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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 $[x^{-1},y^{-1}]$?

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By a result of Ito, a group that is a product of abelian groups is metabelian, so all commutators commute with each other. – Derek Holt Mar 15 '12 at 13:39
Thanks, but I need that result in order to prove Ito's theorem (yes, it's at the beginning of the proof...). I believe the solution could be achieved substituting words like $a^{-1}ba$ with elements like $a^\prime b^\ast$ (or $b^\prime a^\ast$, by the fact that $G=AB=BA$), but I was not able to obtain the exact result. – Oo3 Mar 15 '12 at 16:43
I have looked at the proof of Ito's theorem, which is in German but very short, and I don't understand why you think you need to prove the result you are asking about. Which step exactly in the proof of Ito's Theorem do you not understand? – Derek Holt Mar 15 '12 at 23:05
What I do not understand is that we prove $[a,b]^{xy}$ and $[a,b]^{yx}$ are equal, so it follows that $[a,b]^{xyx^{-1}y^{-1}} = [a,b]$. But this means that $[a,b]$ commutes with $[x^{-1},y^{-1}]$, I wish it was $[x,y]$ instead. – Oo3 Mar 16 '12 at 17:58
But $x$ and $y$ are general elements of $A$ and $B$ here, so you can just replace $x,y$ by $x^{-1}$ and $y^{-1}$. – Derek Holt Mar 16 '12 at 18:31
up vote 3 down vote accepted

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
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I thought that assuming $AB$ as a group $G$ would made $A$ and $B$ automatically permutable. – Oo3 Mar 16 '12 at 18:01
@Oo3: yes, and that assumption is necessary. – Jack Schmidt Mar 16 '12 at 18:32

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