# Is a lie subgroup generated by a normal subgroup necessarily normal?

I always get slightly confused with generated objects where there is more than one structure.

Say $G$ is a lie group, $N<G$ a normal subgroup (not necessarily a lie subgroup), and let $H=\langle N \rangle$ be the lie subgroup generated by $N$.

Is $H$ normal?

All I managed to say is some trivial stuff. Assume by contradiction that $H$ isn't normal, therefore there is some $g\in G$ such that $gHg^{-1} \not \subset H$, and so we have $h\in H$ such that $ghg^{-1} \not\in H$. Necessarily $h\not\in N$ from normality. I'd like to use that to throw away some elements of $H$ and remain with a lie subgroup containing $N$ which would be a contradiction. But how?

This connects to a a possibly larger thing I realize that I don't understand - is it possible to build the generated lie subgroup, in a similar way to the way we build generated subgroups in group theory by products of elements in the generating set?

• Hint: $H$ is the topological closure of $N$ in $G$. Aug 13, 2017 at 12:34

$H$ is the adherence of $H$. Let $g\in G$, define $Ad_g:G\rightarrow G$ by $Ad_g(x)=gxg^{-1}$, if $C$ is a closed subset of $G$ which contains $N$, $C$ contains $H$. This implies that $Ad_g(C)$ contains $Ad_g(H)$. Since $Ad_g$ is a diffeomorphism, we can write $C=Ad_g(Ad_g^{-1}(C))$ and $Ad_g^{-1}(C)$ is a closed subset which contains $N$ and henceforth $H$. We deduce that $C$ contains $Ad_g(H)$. This implies that $Ad_g(H)$ is the adherence of $H$ and $Ad_g(H)=H$.