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A Lie subgroup $H$ of a Lie group $G$ is defined to be an embedded submanifold of $G$ that is also a subgroup. The following exercise helps the reader in showing that every Lie subgroup is closed.

Let $G$ be a Lie group and $H$ a Lie subgroup.

(1) Let $\overline H$ be the closure of $H$ in $G$. Show that $\overline H$ is a subgroup in $G$.

(2) Show that each coset $Hx$ , $x\in\overline H$, is open and dense in $\overline H$.

(3) Show that $H=\overline H$, that is, every Lie subgroup is closed.

Using continuity of the multiplication map, (1) follows readily. Also, since for any fixed $x\in\overline H$, multiplication by $x$ is a diffeomorphism, (2) can be shown relatively easily too. However, I'm stuck at (3). Clearly I should be using (2) but I don't see how. Any hints?

Sha Vuklia
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  • There is a typo in (3): one of the two $H$'s should be $\overline{H}$ – Vincent Jan 16 '21 at 21:39
  • Thanks, fixed it. – Sha Vuklia Jan 16 '21 at 21:41
  • More on point however: is there also a typo in (2) and should the last $H$ be $\overline{H}$? If not (3) follows trivial from (2) as stated. After all for arbitrary $x$ we have $x = ex \in Hx$ and the latter lies in $H$ by the current version of (2). – Vincent Jan 16 '21 at 21:42
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    @Vincent Right, I'm sorry, I copied latex from a pdf, and wasn't being careful in correcting the bits that weren't transferred properly. – Sha Vuklia Jan 16 '21 at 21:44

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Let $x\in \bar H$, write $x=lim_nx_n, x_n\in H$, there exists $N$ such that $n>N$ implies that $x_n\in Hx$ since $Hx$ is open and contains $x$, we deduce that $x_n=hx,$ and $h\in H$ for $n>N$ and $x=h^{-1}x_n\in H$.

Tsemo Aristide
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