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 $k\geq (c+1)/2$, $\gamma _{k}(G)$ is an abelian group, In particular, $\gamma _{k}(G)$ is an abelian characteristic subgroup.

Facts used

1. Lower central series is strongly central: This states that $[\gamma _{m}(G),\gamma _{n}(G)]\leq \gamma _{m+n}(G)$.

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 $k\geq (c+1)/2$.

Proof: By fact (1), $[\gamma _{k}(G),\gamma _{k}(G)]\leq \gamma _{2k}(G)\leq \gamma _{c+1}(G)$, 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.