Quotient of a Lie group by normal lie subgroup is a lie group

Question

If $G$ is a Lie group and $H \leq G$ is a closed normal lie subgroup, how do I show that $G/H$ is a lie group. I know that $f : G \times G \rightarrow G$, by $(g,h) \mapsto gh$ and $h : G \rightarrow G$ by $h \mapsto h^{-1}$ are smooth maps.
How do I show that $(\overline{g}, \overline{h}) \mapsto \overline{gh}$ and $\overline{g} \mapsto \overline{g^{-1}}$ are smooth maps?

Is it true in general that if $f : M \times M \rightarrow M$, and $g : M \rightarrow M$ are smooth maps and $\sim$ is an equivalence relation, then $\overline{f} : (G/\sim) \times (G / \sim) \rightarrow G / \sim$ is a smooth map and similarly for $\overline{h} : G/ \sim \rightarrow G / \sim$, where $\overline{f} , \overline{h}$ are induced maps? If so, how does one prove this?
Is there another way to show that $G/H$ is a lie group, for $H$ normal?

Answer 1

$H$ acts on $G$ by left multiplication smoothly, freely and properly. By the quotient manifold theorem, $G/H$ is a smooth manifold and $\pi : G \rightarrow G/H$ is a smooth map. Let $\pi^{*} : G \times G \rightarrow G/H \times G/H$ be the projection of the product. Then $\pi^{*}$ is a smooth map. Let $G^{*}, (G/H)^{*}$ be the respective multiplication maps. Then $\pi \circ G^{*} = (G/H)^{*} \circ \pi^{*}$, since $\pi \circ G^{*}$ is smooth, we must have that $(G/H)^{*}$. A similar argument can be applied for inversion. Hence, $G/H$ is a lie group.