# Specific basis of a subgroup of a free abelian group

I'm looking for clarification on Fraleigh's "A First Course in Abstract Algebra" Theorem 38.11.

It states: "Let $G$ be a nonzero free abelian group of finite rank $n$, and let $K$ be a nonzero group of $G$. Then $K$ is free abelian of rank $s \le n$. Furthermore, there exists a basis $\{x_1, x_2, ..., x_n \}$ for $G$ and positive integers $d_1, d_2, ..., d_n$, where $d_i$ divides $d_{i+1}$ for $i = 1, 2, ..., s-1$ such that $\{d_1x_1, d_2x_2, ..., d_sx_s \}$ is a basis for $K$."

What I an confused about is his particular proof, where he determines the basis for $K$ utilizing the canonical division algorithm. He does this without use of modules and I am looking for an alternate proof/clarification of the proof that does not use modules.

-
What specifically confuses you? This is the well-known theorem on elementary divisors, which is (almost) equivalent to the structure theorem for finitely generated modules over a principal ideal domain (here $\mathbb{Z}$). You can find its proof everywhere. By the way, the assumptions $G,K \neq 0$ are nonsense. Erase them / inform the author. –  Martin Brandenburg Nov 14 '13 at 8:12
what confuses me is what he means by choosing a basis $Y_1$ that yields a minimal value such that all the nonzero elements of $K$ are written in terms of basis elements in $Y_1$. For example, if you have $Z \times Z \times Z$, a valid basis is $\{(1,0,0), (0,1,0), (0,0,1) \}$ and to obtain the element $(4,0,0)$ you would have a $k_1$ of 4. Is what he means to choose another basis, i.e. something with a first $(4,0,0)$ term to get $k_1 = 1$. In general, why wouldn't $k_1$ always be 1? –  user17635 Nov 14 '13 at 14:41