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 $G$? 
 A: As Yves Cornulier already said: Your presumed statement is wrong.
Any connected, linear, solvable Lie group over the reals is the semi-direct product of a compact abelian subgroup and a simply connected normal subgroup. 
(This holds more general for algebraic, connected, solvable lie groups ober a field of characteristic 0, what can be found in Chevalley's "Théorie des groupes de Lie")
This is in general false for non-linear Lie groups, what explains Yves Cornulier's counter example as the quotient of the heisenberg group with it's central discrete cyclic subgroup is not linear.
(The non-linearity of this group is e.g. proved in "The Structure of Compact Groups: A Primer for Students, a Handbook for the Expert" of Hofmann and Morris on page 169.)
A: 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")
