Equivalence of definitions of CN-group

This article gives a proof/explanation of the equivalence of multiple definitions for the term CN-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 nilpotent	the centralizer of any non-identity element is a nilpotent 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 a nilpotent subgroup of $G$.
2	subgroup centralizers are nilpotent	the centralizer of any nontrivial subgroup is a nilpotent subgroup.	for every nontrivial subgroup $H$ of $G$, the centralizer $C_{G}(H)$ is a nilpotent subgroup of $G$.
3	subgroups: nilpotent or centerless	every subgroup of the group is either a nilpotent group or a centerless group.	for every nontrivial subgroup $H$ of $G$, either $H$ is nilpotent or $H$ is centerless, i.e., the center of $H$ is trivial.

Facts used

1. Nilpotency 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 nilpotent.

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

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 nilpotent by assumption, hence, by Fact (1), $C_{G}(H)$ is nilpotent as well.

(2) implies (3)

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

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

Proof: $K\subseteq C_{G}(Z(K))$. The group $C_{G}(Z(K))$ is nilpotent by assumption, so by Fact (1), $K$ is nilpotent.

(3) implies (1)

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

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

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 nilpotent, completing the proof.