TITLE: Finite Generated Abelian Torsion-Free Group is a Free Abelian Group
QUESTION [5 upvotes]: I am trying to prove that every Finite Generated Abelian Torsion-Free Group is a Free Abelian Group. In order to do this, I am trying to show that if $\{x_1, \dots, x_n\}$ is a minimal generator of the group and if $n_1x_1 + \dots n_n x_n=0$ then $n_1=\dots = n_n=0$. I'm stuck here.

REPLY [4 votes]: Woo-hoo! Here's the proof (without the structure theorem) that I was trying for yesterday.
Notation: If $w=(w_1,\dots,w_n)\in\Bbb Z^n$ we will write $$\gcd(w)=\gcd(w_1,\dots,w_n).$$ If $x_1,\dots,x_n$ is your minimal generating set we will write $$x=(x_1,\dots,x_n).$$The reason being that now if $w\in\Bbb Z^n$ we will want to write $$w\cdot x=w_1x_1+\dots+w_nx_n$$in order to make various things readable.
Here's the theorem I conjectured yesterday (based on the case $n=2$):
Theorem Suppose $m\in\Bbb Z^n$ and $\gcd(m)=1$. There exists a set $v_1,\dots, v_n$ of generators for $\Bbb Z^n$ with $m=v_1$.
You can find a proof of that at Generating Sets for Subgroups of $(\Bbb Z^n,+)$.. The theorem makes the result you want trivial:
Suppose $x_1,\dots,x_n$ is a minimal set of generators for your torsion-free abelian group. Suppose that there exists $m\in\Bbb Z^n$ with $m\ne0$ and $$m\cdot x=0.$$
(See above for the notation.)
As noted previously, if there exist $d\in\Bbb Z$ and $m'\in\Bbb Z^n$ with $m=dm'$ then $d(m'\cdot x)=0$ and hence $m'\cdot x=0$. So we may assume $\gcd(m)=1$. Let $v_1,\dots,v_n$ be a set of generators for $\Bbb Z^n$ with $v_1=m$.
Now if $w\in\Bbb Z^n$ there exist integers $k_j$ with $$w=k_1m+k_2v_2+\dots+k_nv_n.$$Hence $$w\cdot x=k_2(v_2\cdot x)+\dots+k_n(v_n\cdot x).$$So $v_2\cdot x,\dots,v_n\cdot x$ generate your group, contradicting the minimality of $m$.