Equivalence of definitions of maximal among abelian subgroups

This article gives a proof/explanation of the equivalence of multiple definitions for the term maximal among abelian subgroups
View a complete list of pages giving proofs of equivalence of definitions

Statement

The following are equivalent for a subgroup $H$ of a group $G$:

1. $H=C_{G}(H)$.
2. $H$ is an abelian subgroup of $G$ and $C_{G}(H)\leq H$ (i.e., $H$ is a self-centralizing subgroup of $G$).
3. $H$ is an abelian subgroup of $G$ and it is not contained in any bigger abelian subgroup of $G$.

Proof

Equivalence of (1) and (2)

(1) says that $H=C_{G}(H)$. This is equivalent to saying that $H\leq C_{G}(H)$ (i.e., $H$ is abelian) along with $C_{G}(H)\leq H$ (i.e., $H$ is self-centralizing).

(1) implies (3)

Suppose $K$ is an abelian subgroup of $G$ containing $H$. Then, $K$ centralizes $H$, so $K\leq C_{G}(H)$. But $H=C_{G}(H)$, so $K\leq H$.

(3) implies (1)

Consider the group $C_{G}(H)$. Since $H$ is abelian, $H\leq C_{G}(H)$.

Suppose $H$ is properly contained in $C_{G}(H)$. Then, there exists $x\in C_{G}(H)\setminus H$. Consider the group $K=\langle H,x\rangle$. Since $x$ centralizes $H$, and $H$ is abelian, $K$ is abelian. Thus, $K$ is an abelian subgroup of $G$ properly containing $H$. This contradicts the assumption that $H$ is maximal among abelian subgroups.

Thus, $H=C_{G}(H)$.