Is $\{ (e^{2\pi i a n},e^{2\pi i b n }) : n \in \mathbb{Z} \}$ dense in the torus, where $a,b$ irrationals such that $a/b$ irrational? I know that the set of points of the form $(e^{2\pi i t},e^{2\pi i a t })$ where $t \in \mathbb{R}$ is dense in the torus $S^1 \times S^1$ when $a \in \mathbb{R}$ is irrational. (This problem is asked here.)
I also know that the set of points of the form $e^{2\pi i a n}$ where $n \in \mathbb{Z}$ is dense in $S^1$, when $a$ is irrational. (This problem is asked here.)
I believe it is true that the set of points of the form
$$
(e^{2\pi i an},e^{2\pi i b n })
$$
is also dense in the torus, where $n \in \mathbb{Z}$ and $a,b$ are irrationals such that $a/b$ is also irrational. But I am unable to show this, and would appreciate any hint on how to proceed.

I tried viewing the torus as the unit square $[0,1] \times [0,1]$ with edges identified appropriately. So, I need to show that the set of points $$(an \pmod 1, bn \pmod 1)$$ is dense in the unit square.
In showing that the line $(e^{2\pi i t},e^{2\pi i a t })$ is dense in the torus, it suffices to show that it intersects every ball with centre on the vertical edge $\{ 0 \} \times [0,1]$. However, to apply the same idea here, I need to find an integer $n$ such that $an \pmod 1$ and $bn \pmod 1$ are both arbitrarily close to $0$.
I can find integers $n_1$ and $n_2$ such that $an_1 \pmod 1$ and $bn_2 \pmod 1$ are both arbitrarily close to $0$, by the same argument used to show that $e^{2 \pi i a n}$ is dense in $S^1$. But I'm unable to arrive at a single integer that does the job for $a$ and $b$ simultaneously.
I'm possibly missing only some small idea. Can anyone help me?
 A: Based on @kimchilover's hint, the problem can be solved in the following manner.
The closed subgroups of the torus belong to the following isomorphism classes: $S^1 \times S^1$, $S^1 \times \mathbb{Z}_m$, $\mathbb{Z}_n \times S^1$ or $\mathbb{Z}_n \times \mathbb{Z}_m$, for any $n,m \in \mathbb{N}$. Let
$$
A = \{ (e^{2\pi i a n},e^{2 \pi i b n}) : n \in \mathbb{Z} \}.
$$
Note that the closure of a subgroup is also a subgroup, so $\bar{A}$ must be of one of the above types.
Since $A$ is infinite, $\bar{A} \not\cong \mathbb{Z}_n \times \mathbb{Z}_m$.
Since $a/b$ is irrational, $\bar{A} \not\cong S^1$. So, we have ruled out the cases that $\bar{A}$ is isomorphic to $S^1 \times \{e\}$ or $\{e \} \times S^1$, where $\{ e \} = \mathbb{Z}_1 = \mathbb{Z}/1\mathbb{Z}$.
Let $m > 1$ and suppose that $\varphi : \bar{A} \to S^1 \times \mathbb{Z}_m$ is an isomorphism. Then $\varphi |_A : A \to S^1 \times \mathbb{Z}_m$ is an injective homomorphism. Since $\mathbb{Z}_m$ has the discrete topology, it must be that $\varphi |_A$ is surjective onto the second component $\mathbb{Z}_m$ for $\varphi$ to be an isomorphism. Hence,  $A$ must have some torsion, which is a contradiction.
We can similarly rule out the cases that $\bar{A}$ is isomorphic to $\mathbb{Z}_n \times S^1$ for any $n > 1$.
Thus, $\bar{A} = S^1 \times S^1$.
