TITLE: Subgroups of $\mathbb{Z}^n$
QUESTION [17 upvotes]: I hope that the following problem isn't actually elementary (at least, for the sake of the fact that I'm posting it here), and I apologize if it is.  I did try hard to solve it first.
Let $V$ be a $\mathbb{Q}$-vector subspace of $\mathbb{Q}^n$, and let $G = V \cap \mathbb{Z}^n$.
Does there exist a linearly independent generating set for $G$ (i.e. a subset of $G$ such that every element of $G$ can be expressed uniquely as as $\mathbb{Z}$-linear combination of elements of this subset)?
Is there an algorithm to find it (given a basis for $V$)?

REPLY [14 votes]: EDIT: I typed this up before everyone piled on to say that this is too elementary. Moving to CW rather than deleting, as I distinctly remember not seeing this during my education on groups and rings (but then maybe I missed class that day). And don't even go down the line of "grad school courses"...
Feel free to downvote if you feel like it, of course.

Interesting question! The following is not a full answer, but seemed worth a bit more than a comment.
If I understand correctly, the first part asks if G is free abelian, in which case the answer is apparently yes, see this MO question from someone reputable. (I confess this is not something I knew, although I did have my suspicions.)
My guess is that some modification of Gauss-Jordan elimination should provide an algorithm for extracting a "Z-basis" from a Z-generating set, but it's not clear to me right now how one would get a Z-generating set from a given Q-basis of your original V.