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:

$$\ell \le \log_2 c + 1$$.

Related facts

 * Nilpotent implies solvable
 * Derived length gives no upper bound on nilpotency class: Knowing the derived 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.

Applications

 * Nilpotency class uniquely determines derived length only for class at most three

Facts used

 * 1) uses::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 $$\log_2 c + 1$$.