Nilpotent not implies generated by abelian normal subgroups

Statement
A nilpotent group need not be generated by Abelian normal subgroups.

Example of the dihedral group
Let $$G$$ be the dihedral group of order $$16$$:

$$G = \langle a,x \mid a^8 = x^2 = e, xax^{-1} = a^{-1} \rangle$$.

$$G$$ is a group of prime power order, hence it is nilpotent. On the other hand, $$G$$ is not generated by Abelian normal subgroups. Here's the reasoning:


 * Let $$H$$ be the cyclic subgroup generated by $$a$$.
 * Consider the elements of $$G$$ outside $$H$$. These elements fall in two conjugacy classes of size four each. The normal closure of any element outside $$H$$ must thus contain all its conjugates, and a quick inspection shows that this normal closure must be the dihedral group of order eight, which is not Abelian.
 * Hence, every Abelian normal subgroup of $$G$$ is contained in $$H$$.