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

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Statement

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

$G_0 = G, \qquad G_{m+1} = [G_m, G]$ .

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

Facts used

Lower central series is strongly central: This states that $[G_m,G_n] \le G_{m+n}$ .

Applications

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 $k t h$ 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 nilpotence class $c$ .

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

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