q\in\mathbb{Q}\setminus\{0\}$. Show that for the cyclic groups $\langle p\rangle$ and $\langle q\rangle$, we have that $$ \langle p\rangle\cap\langle q\rangle\neq\{0\}.\tag{1} $$
I think I am only missing the very last step.
I know that there exist $k,\ell\in\mathbb{Z}$ with $k,\ell\neq 0$ such that
$$
kp=\ell q.\tag{2}
$$
What I am missing is the reason why this implies $(1)$.
I only see that $(2)$ implies $p=\frac{\ell}{k}q$ and $q=\frac{k}{\ell}p$, respectively. Hence, I get $$ \langle p\rangle = \{p^n: n\in\mathbb{Z}\}=\left\{\left(\frac{\ell}{k}\right)^n q^n: n\in\mathbb{Z}\right\},\qquad\langle q\rangle=\{q^n: n\in\mathbb{Z}\}=\left\{\left(\frac{k}{\ell}\right)^n p^n: n\in\mathbb{Z}\right\} $$
The group is additive not multiplicative hence $$\langle p \rangle = \{np:n\in \mathbb{Z}\}$$ and the desired result is immediate by $(2)$.