Lie Subgroup Example - Explanation? I'm currently working through Jeff Lee's 'Manifolds and Differential Geometry'. He defines a Lie Subgroup, $H$, to be an abstract subgroup of a Lie Group $G$, such that the inclusion map $H\hookrightarrow G$ is an immersion. He then lists an example of a Lie Subgroup which I am trying to understand. 
Defining $S^1$ to be the unit circle in $\mathbb{C}$, he writes:
'The image in the torus $T^2=S^1\times S^1$ of the map $\mathbb{R}\rightarrow S^1\times S^1$ given by $t\mapsto (e^{2\pi i t},e^{2\pi i a t})$ is a Lie Subgroup. This map is a homomorphism. If $a$ is rational, then the image is an embedded copy of $S^1$ wrapped around the torus several times depending on $a$. If $a$ is irrational, then the image is still a Lie subgroup but it is now dense in $T^2$.
I am struggling to understand this example at all. I can see why the map is a homomorphism, but I do not understand his argument about the rationality/irrationality of $a$. I also wouldn't know how to show the image of the map is a Lie subgroup. That is, I wouldn't know how to show the inclusion map is a submersion.
Any help with this would be very much appreciated!
 A: Draw the torus as a square and start drawing the lines (the embedding of $S^1$. The parameter $a$ gives you the slope of a line in the plane where the square is embedded. Start at zero with that slope. Once you reach the border, use the (usually) unique other boundary point to start over again drawing a new line with the same slope.
Now: If you have a rational slope, you will end up again at you starting point after some (how many?) repitions of drawing. If you have a irrational slope you will keep on drawing new lines and you will figure, that you come arbritrary close to every point in that square.
A: First notice that  $e^{i2\pi a m}\ne e^{i2\pi a n}$ for $n,m \in \mathbb{Z}$ distinct. This is because equality is possible only if exist $k\in \mathbb{Z}$ such that
$$am+k=an \implies a(m-n)=k.$$
This contradicts the irrationality of $a$.
The above result implies that the set $A:=\{e^{i2\pi na}:n\in \mathbb{Z}$} is not finite for $a$ irrational. We will use this to show that $A$ is a dense set in $S^1$. Partition $S^1$ to $N$ sets of length $\frac{2\pi}{N}$. By the pigeonhole principle, at least 2 distinct elements of $A$ exist in one of the length $\frac{2\pi}{N}$ sets. Abbreviate them by $r_1,r_2$. Notice that $r_1 \cdot r_2^{-1}$ is a rotation with an angle less than $$\frac{2\pi}{N},$$ hence $\{(r_1 \cdot r_2^{-1})^n: n\in \mathbb{Z}\}\subset A$ has a nonempty intersection with each of the sets in length $\frac{2\pi}{N}$ partition of $S^1$. As $N$ was arbitrary, this shows that $A$ is dense in $S^1$.
It was essential in our proof that $a$ is irrational; otherwise, $A$ is not infinite, and we can't use the pigeonhole principle for all $N\in \mathbb{N}$.
Back to your problem. Let $(a,b)\in T^2$ then there exists $t_0$ such that $e^{i2\pi (t_0+n)}=a$ for all $n\in \mathbb{N}$. We have that $e^{i2\pi(t_0+n)a}=e^{i2\pi t_0a}\cdot e^{i2\pi n a}.$  By what we have shown above there exists $n$ such that $ e^{i2\pi n a}$ is arbitrarily close to $e^{-i2\pi t_0a}\cdot b$. Hence for $t=t_0+n$ we have $\left(e^{2 \pi i t}, e^{2 \pi i a t}\right)=(a,e^{2 \pi i a t})$ is arbitrarily close to $(a,b)$.
To show that the map is an immersion, notice that the $t\mapsto e^{it}$ is the exponential map of the lie group $S^1$. The exponential map is known to be an immersion, so a composition of it with another immersion, such as $t \mapsto 2 \pi t$ will also be an immersion. It follows that $t \mapsto \left(e^{2 \pi i t}, e^{2 \pi i a t}\right)$ is an immersion.
