I am currently trying to understand why for a subgroup $g\subset \mathbb{Z}^n$ there is a basis of this subgroup $b_1, \dots, b_i$, such that the $b_j$ are multiples of a basis in $\mathbb{Z}^n$.
From an "experimental" standpoint this is clear - if I try some subgroups of $\mathbb{Z}^n$ I can find the corresponding basis pretty easy. E.g. $\langle (2,2,0), (0,0,4)\rangle$ is a subgroup that is generated by $(2,2,0)$ and $(0,0,4)$ and who's generators are multiples from elements from the basis $\{(1,1,0), (0,0,1), (1,0,0)\}$. But how do I proof that I can always find such a basis?
I was told that to proof this I can use that it is equivalent to the fact that every matrix over $\mathbb{Z}$ can be converted into Smith Normal Form. I do not understand how these statements are equivalent.
Is there a naive way (without deeper theory) to understand this connection or does it lie deeper? Do you know a proof that does not use the smith normal form? Is there a good source where the statement is proved?
This is a property of principal ideal domains in general, much like the existence of the Smith normal form itself.
To see how it works over $ \mathbf Z $ in particular, imagine you're given a subgroup $ H \subset \mathbf Z^n $. The first coordinates of the elements of $ H $ form an ideal of $ \mathbf Z $, which is principally generated by some $ c $, say. It follows that there exists some element $ x \in H $ of the form $ x = (c, x_2, \ldots, x_n) $, and that you can write any element $ y \in H $ in the form $ y = qx + (0, z_2, \ldots, z_n) $. Iterate the procedure on $ (0, z_2, \ldots, z_n) $ by induction and you end up with a basis for $ H $ that has size at most $ n $.