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I am trying to understand the proof of the following Lemma:

If $G$ is a solvable Lie Group then there exits a closed connected non-trivial normal subgroup of $G$ of codimension 1.

For the proof, observe that since $G$ is solvable then there exist closed connected normal subgroups such that $G>G_1>G_2...>G_k=\{e\}.$ Consider a closed subgroup $H<G/G_1$ of codimension $1.$ If this was trivial then $G_1$ would be our candidate and we would be done. Then using the projection $p: G\to G/G_1$ we can pull back $H$ and obtain a closed connected subgroup $H_1 =p^{-1}(H)<G$.

My question is why is this subgroup $H_1$ a normal subgroup of $G$?

Student
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  • Another idea is to pass to Lie algebras (for connected and simply connected $G$). There is it well-known, see this post. Also, on the Lie algebra level one shows that it is an ideal: "because any subspace containing the derived algebra is an ideal. " The same idea should work here for groups. – Dietrich Burde Jan 18 '22 at 12:05

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If $H_1$ is not normal, then $gH_1 \neq H_1g$ for some $g \in G$. But then $p(g)H \neq Hp(g)$, contradiction with $G/G_1$ being abelian.

In general, given a quotient $p : G \to G/G_1$, a subgroup $H \supset G_1$ is normal if and only if $p(H)$ is normal in $G/G_1$.

Luis
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