While trying to proof the following statement I got stuck.
The factors of the upper central series of a torsion-free nilpotent group are also torsion-free.
The problem is quite similar to the following question and among the answers of the related question there is also a short proof regarding my question. Are lower central factors of torsion-free nilpotent groups torsion-free?
Question 1: Why does the problem (according to the answer in the link) reduce to showing that $G/Z(G)$ is torsion-free?
Now let me cite a different statement which is quite similar to the above answer given but leaves me with 2 more technical questions. Its from the book 'a course in the theory of groups' by Derek J.S. Robinson, page 137, theorem 5.2.19.
5.2.19 If the center of a group $G$ is torsion-free, each upper central factor is torsion-free.
Proof. Let $\zeta$G = $ \zeta_1G$ be torsion free. It is enough (Why?) to prove that $\zeta_2G/\zeta_1G$ is torsion-free. Suppose that $x \in \zeta_2G$ and $x^m \in \zeta_1G$ where $m>0$. [...] We have $[x,g]^m = [x^m,g] = 1$ , because $[x,g] \in \zeta_1G$ (Why is $[x,g] \in \zeta_1G$ and how does the equation follow?). Since $\zeta_1G $ is torsion-free, $[x,g] = 1 $ for all $ x \in G$, and $x \in \zeta_1G$.
So the way I see it, by reducing the proof to showing $G/Z(G)$ is torsion-free I can further reduce it to showing $\zeta_2G/\zeta_1G$ is torsion-free.
I hope to have clarified the question in a sufficient way. I'd be really thankful for any answer/hint/advice as I don't seem to find other related info on this one. Thanks for taking the time reading the question!