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. 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 ${\log }_{2}c+1$.