Let $G$ be a free abelian group of rank $r$, and $H$ a subgroup of $G$. Then $G/ H$ is finite if and only if the ranks of $G$ and $H$ are equal. If this is the case, and if $G$ and $H$ have $\Bbb Z$-bases $x_1, \cdots , x_r$ and $y_1,\cdots , y_r$ with $y_i=\sum_j a_{ij}x_j$, then $|G/H|=|\det(a_{ij})|$. This is Theorem Theorem 1.1 7. in the book Algebraic-Number Theory by Ian Stewart and David Tall, on page 30.
in the proof it is written that -
Let $H$ have rank $s$, and use Theorem $1.16$ to choose $\mathbb Z$-bases $u_1, ... , u_r$ of $G$ and $v_1, ... , v_s$ of $H$ with $v_i =a_iu_i$ for $1 \leq i \leq s$. Clearly $G/ H$ is the direct product of finite cyclic groups of orders $\alpha_1 \cdots \alpha_s$ and $r - s$ infinite cyclic groups. Hence $|G/H|$ is finite if and only if $r = s$...
Note that according to Theorem 1.16. Every subgroup $H$ of a free abelian group $G$ of rank $n$ is free of rank $s ≤ n$. Moreover there exists a basis $u_1, . . . , u_n$ for $G$ and positive integers $α_1, . . . , α_s$ such that $α_1u_1, . . . , α_su_s$ is a basis for $H$.
The following line is not clear to me -
Clearly $G/ H$ is the direct product of finite cyclic groups of orders $\alpha_1 \cdots \alpha_s$ and $r - s$ infinite cyclic groups.
How we know clearly $G/ H$ is the direct product of finite cyclic groups of orders $\alpha_1 \cdots \alpha_s$ and $r - s$ infinite cyclic groups?
Note that there is a theorem that states similar thing (see 38.12 Theorem on page 338 of "A First Course in Abstract Algebra" by John B. Fraleigh, 7th Edition), but I want to what obvious thing given in the book of Ian Stewart and David Tall make it obvious that $G/ H$ is the direct product of finite and infinite cyclic groups.