Second half of lower central series of nilpotent group comprises abelian groups

Statement
Suppose $$G$$ is a nilpotent group of  nilpotency class $$c$$. Define the lower central series of $$G$$ as follows:

$$\gamma_1(G) = G, \qquad \gamma_{m+1}(G) = [\gamma_m(G),G]$$.

Then, for $$k \ge (c + 1)/2$$, $$\gamma_k(G)$$ is an abelian group, In particular, $$\gamma_k(G)$$ is an abelian characteristic subgroup.

Facts used

 * 1) uses::Lower central series is strongly central: This states that $$[\gamma_m(G),\gamma_n(G)] \le \gamma_{m+n}(G)$$.

Applications

 * Penultimate term of lower central series is abelian in nilpotent group of class at least three
 * Derived length is logarithmically bounded by nilpotency class
 * Nilpotent and every abelian characteristic subgroup is central implies class at most two

Breakdown for upper central series
The first half of the upper central series of a nilpotent group need not comprise Abelian groups. In fact, even the second term of the series need not be Abelian, however large the nilpotence class. More specifically:


 * Upper central series may be tight with respect to nilpotence class: For any natural number $$c$$, we can construct a nilpotent group such that the $$k^{th}$$ term of the upper central series of the group has nilpotence class precisely $$k$$ (note: the nilpotence class clearly cannot be greater than $$k$$, and this result says that tightness may hold.
 * Second term of upper central series not is Abelian

Proof
Given: A group $$G$$ of nilpotency class $$c$$.

To prove: $$\gamma_k(G)$$ is Abelian for $$k \ge (c + 1)/2$$.

Proof: By fact (1), $$[\gamma_k(G), \gamma_k(G)] \le \gamma_{2k}(G) \le \gamma_{c+1}(G)$$, and since $$G$$ has class $$c$$, $$\gamma_{c+1}(G)$$ is trivial. Thus, $$[\gamma_k(G), \gamma_k(G)]$$ is trivial, and thus, $$\gamma_k(G)$$ is an abelian group.