# CA not implies nilpotent

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., CA-group) need not satisfy the second group property (i.e., nilpotent group)
View a complete list of group property non-implications | View a complete list of group property implications

## 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:

## Proof

### Finite example

Further information: symmetric group:S3, subgroup structure of symmetric group:S3

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

Further information: infinite dihedral group

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.