Derived length gives no upper bound on nilpotency class

Statement

For $l>1$, there exist Nilpotent group (?)s of Solvable length (?) $l$ and arbitrarily large Nilpotence class (?).

Related facts

• Solvable not implies nilpotent

Converse

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

Proof

Dihedral groups

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

For $n\geq 3$, 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 $l$.