Derived length is logarithmically bounded by nilpotency class
From Groupprops
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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 l denotes the derived length of G, we have:
.
Related facts
- Nilpotent implies solvable
- Derived length gives no upper bound on nilpotency 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 Gk denotes the kth term of the lower central series of G, then Gk is Abelian for
.
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 log2c + 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, Gk is an Abelian group, and G / Gk is a group of class k − 1, which is at most c / 2.
By the induction assumption, we have:
.
Thus, G has an Abelian normal subgroup such that the solvable length of the quotient is at most log2c. This yields that the solvable length of G is at most log2c + 1.