Derived length is logarithmically bounded by nilpotency class

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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

Applications

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.