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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.

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