Abelian algebraic groups?

Question

Is is true that every abelian complex algebraic group has the form $$\mathbb{C}^n/\Lambda$$ where $\Lambda\subseteq\mathbb{C}^n$ is a lattice (not necessarily full rank)?

Answer 1

I just wanted to provide a justification of what was already said in the comments for compact groups.

Suppose that $G$ is a compact and complex algebraic (not necessarily abelian) of dimension greater than $0$ i.e. $G$ is not a finite group, then the $Ad: G \to \mathfrak{g}$ map is holomorphic (regular) and therefore constant, this implies that $G$ is abelian, note that as a complex manifold the exponential map $\exp: \mathfrak{g} \to G$ is a surjective homomorphism by Hopf-Rinow Theorem and therefore $G \cong \frak{g} / \Lambda$ for some subgroup $\Lambda$ which is of the form $\mathbb{Z}^n$, since the space is compact it must be a lattice.