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 $[x^{-1},y^{-1}]$?
Commutator property in product of groups
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0@Oo3: luckily$x$and$y$are arbitrary. – 2012-03-16
1 Answers
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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0@Oo3: yes, and that assumption is necessary. – 2012-03-16