When a connected solvable Lie group has a normal maximal torus, is it true that this torus is a subgroup of the center? When $G$ is a linear algebraic connected solvable Lie group and has a normal maximal torus $T$ (in fact, $T$ is the unique maximal torus of $G$), it follows that $T$ is a subgroup of the center of $G$. 
In fact, since $G$ is a linear algebraic connected solvable Lie group, it follows that $G = TU$, where $U$ is a normal simply-connected subgroup and $T \cap U = 1$. Since $T$ is also normal, $G = TU$ is a direct product and, since $T$ is abelian, it follows that $T$ is a subgroup of the center of $G$. 
And when $G$ is a connected solvable Lie group which has a normal maximal torus $T$, is it true that $T$ is a subgroup of the center of $G$?
 A: This is an elaboration of nfdc23's comment. Claim: A normal (compact) torus $T$ in a connected Lie group (solvable or not) is central.
Proof: Let $\exp:\mathfrak t=\text{Lie}\ T\to T$ be the exponentail map. Its kernel $\Gamma$ is a lattice in $\frak t$. Since $T$ is normal, the group $G$ acts on $T$ and therefore $\mathfrak t$ by conjugation. Moreover, it preserves $\Gamma$. For any $\gamma\in\Gamma$, the set $\{g\gamma\mid g\in G\}$ is connected (since $G$ is) and contained in the discrete set $\Gamma$. So it equals $\{\gamma\}$ which means that $G$ acts trivially on $\Gamma$. But then $G$ acts also trivially on $\mathfrak t$ and therefore on $T$, i.e., $T$ is central.
A: This is a long comment rather than an answer, since I don't understand precisely what the question really means.     
I'm still confused about the formulation here, though Knop and others have contributed what they can to sorting it out.   For example, the initial wording "linear algebraic connected solvable Lie group" doesn't immediately make sense to me, since Lie groups are originally real manifolds whereas linear algebraic groups usually require an algebraically closed ground field.   I guess you might be looking at the real points of a connected solvable linear algebraic group, but if so then your further comments suggest that you intend the group to be $\mathbb{R}$-split.   Otherwise the rational points of a maximal $\mathbb{R}$-torus (in the algebraic group sense) might be of any size, even trivial.     
What do you mean here by "maximal torus" of a connected solvable Lie group?   The study of solvable Lie groups (for instance by Hochschild and Mostow) is fairly delicate, and the role of topological tori is not obvious outside the context of reductive or semisimple Lie groups.  The use of the term "torus" in both algebraic and topological settings is perhaps unfortunate, but you need to be clear about the distinction.     The Jordan decomposition for linear algebraic groups doesn't work well for all Lie groups, since the notions of "semisimple" and "unipotent" for a Lie group (or "nilpotent" in its Lie algebra) are not intrinsic.   
