algebraic closure of Lie groups in

Let $G$ be a connected, simply connected, solvable, complex Lie group with a discrete subgroup $\Gamma$. Let also $G_a$ be Hochshild-Mostow hull of $G$, i.e., there exists a solvable linear algebraic group $G_a =({\mathbb C}^*)^k \ltimes G$ such that $G_a$ contains $G$ as a Zariski dense, topologically closed, normal complex subgroup.

Is it true that algebraic closure of $\Gamma$ and $G$ in $G_a$ are the same?

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What is the "algebraic closure" of a group? (I guess you must mean Zariski closure?) – John Pardon Aug 4 '12 at 5:03
Could I take $\Gamma$ to be $\{1\}$? – Ben McKay Aug 4 '12 at 10:07
$\Gamma$ is a lattice!! – user13559 Aug 4 '12 at 20:47
It would be helpful is your title were a complete sentence! – Mariano Suárez-Alvarez Dec 24 '12 at 0:23

Although you have not stated it this way, I will assume that $\Gamma$ is a lattice in a connected linear complex solvable Lie group $G$. If $\rho G \rightarrow GL_n({\mathbb C})$ is a holomorphic representation of $G$, it can be proved that the Zariski closure of $G$ and $\Gamma$ are the same. Suppose that the Zariski closures are $G'$ and $H'$ resp.
$G'/H'$ is affine and you cannot have a non-constant holomorphic map from the compact complex manifold $G/\Gamma$ into an affine space (by the maximal modulus principle).
You don't really need to use that $G'/H'$ is affine, and can argue by induction on the dimension of $G'/H'$ (using the solvability of $G'$)
Actually, it turns out that if $G$ is a connected linear complex Lie group and $\Gamma \subset G$ is a lattice, then the Zariski closure of $\Gammma$ and of $G$ are the same in any linear representation of $G$. This follows from the solvable case as above and semi-simple case by the Borel density theorem (easier for the complex lie group case). One can put these cases together thanks to a theorem of Auslander on lattices in a semi-direct product of a solvable and a semi-simple group. – Venkataramana Dec 10 '12 at 2:31