Commutator property in product of groups Consider the group $G=AB$ with $A$ and $B$ abelian. Is it true that $[a,b] \in [A,B]$ commutes with $[x,y]$ (with $x\in A$, $y \in B$) implies $[a,b]$ commutes with $[x^{-1},y^{-1}]$?
 A: If it helps, you do need to assume A and B permute.
For example, take $a=x$ and $b=y$.  Then obviously $\{a,x\}$ comes from an abelian group, and $\{b,y\}$ comes form an abelian group, and $[a,b]=[x,y]$ commutes with itself.  However, taking: $$a=\begin{bmatrix}1&1\\0&1\end{bmatrix}, \quad b=\begin{bmatrix}0&-1\\1&0\end{bmatrix}, \quad [a,b]=\begin{bmatrix}2&-1\\-1&1\end{bmatrix}, \quad [a^{-1},b^{-1}]=\begin{bmatrix}2&1 \\ 1&1\end{bmatrix}$$
and the last two do not commute.
Of course $ba$ cannot be written in the form $a^i b^j$ (look at the zero pattern) and so the subgroups A and B do not permute (AB is not a group).

To prove Ito's theorems along your lines, it seems easiest just to show (as Derek Holt indicated) that all commutators commute.  I believe this is done by by writing $b^x = a_1 b_1$ and $a^y=a_2b_2$ and then computing $[a,b]^{xy} = [a,b]^{yx}$ (simply by expanding) so that $[a,b]^{[x^{-1},y^{-1}]} = [a,b]$ and the two commutators commute.  Since $x,y$ are arbitrary, it does not matter if they have an inverse or not.  Similarly, since $a,b$ are arbitrary, $xy$ and $yx$ act as the same operator, so the commutator $[x,y]$ acts trivially.  Ito's (1) and (2) help to show $[AB,AB]=[A,B]$ so that one gets an action on a nice object.
Ito's original proof is extremely similar (equations (3) and (4) on the first page):


*

*Itô, Noboru.
"Über das Produkt von zwei abelschen Gruppen."
Math. Z. 62 (1955), 400–401.
MR71426
DOI:10.1007/BF01180647
