Equivalence of definitions of maximal among abelian subgroups

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 $$C_G(H) \le H$$ (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 $$G$$.

Equivalence of (1) and (2)
(1) says that $$H = C_G(H)$$. This is equivalent to saying that $$H \le C_G(H)$$ (i.e., $$H$$ is abelian) along with $$C_G(H) \le H$$ (i.e., $$H$$ is self-centralizing).

(1) implies (3)
Suppose $$K$$ is an abelian subgroup of $$G$$ containing $$H$$. Then, $$K$$ centralizes $$H$$, so $$K \le C_G(H)$$. But $$H = C_G(H)$$, so $$K \le H$$.

(3) implies (1)
Consider the group $$C_G(H)$$. Since $$H$$ is abelian, $$H \le C_G(H)$$.

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, $$H = C_G(H)$$.