In my book on lie algebras, it is written the following stuff :
Let $G$ be a Lie group, $H$ a Lie subgroup of G.
$G/H$ is then a manifold.
Is it hard to prove ?
It is a little abstract for me : $G/H$ could not be "smooth" in my vision (we could have like a discrete set of elements, so we could'nt find open sets to define the charts). But it seems it's not the case.
But I don't have any idea to how to prove this.
(I'm a huge beginner in group theory and in Lie groups, I just started to learn this)
It follows from the Quotient Manifold Theorem:
Note that if $H$ is a closed Lie subgroup of $G$ then $H$ acts smoothly, freely and properly on $G$. Being closed is necessary for the action to be proper, if it is not closed then the quotient is not even $T_1$. Finally the quotient space under this action is equal to the space of cosets. BTW This also implies that $G/H$ is always a manifold, even when $H$ is not normal.
The proof of the Quotient Manifold Theorem (which is not trivial at all) can be found for example in John M. Lee "Introduction to Smooth Manifolds".