Equivalence of definitions of CA-group

This article gives a proof/explanation of the equivalence of multiple definitions for the term CA-group
View a complete list of pages giving proofs of equivalence of definitions

The definitions that we have to prove as equivalent

No.	Shorthand	A group is termed a CN-group if ...	A group $G$ is termed a CN-group if ...
1	element centralizers are abelian	the centralizer of any non-identity element is an abelian subgroup.	for every non-identity element $x \in G$ , the centralizer $C_{G} (x)$ (i.e., the set of all elements of $G$ that commute with $x$ ) is an abelian subgroup of $G$ .
2	subgroup centralizers are abelian	the centralizer of any nontrivial subgroup is an abelian subgroup.	for every nontrivial subgroup $H$ of $G$ , the centralizer $C_{G} (H)$ is an abelian subgroup of $G$ .
3	subgroups: abelian or centerless	every subgroup of the group is either an abelian group or a centerless group.	for every nontrivial subgroup $H$ of $G$ , either $H$ is abelian or $H$ is centerless, i.e., the center of $H$ is trivial.

Related facts

Equivalence of definitions of CN-group

Facts used

Abelianness is subgroup-closed

Proof

(1) implies (2)

Given: A group $G$ such that, for every non-identity element $x \in G$ , the centralizer $C_{G} (x)$ is abelian.

To prove: $C_{G} (H)$ is abelian.

Proof: Since $H$ is nontrivial, there exists a non-identity element $x \in H$ . We have by definition of centralizer that $C_{G} (H) \subseteq C_{G} (x)$ . The latter is abelian by assumption, hence, by Fact (1), $C_{G} (H)$ is abelian as well.

(2) implies (3)

Given: A group $G$ such that for every nontrivial subgroup $H$ of $G$ , $C_{G} (H)$ is abelian. A subgroup $K$ of $G$ .

To prove: If the center $Z (K)$ of $K$ is nontrivial, then $K$ is abelian.

Proof: If the center $Z (K)$ is nontrivial, then $K \subseteq C_{G} (Z (K))$ . The group $C_{G} (Z (K))$ is abelian by assumption, so by Fact (1), $K$ is abelian.

(3) implies (1)

Given: A group $G$ with the property that for every subgroup $K$ of $G$ , $K$ is either abelian or centerless. A non-identity element $x \in G$ .

To prove: $C_{G} (x)$ is abelian.

Proof: Let $K = C_{G} (x)$ . Note that $x \in K$ , hence $x \in Z (K)$ . Thus, $K$ is not centerless. This forces it to be abelian, completing the proof.