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

### 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.