# Understanding: The factors of the upper central series of a torsion-free nilpotent group are also torsion-free.

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.
• Question 1 is a straightforward induction on the length of the upper central series. Assuming that we can prove that $G/Z(G)$ is torsion-free, we can apply the inductive hypothesis to deduce that the factors of the upper central series of $G/Z(G)$ are torsion-free. But the factors of the upper central series of $G$ are the same together with $Z(G)$, and $Z(G)$ is torsion-free because $G$ is. $[x,g] \in \zeta_1G$ follows immediately from $\zeta_2 G/\zeta_1 G = Z(G/\zeta_1 G)$. – Derek Holt Sep 20 '18 at 14:53
• Thanks, understood. Still not sure how $[x,g] \in \zeta_1G$ follows and how exactly $[x,g]^m = [x^m,g] = 1$ is obtain through this. – Arwid Sep 20 '18 at 15:34
• In general $g \in Z(G) \Rightarrow [g,h]=1$. $[x,g]^m = [x^m,g]$ follows from the standard commutator identities. – Derek Holt Sep 20 '18 at 16:57