Finite group generated by abelian normal subgroups may have arbitrarily large nilpotency class

Statement
It is possible to have a finite group (more specifically, a group of prime power order for any chosen prime number $$p$$) that has arbitrarily large nilpotency class but is a group generated by abelian normal subgroups.

In fact, the examples that we choose also have the property that every group of prime power order can be embedded into one of these groups, hence we get the stronger statement that every finite nilpotent group is isomorphic to a subgroup of a finite group generated by abelian normal subgroups.

Proof
The idea is to take the group $$U(n,p)$$ of upper triangular matrices with $$1$$s on the diagonal and entries in the prime field $$\mathbb{F}_p$$. This group has nilpotency class $$n - 1$$. Moreover, it is generated by abelian normal subgroups parameterized by top right rectangles: for every $$k \times (n - k)$$ top right rectangle, define the corresponding subgroup as the subgroup where the diagonal has 1s, entries in the rectangle are arbitrary, and all other entries are $$0$$. This subgroup is elementary abelian of order $$p^{k(n - k)}$$, and these subgroups together generate the whole group.