finitely generated subgroup of $\mathbb{Q}^n$

Question

For such a seemingly standard problem, I can't seem to find a reference for it...

Prove that if $A$ is a finitely generated subgroup of $\mathbb{Q}^n$ then it has the form $\{\sum_{i=1}^k n_ia_i:n_i \in \mathbb{Z}\}$ where $a_1, \dots, a_k$ is a linearly independent set of elements of the vector space $\mathbb{Q}^n$ (in particular, $k \le n$ and $A \cong \mathbb{Z}^k$).

Answer 1

$A$ is a finitely generated torsion-free abelian group. $($$a \in A$ is called a torsion element if $o(a) < \infty$ or there exists $n > 0$ such that $na = 0$.$)$ Since $A \subseteq \mathbb{Q}^n$ and $a \neq 0$, this means that $na \ne 0$ for all $n \in \mathbb{Q}$, $n \neq 0$ as $a$ is linearly independent in $\mathbb{Q}^n$. By the fundamental theorem of finitely generated abelian groups, $A \cong \mathbb{Z}^k$ for some $k \ge 0$. That is, there exists $a_1, \dots, a_k \in A$ such that $$\mathbb{Z}^k \to A,\text{ }(n_1, \dots, n_k) \mapsto n_1a_1 + \dots + a_ka_k$$is an isomorphism. In particular, $$A = \{n_1a_1 + \dots + n_ka_k\text{ }|\text{ }n_i \in \mathbb{Z}\} = \mathbb{Z}a_1 \oplus \dots \oplus \mathbb{Z}a_k$$ and $n_1a_1 + \dots + n_ka_k = 0$, $n_i \in \mathbb{Z}$, which is only the case if and only if $$n_1 = n_2 = \dots = n_k = 0.$$We now prove $k \le n$. Since $\mathbb{Q}^n$ is an $n$-dimensional vector space over $\mathbb{Q}$, any linearly independent set has at most $n$-elements. So it is enough to show $\{a_1, \dots, a_k\} \subseteq \mathbb{Q}^n$ is a linearly independent set. Assume$${{r_1}\over{m}}a_1 + \dots + {{r_k}\over{m}}a_k = 0$$ for $r_i, m \in \mathbb{Z}$. This means that$$r_1a_1 + \dots + r_ka_k = 0 \implies r_i = 0.$$Hence, $a_1, \dots, a_k$ are linearly independent.

Answer 2

Every finitely generated torsion free abelian group is isomorphic to $\mathbb{Z}^k$, for some positive integer $k$, by the Structure Theorem for Finitely Generated Abelian Groups.