I know that this question has been stated in a little bit of a different form but my question is not about how to proof it but about a part in the proof instead. So we need to show that:
Lemma: A finitely generated torsion-free abelian group H is free-abelian
- finitely generated means $H=(a_1,a_2,...,a_n)$
- torsion-free means $T(H)=\{a\in H: \text{ord}(a)<\infty\} = \{e\}$, which in other word means that for every $a\in H, a\neq e \Rightarrow \text{ord}(a)=\infty$
- free-abelian group is an abelian group generated by a free system, that means by a (finite) system $(b_1,...b_m)$ with the property that (denoting multiplication as the group-operation): $$ b_1^{k_1}\cdot...\cdot b_m^{k_m}=e \Rightarrow k_1,...,k_m=0 $$
Proof: Let $H=(a_1,a_2,...,a_n)$. First we show that with a suitable nummeration we can chose $(a_1,a_2,...,a_r),r\leq n$ with the following properties:
(i) $a_1^{k_1}\cdot...\cdot a_r^{k_r}=e \Rightarrow k_1,...,k_r=0$
(ii) for all $i\in\{r+1,...,n\}$ there is a representation $a_1^{k_1}\cdot...\cdot a_r^{k_r} \cdot a_i^{k_i}=\{e\}$ with on $i$ dependent $k_1,...,k_r,k_i\in \mathbb{Z}$ and $k_i\neq 0$.
It is clear for me that for $H\neq \{e\}$ we can chose an $a_1\neq e$ and because $H$ torsion-free $a_1^{k_i}=e \Rightarrow k_i=0$. Now we can add such $a_i$ that property (i) stays true and with suitable nummeration we get our maximal $(a_1,a_2,...,a_r),r\leq n$. What I don't understand is how (ii) follows from that or from (i). My guess is that (ii) is meant to say that we chose the biggest such system suitable nummeration we can chose $(a_1,a_2,...,a_r),r\leq n$ and that for every $a_i$ with $i\in\{r+1,...,n\}$ there is $a_1^{k_1}\cdot...\cdot a_r^{k_r} \cdot a_i^{k_i}=\{e\}$ then $k_i\neq 0$ because it is not in the system. But still I can't find how to formaly show that this property holds with (i) or $(a_1,a_2,...,a_r),r\leq n$ being maximal with property (i).
I would appreciate your help very much.
You seem to have understood the point.
Define $r$ to be the maximal integer $k\leqslant n$ such that there is an enumeration of the $a_i$ satisfying $(i)$. As you noted, $r\geqslant 1$ when the group is nontrivial. If $r=n$ then you are done, you found a basis of $H$.
Otherwise, for any $i>r$, $(a_1,\dots,a_r,a_i)$ cannot satisfy $(i)$ since otherwise $r$ would not be maximal. But the negation of $(i)$ is that there are some $k_j$ such that $$a_1^{k_1}\cdots a_r^{k_r}\cdot a_i^{k_i}=e$$ but not all $k_j$ are $0$. But now you cannot have $k_i=0$ since otherwise you would have $$a_1^{k_1}\cdots a_r^{k_r}=e$$ and one of the $k_j$ with $1\leqslant j\leqslant r$ would be nonzero, which is impossible since $(a_1,\dots,a_r)$ satisfies $(i)$. So we must have $k_i\neq 0$, which is property $(ii)$.