Derived length gives no upper bound on nilpotency class

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

Related facts

 * Solvable not implies nilpotent

Converse

 * Nilpotent implies solvable
 * Solvable length is logarithmically bounded by nilpotence class

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 $$l$$.