# Derived length is logarithmically bounded by nilpotency class

## Statement

Let $G$ be a nilpotent group and let $c$ be the nilpotency class of $G$.Then, $G$ is a solvable group, and if $\ell$ denotes the derived length of $G$, we have: $\ell \le \log_2 c + 1$.

## Facts used

1. Second half of lower central series of nilpotent group comprises abelian groups: If $G$ is nilpotent of class $c$, and $\gamma_k(G)$ denotes the $k^{th}$ term of the lower central series of $G$, then $\gamma_k(G)$ is abelian for $k \ge (c + 1)/2$.

## Proof

We prove this by induction on the nilpotence class. Note that the statement is true when $c = 1$ or $c = 2$.

Given: A finite nilpotent group $G$ of class $c$.

To prove: The derived length of $G$ is at most $\log_2 c + 1$.

Proof: Let $k$ be the smallest positive integer greater than or equal to $(c + 1)/2$. In other words, either $k = (c+1)/2$ or $k = c/2 + 1$, depending on the parity of $c$. Then, $\gamma_k(G)$ is an abelian group, and $G/\gamma_k(G)$ is a group of class $k - 1$, which is at most $c/2$.

By the induction assumption, we have: $\ell(G/\gamma_k(G)) \le \log_2(k) + 1 \le \log_2 (c/2) + 1 = \log_2 c$.

Thus, $G$ has an abelian normal subgroup such that the derived length of the quotient is at most $\log_2 c$. This yields that the derived length of $G$ is at most $\log_2 c + 1$.