Finite Generated Abelian Torsion-Free Group is a Free Abelian Group

Question

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.

Answer 1

Presumably you don't have the structure theorem in your bag of tools. Then you can proceed as follows. If the coefficients $n_1,n_2,\ldots,n_n$ have a common factor, say $d$, you can cancel that from the equation, because otherwise $(n_1/d)x_1+(n_2/d)x_2+\cdots+(n_n/d)x_n$ would be a non-zero torsion element.

The idea is that if $\{x_1,x_2,\ldots,x_n\}$ is a minimal generating set, then so is the set gotten from this by replacing $x_i$ with $x_i'=x_i+ax_j$ for some $j\neq i$ and integer $a$ (prove this as an exercise if you don't see it right away). Furthermore, the putative linear dependency relation can then be rewritten to read $$ n_1x_1+\cdots+n_ix_i'+\cdots+(n_j-a n_i)x_j+\cdots +n_nx_n=0. $$ Basically this allows us to run Euclid's algorithm on the set of coefficients - always select $a$ in such a way that the altered coefficient $n_j-an_i$ becomes as small as possible (in absolute value). Rinse. Repeat. Because we earlier saw that the no-torsion requirement implies that $\gcd(n_1,n_2,\ldots,n_n)=1$, we eventually get a linear dependency relation, where one of the coefficients is $=\pm1$. This means that one of the generators can then be written as a $\Bbb{Z}$-linear combination of the others and thus disposed of as a generator - in violation of the minimality of the generating set you started with.

Answer 2

It follows directly from the structure theorem, see under "corollaries": "A corollary to the structure theorem is that every finitely generated torsion-free abelian group is free abelian". The torsionfree group $\mathbb{Z}^n$ is free-abelian of rank $n$.

Answer 3

Let $M$ be such a group. By the structure theorem of modules over $\mathbb{Z}$, we can find $r \geq 0$ and $a_{1} | a_{2}| \cdots |a_{t}$ (non zero positive integers) such that $M \cong \mathbb{Z}^{r} \times \mathbb{Z}/a_{1}\mathbb{Z}\times \mathbb{Z}/a_{2}\mathbb{Z} \times \cdots \times \mathbb{Z}/a_{t}\mathbb{Z}$. Now since $M$ is torsion-free, the $a_{i}$ are 1, so $M \cong \mathbb{Z}^{r}$

Answer 4

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 $m$.