Derived length gives no upper bound on nilpotency class

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Statement

For $l > 1$ , there exist nilpotent groups of solvable length $l$ and arbitrarily large nilpotence class.

Related facts

Solvable not implies nilpotent

Converse

Proof

Dihedral groups

We first show that for $l = 2$ , there exist groups of arbitrarily large nilpotence class.

For $n \ge 3$ , the dihedral group $D_{2^n}$ , given by the presentation:

$\langle a,x \mid a^{2^{n-1}} = x^2 = e, xax = a^{-1} \rangle$ ,

has nilpotence class $n - 1$ , but solvable length $2$ , since it has an abelian normal subgroup $\langle a \rangle$ such that the quotient is also an abelian group.

To get an example of a group of length exactly $l$ for $l > 2$ that has arbitrarily large nilpotence class, take the direct product of $D_{2^n}$ with any nilpotent group of solvable length precisely $l$ .