CA not implies nilpotent

Statement
It is possible to have a CA-group $$G$$ that is not a nilpotent group. Here, CA means that the centralizer of every non-identity element is abelian. Equivalently, it means that every proper subgroup is either a centerless group or an abelian group.

This also shows that:


 * A CN-group need not be a nilpotent group.
 * A CA-group need not be an abelian group.

Finite example
The group symmetric group:S3, defined as the group of permutations on $$\{ 1,2,3 \}$$, is a CA-group that is nontrivial and centerless, and therefore not nilpotent. To see that it is CA, note that it is centerless, and all its proper subgroups are abelian.

Infinite example
Consider the infinite dihedral group:

$$G := \langle a,x \mid x = x^{-1}, xax^{-1} = a^{-1} \rangle$$

This is centerless, and therefore, not nilpotent. On the other hand, the centralizer of every non-identity element is abelian. To see this, note that:


 * The centralizer of any non-identity element inside $$\langle a \rangle$$ is precisely $$\langle a \rangle$$. This is abelian. In fact, it is isomorphic to the group of integers.
 * For any element outside $$\langle a \rangle$$, the element has order two and its centralizer is the cyclic subgroup it generates. This is abelian, and is isomorphic to cyclic group:Z2.