Derived length is logarithmically bounded by nilpotency class

Statement

Let $G$ be a Nilpotent group (?) and let $c$ be the Nilpotence class (?) of $G$ .Then, $G$ is a Solvable group (?), and if $l$ denotes the Solvable length (?) of $G$ , we have:

$l \leq \log_{2} c + 1$ .

Related facts

Nilpotent implies solvable
Solvable length gives no upper bound on nilpotence class: Knowing the solvable length of a nilpotent group gives no bound on its nilpotence class. In fact, there are metabelian groups of arbitrarily large class, such as the dihedral $2$ -groups and the generalized quaternion groups.

Facts used

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

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 solvable 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, $G_{k}$ is an Abelian group, and $G / G_{k}$ is a group of class $k - 1$ , which is at most $c / 2$ .

By the induction assumption, we have:

$l (G / G_{k}) \leq \log_{2} (k) + 1 \leq \log_{2} (c / 2) + 1 = \log_{2} c$ .

Thus, $G$ has an Abelian normal subgroup such that the solvable length of the quotient is at most $\log_{2} c$ . This yields that the solvable length of $G$ is at most $\log_{2} c + 1$ .