Theorem 38.11 in First Course in Abstract Algebra, Fraleigh.
Let $G$ be a nonzero free abelian group of finite rank $n$, and let $K$ be a nonzero subgroup of $G$.
Then $K$ is a free abelian group of rank $s\le n$. Furthermore, there exists a basis $\{x_1, x_2, \dots, x_n \}$ for $G$ and positive integers, $d_1, d_2, \dots, d_s$ where $d_i$ divides $d_{i+1}$ for $i=1, 2, \dots, s-1$, such that $\{d_1x_1, d_2x_2, \dots, d_sx_s \}$ is a basis for $K$.
In this proof, I don't understand existence of $Y_1$. While we cannot gaurantee that we have finite number of bases for $G$, how can we define $Y_1$ having some nonzero element of $K$ with the least such nonzero value $|k_i|$?