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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?

user100101212
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    First, you need to equip G/H with a suitable smooth manifold structure (say, using quotient mapping theorem). – Moishe Kohan Feb 29 '20 at 21:40
  • What is the quotient mapping theorem? – user100101212 Feb 29 '20 at 21:45
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    Sorry, I meant quotient manifold theorem. Just Google it. Lee's book is a good reference. – Moishe Kohan Feb 29 '20 at 21:57
  • Yes, so $H$ acting on $G$ in the natural way is a smooth, free and proper action, and hence $G/H$ is a smooth manifold, how do I show that multiplication and inversion on $G/H$ is continuous. Suppose we fix representatives of $G/H$, say $r_{1}, \dots , r_{n}$. Is the map $f: G/H \rightarrow G$ given by $f(gH) = r(g)$, where $r(g)$ is a unique representative of $gH$ a smooth map from $G/H$ to $G$? – user100101212 Feb 29 '20 at 22:01
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    No, what you wrote will not work (remember, $G/H$ is typically infinite). Instead, you should also use the fact that $G\to G/H$ is a submersion (in fact, a locally trivial fiber bundle): This is also a part of the quotient manifold theorem. This will give you smooth local sections $U\subset G/H\to G$ of the bundle $G\to G/H$. Then you use the fact that the product map on $G$ is smooth and that composition of smooth maps is smooth. Try to write an answer to your question given these hints. You can then even accept your answer. – Moishe Kohan Mar 01 '20 at 04:10
  • @MoisheKohan I have written an answer. Please let me know if anything is incorrect. – user100101212 Mar 01 '20 at 16:13

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$H$ 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.

user100101212
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  • You should proofread; also, I suggest to use $\mu_G$ for the multiplication map in $G$ (rather than $G^$, which is very nonstandard). Lastly, when you say "since $\pi \circ G^{}$ is smooth, we must have that $(G/H)^{*}$." is incomplete: The argument that I know, uses the existence of smooth local sections. For topological groups, one can verify continuity of multiplication for $G/H$ without constructing sections. – Moishe Kohan Mar 01 '20 at 18:06