What is the dimension of the quotient of Lie groups? Let $G$ be a $g$-dimensional Lie group and $H$ be an $h$-dimensional normal Lie subgroup of $G$. Is the dimension of the quotient group $G / H$ always $g-h$? This seems intuitively true, but I can't find such a statement anywhere.
For finite groups, we have that the order of $G/H$ is the quotient of the order of $G$ and the order of $H$, which suggests that there should be a simple relationship in the Lie case as well - but of course a nontrivial Lie group doesn't have finite order, so the result doesn't directly carry over.
 A: The subgroup $H$ needs to be closed to ensure that $G/H$ is a Lie group. For example, let $G = S^1\times S^1$ and $H = \{(e^{2\pi i\theta}, e^{2\pi i\alpha\theta}) \in S^1\times S^1 \mid \theta \in \mathbb{R}\}$ where $\alpha$ is a fixed irrational number. While $H$ is a Lie subgroup of $G$, the quotient $G/H$ is not Hausdorff (the cosets of $H$ are dense in $G$ so the quotient topology on $G/H$ is the trivial topology) and therefore cannot be a Lie group.
It is worth noting that $G/H$ is a smooth manifold of dimension $g - h$ even if $H$ is not normal (as long as it is closed); of course, $G/H$ is not a Lie group in this case. The following is from Lee's Introduction to Smooth Manifolds (second edition).

Theorem $\mathbf{21.17}$ (Homogeneous Space Construction Theorem). Let $G$ be a Lie group and let $H$ be a closed subgroup of $G$. The left coset space $G/H$ is a topological manifold of dimension equal to $\dim G - \dim H$, and has a unique smooth structure such that the quotient map $\pi : G \to G/H$ is a smooth submersion. The left action of $G$ on $G/H$ given by $g_1\cdot (g_2H) = (g_1g_2)H$ turns $G/H$ into a homogeneous $G$-space.

A: Yes, it is true. If $\mathfrak g$ is the Lie algebra of $G$ and $\mathfrak h$ is the Lie algebra of $H$, then the Lie algebra of $G/H$ is isomorphic to $\mathfrak g/\mathfrak h$. And therefore\begin{align}\dim(G/H)&=\dim(\mathfrak g/\mathfrak h)\\&=\dim\mathfrak g-\dim\mathfrak h\\&=\dim G-\dim H.\end{align}
