Nilpotent not implies generated by abelian normal subgroups

This article gives the statement and possibly, proof, of a non-implication relation between two group properties. That is, it states that every group satisfying the first group property (i.e., nilpotent group) need not satisfy the second group property (i.e., group generated by Abelian normal subgroups)
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Statement

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

Proof

Example of the dihedral group

Further information: dihedral group:D16

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

$G=⟨a,x\mid a^8=x^2=e,xa{x}^{-1}=a^{-1}⟩$.

$G$ 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 $H$.