Foliation dense if $G = \textbf{R}$, where $G$ is a subgroup of a Lie group $G'$.

Question

I have the following statement:

Let $G$ be a subgroup of a lie group $G'$, and the action is left multiplication. The leaves are then the left cosets of $G$ in $G'$. If for example, we let $G = \textbf{R}$ be a noncompact 1-parameter subgroup of a torus (a line of irrational slope), then every leaf of the resulting foliation is dense.

I understand the concept of foliation, but I cannot understand why the foliation is dense if $G = \textbf{R}$, what is the intuition behind it?

Answer 1

It is certainly not true in general that a $\Bbb R$ subgroup is dense. What you want, for the general story, is the following.

1) The closure of an abelian Lie subgroup of $G'$ is an abelian Lie subgroup of $G'$.

2) Every compact abelian Lie group is a torus.

From this, we see that the closure if your $\Bbb R$ is a torus. It can't be a circle, so in the case $G' = T^2$, its closure is the whole torus.