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.
Abstract Algebra – Finite Generated Abelian Torsion-Free Group
abstract-algebra
Best Answer
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.