Abelian implies nilpotent

Verbal statement
Any abelian group is a nilpotent group. In fact, abelian groups are precisely the nilpotent groups of nilpotency class equal to 1.

Using upper central series
Given: An Abelian group $$G$$

To prove: The upper central series of $$G$$, terminates, after finitely many steps, at $$G$$

Proof: The first member $$Z^{(1)}(G)$$ of the upper central series is the center of $$G$$, which is the whole of $$G$$ since $$G$$ is Abelian. Thus the upper central series terminates in 1 step, and $$G$$ is nilpotent of nilpotence class 1.

Using lower central series
Given: An abelian group $$G$$

To prove: The lower central series of $$G$$ terminates in finitely many steps at the trivial subgroup

Proof:The first member of the lower central series of $$G$$ is the derived subgroup of $$G$$, which is trivial because $$G$$ is abelian. Thus the upper central series terminates in 1 step, and $$G$$ is nilpotent of class 1.

Converse
The converse of this statement is not true: nilpotent not implies abelian.

For finite groups

 * Dedekind group: This is a group where every subgroup is normal.
 * ACIC-group: This is a group where every automorph-conjugate subgroup is characteristic. For finite groups, ACIC implies nilpotent. In general, it does not, but Abelian always implies ACIC.
 * Group generated by abelian normal subgroups: This property is weaker than Abelianness. If we require only finitely many Abelian normal subgroups to generate the group, then it is nilpotent. In particular, for finite groups, such a group is always nilpotent.
 * Finite group that is 1-isomorphic to an abelian group: This is a finite group for which there exists a bijection to an abelian group (and hence, a finite abelian group) with the property that the bijection is a 1-isomorphism, i.e., it restricts to an isomorphism on cyclic subgroups.
 * Finite group that is order statistics-equivalent to an abelian group: This is a finite group whose order statistics match those of a finite baelian group.