Equivalence of definitions of CA-group

Related facts

 * Equivalence of definitions of CN-group

Facts used

 * 1) uses::Abelianness is subgroup-closed

(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.