# Decomposition of solvable Lie group

Suppose $G$ is a connected Lie group whose radical is $R$. It is known that the solvable group $R$ can always be decomposed as $R=UT$ where $U$ is a simply-connected normal subgroup of $R$ and $T$ is a compact abelian subgroup of $R$ with $U\cap T = 1_G$. We know that $R$ is a normal subgroup of $G$.

Question: Is $U$ necessarily a normal subgroup of $G$?

• Your statement "it is known..." is not true. Example: $G=R$ the quotient of the 3-dimensional Heisenberg group by an infinite discrete central subgroup. – YCor Mar 25 '13 at 17:40
• Even assuming that it holds (a sufficient condition being that $G$ is linear; a weaker one being that $R$ is linear), the question is unclear. Indeed, there exists a decomposition $UT$... is $U$ normal in $G$? normal for every such decomposition? for at least one decomposition? $U$ is not unique in general. – YCor Apr 1 '19 at 7:36

In the case of connected linear algebraic groups it is true: Any inner automorphism of $G$ is an algebraic group automorphism of $R$. And so it carries all the unipotent elements of $R$ to unipotent elements, (See Section 19. Connected Solvable Groups in J.E. Humphreys textbook "Linear Algebraic Groups")

• In general, $U$ is not the unipotent radical. For instance, if $G$ is the affine group $\mathbf{R}\ltimes\mathbf{R}$, then $U=G$. – YCor Apr 1 '19 at 7:38