Example 15.9 is showed as follows in Loring Tu's An Introduction to manifolds:
Example 15.9 (Lines with irrational slope in a torus). Let $G$ be the torus $\mathbb{R}^2/\mathbb{Z}^2$ and $L$ a line through the origin in $\mathbb{R}^2$. The torus can also be represented by the unit square with the opposite edges identified. The image $H$ of $L$ under the projection $\pi:\mathbb{R}^2\longrightarrow\mathbb{R}^2/\mathbb{Z}^2$ is a closed curve if and only if the line $L$ goes through another lattice point, say $(m,n)\in\mathbb{Z}^2$. This is the case if and only if the slope of $L$ is $n/m$, a rational number or $\infty$; then $H$ is the image of finitely many line segments on the unit square. It is a closed curve diffeomorphic to a circle and is a regular submanifold of $\mathbb{R}^2/\mathbb{Z}^2$ (Figure 15.1).
If the slope of $L$ is irrational, then its image $H$ on the torus will never close up. In this case the restriction to $L$ of the projection map, $f=\pi|_L:L\longrightarrow\mathbb{R}^2/\mathbb{Z}^2$, is a one-to-one immersion. We give $H$ the topology and manifold structure induced from $f$.
It can be shown that $H$ is a dense subset of the torus [3, Example 111.6.15, p. 86]. Thus, $H$ is an immersed submanifold but not a regular submanifold of the torus $\mathbb{R}^2/\mathbb{Z}^2$.
Whatever the slope of $L$, its image $H$ in $\mathbb{R}^2/\mathbb{Z}^2$ is an abstract subgroup of the torus, an immersed submanifold, and a Lie group. Therefore, $H$ is a Lie subgroup of the torus.
I see that $H$ is an immersed submanifold, but I cannot understand how $H$ is an immersed submanifold but not a regular submanifold of the torus $\mathbb R^2/\mathbb Z^2$ is deduced from previous context. Can anyone explain the details for me?
