# Are these enough conditions for the subgroup to be a full latice?

I was wondering how to prove the following, or if you like, whether it is true, although I am almost certain it is. Let $$V$$ be a finite dimensional vector space over $$\mathbb{C}$$, say of dimension $$n$$. Let $$\Lambda \subset V$$ be a discrete subgroup, such that the quotient $$V/\Lambda$$ is compact as a topological space. Then $$\Lambda$$ is free of rank $$2n$$, as a $$\mathbb{Z}$$-module.

Motivation:

I'm doing exercises in Lie theory (that's where this comes from), where you have that if $$G$$ is a complex connected compact lie group, then the exponential map induces an isomorphism

$$\mathfrak{g}/ \ker(\exp) \cong G$$ I already know $$\ker(\exp)$$ is discrete, but I have to prove it is a 'full lattice'.

Ideas so far:

I'd like to take some maximal subset of $$\Lambda$$ that is linearly independent, then making all vectors in it as small as possible and then prove this spans $$\Lambda$$ as a $$\mathbb{Z}$$-module and is of cardinality $$2n$$ (it is at most $$2n$$, this can be seen immediately).

Another idea in case this does not work is taking some element in $$\Lambda$$ of minimal norm after fixing a norm, then looking at all elements of $$\Lambda$$ that are not in the $$\mathbb{Z}$$-span of this element and taking an element of minimal norm there, etc.

Thanks in advance for any ideas or solutions.