If $A$ and $B$ are subgroups of $G$ such that $[[A, B], B] = 1$, then $[A, B]$ is abelian. I am stuck on the following question:
If $A$ and $B$ are subgroups of $G$ such that $[[A, B], B] = 1$, then $[A, B]$ is abelian, where $[X, Y]$ denotes the commutator subgroup of $X$ and $Y$ as usual.
I have seen that $[A, B]$ is in the intersection of all the centralizer of each $b\in B$. And I am trying to show that $B$ is a normal subgroup of $G$, then it follows that $[A, B] \subseteq Z(B)$, where $Z(B)$ denotes for the centre of $B$. Then the result follows.
But I have no idea how to show that $B$ is normal in $G$ or maybe I am completely on the wrong way.
Anyone can help?
 A: Lemma Let $A,B$ be subgroups a of group $G$, then both $A$ and $B$ normalize $[A,B]$.
Proof By symmetry we only prove this for $A$: let $a,a_1 \in A, b_1 \in B$, then
$$a^{-1}[a_1,b_1]a=(a_1a)^{-1}b_1^{-1}(a_1a)b_1.b_1^{-1}a^{-1}a_1^{-1}a_1b_1a=[a_1a,b_1][b_1,a] \in [A,B].$$
Proposition Let $A,B$ subgroups of a group $G$ and let $A$ or $B$ centralize $[A,B]$, then $[A,B]$ is abelian.
Proof Again, by symmetry, we only need to prove this if $[[A,B],B]=1$. Let $a_1,a_2 \in A$ and $b_1,b_2 \in B$, it is enough to show that the commutators $[a_1,b_1]$ and $[a_2,b_2]$ commute. Observe that by the Lemma $[a_2,b_2]^{a_1^{-1}} \in [A,B]$:
$$[a_1,b_1][a_2,b_2]=a_1^{-1}b_1^{-1}a_1b_1[a_2,b_2]\underset{b_1 \text { commutes with } [a_2,b_2]}=a_1^{-1}b_1^{-1}a_1[a_2,b_2]b_1=a_1^{-1}b_1^{-1}[a_2,b_2]^{a_1^{-1}}a_1b_1 =\underset{\text{ using the Lemma and } b_1^{-1} \text{ centralizes }[A,B]}=\\ a_1^{-1}[a_2,b_2]^{a_1^{-1}}b_1^{-1}a_1b_1=[a_2,b_2][a_1,b_1].$$ 
Note (added March 24th 2021) There is still another elegant way to show that $[A,B]$ is abelian, and that is by an appeal to the Three Subgroups Lemma:
Lemma Let $X,Y,Z$ be subgroups of a group $G$. If $[X,[Y,Z]]=1$ and $[Z,[X,Y]]=1$ then $[Y,[Z,X]]=1$.
Note that the subgroups/letters are shifted cyclically. For a proof and a slight generalization (replacing $1$ by containment in a normal subgroup $N$) see for example here. Proofs usually apply the so-called Hall-Witt Identity. 
Now let's show that if $[A,B]$ is centralized by $B$, $[A,B]$ must be abelian. By the first Lemma, $A$ normalizes $[A,B]$, so $[A,[A,B]] \subseteq [A,B]$, implying $[B,[[A,B],A]]=1$. Further, $[A,[B,[A,B]]]=[A,1]=1$. The Three Subgroups Lemma now yields, $[[A,B],[A,B]]=1$, that is $[A,B]$ is abelian.
A: First of all, each $b \in B$ commutes with $[A, B]$.
$[a_1,b_1][a_2,b_2]=(a_1^{-1}b_1^{-1}a_1)[a_2,b_2]b_1=(a_1^{-1}b_1^{-1}a_1)[a_2,b_2](a_1^{-1}b_1^{-1}a_1)^{-1}[a_1,b_1]=[a_2,b_2][a_1,b_1]$
as
$(a_1^{-1}b_1^{-1}a_1)[a_2,b_2](a_1^{-1}b_1^{-1}a_1)^{-1}=[a_2,b_2]$ as $b_1$ commutes with any $[a, b]$.
Sorry I don't have any mathtype software please don't be upset.
