Normalizer criterion for maximal among abelian subgroups

Statement
Suppose $$G$$ is a group, $$A$$ is fact about::maximal among abelian subgroups in $$G$$ (in other words, $$A$$ is an abelian subgroup of $$G$$ not cintained in any bigger abelian subgroup. Suppose $$B$$ is a subgroup of $$G$$. Then, we have:

$$B \le C_G(A) \iff [[B,A],A] = 1$$.

Similarly, if $$H \le \operatorname{Aut}(G)$$, we have that:

$$H \le C_{\operatorname{Aut}(G)}(A) \iff [[H,A],A] = 1$$.

Note that this holds in particular when $$G$$ is a group of prime power order and $$A$$ is an fact about::abelian subgroup of maximum order.

Facts used

 * 1) uses::Equivalence of definitions of maximal among abelian subgroups: A subgroup that is maximal among abelian subgroups equals its own centralizer in the whole group.

Proof
Given': A subgroup $$A$$ that is maximal among abelian subgroups of $$G$$, a subgroup $$B$$ of $$G$$.

To prove: $$[[B,A],A] = 1 \iff B \le N_G(A)$$.

Proof:


 * 1) $$[[B,A],A] = 1 \iff [B,A] \le C_G(A)$$: This is by definition of the commutator.
 * 2) $$C_G(A) = A$$: This follows from fact (1).
 * 3) $$[B,A] \le A \iff B \le N_G(A)$$: This is by definition of normalizer.
 * 4) steps (1)-(3) combine to give the desired conclusion.