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Every smooth manifold is assumed to be Hausdorff and second-countable.

Suppose $G$ is a Lie group, $H$ is a Lie subgroup of $G$, $N$ is a closed Lie subgroup of $G$ such that $N$ is normal, $N\cap H=\{e\}$, and $NH=G$, where $NH=\{ab:a\in N, b\in H\}$.

Is $H$ closed in $G$?

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    $\begingroup$ Can you define "Lie subgroup"? Some authors mean by this a closed subgroup. Do you mean the image of connected Lie group by a continuous homomorphism? or something weaker without connectedness? $\endgroup$
    – YCor
    Commented Oct 6, 2019 at 20:57
  • $\begingroup$ Hence, according to the definition, every subgroup $H$ of every Lie group $G$ is a Lie subgroup (being the image in $G$ of $H$ endowed with the discrete topology). Since your question does not refer to the additional structure on $H$, it's just the same as assuming that $H$ is an arbitrary subgroup. $\endgroup$
    – YCor
    Commented Oct 7, 2019 at 7:10
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    $\begingroup$ In any case, take $G$ to be the 2-torus, $N$ a closed circle subgroup, $H$ a dense line. Then $NH=G$ and $H$ is not closed. This is the simplest example of a non-closed connected immersed Lie subgroup... $\endgroup$
    – YCor
    Commented Oct 7, 2019 at 7:12
  • $\begingroup$ @YCor Every smooth manifold is assumed to be second countable. $\endgroup$
    – Born to be proud
    Commented Oct 7, 2019 at 11:52
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    $\begingroup$ @Ycor I really like your example of a non-closed Lie subgroup - it is easy to visualize and insightful. However I don't believe it satisfies the condition $H \cap N = \{e\}$ from the original post. Doesn't the line intersect the circle infinitely many times? $\endgroup$
    – Vincent
    Commented Oct 7, 2019 at 12:49

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Yes it's true, as a general fact on topological groups.

Let $G,H$ be $\sigma$-compact locally compact groups, $N$ a closed normal subgroup of $G$ and $i$ an injective continuous homomorphism $H\to G$. Suppose that as an abstract group one has $G=N\rtimes H$ (that is, $N\cap i(H)=\{e\}$ and $Ni(H)=G$). Then $i(H)$ is closed.

Indeed, by assumption the canonical map $N\rtimes H\to G$, $n.h\mapsto ni(h)$, is continuous and is an abstract group isomorphism. Hence, by the next lemma, it is a topological isomorphism (i.e., its inverse is continuous), and in particular $i(H)$ is closed.

Lemma Let $f:G\to H$ be a continuous bijective homomorphism between locally compact groups, with $G$ $\sigma$-compact. Then $f$ is a topological isomorphism, i.e., $f^{-1}$ is continuous.

For a proof of the latter (which is a simple application of Baire's theorem), see this MathSE answer.

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