Any abelian Lie subgroup containes a connected Lie subgroup of codimension 1 [closed]

Question

I am trying to understand the proof of the following claim (see A.L. Onishchik, E.B. Vinberg (Eds.) Lie Groups and Lie Algebras III, p.50, Theorem 3.1).

Theorem 3.1 (ii) If the Lie group $G$ is solvable, then it has a connected normal Lie subgroup of codimension 1.

Proof.

Consider the Lie subgroup $G'_1:= \overline{[G,G]}$, the closure of the commutator group $[G,G]$. The solvability of the Lie group $G$ implies that $G_1' \ne G$. The group $G/G_1'$ is abelian, and evidently has a connected Lie subgroup $A'$ of codimension $1$. If $\pi: G \to G/G_1'$ is the natural epimorphism, then $G_1: = \pi^{-1}(A')$ is the desired normal Lie subgroup of $G$. Q.E.D.

So, I don't understand why any abelian Lie group has a connected Lie subgroup of codimension $1$. I tried to think as follows: maybe there is a nice Lie homomorphism $f:A \to A$ that has a constant rank $1$ and thus its kernel is a subgroup of the desired type but I cannot see this homomorphism in general.

Answer 1

I think I've got it. As I understood any $n$-dimensional connected abelian Lie group $A$ has a form $\mathbb{T}^k \times \mathbb{R}^{n-k}$ and thus we get set $B:= \mathbb{T}^k \times \mathbb{R}^{n-k-1}$ if $n-k >0$ and $B:=\mathbb{T}^{n-1}\times\{e\}$ in otherwise and we are done, i.e,. $B$ is a subgroup of codimension $1$.