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I need help in folloving lemma.

Lemma: Any product of $k$ commutators is expressible in the form $a_1^{-1}a_2^{-1}...a_{2k}^{-1}a_1a_2...a_{2k}$

I guess author means that any product

$[x_1,y_1][x_2,y_2]...[x_k,y_k]$ can be expressible in that form $a_1^{-1}a_2^{-1}...a_{2k}^{-1}a_1a_2...a_{2k}$ where $[x_i,y_i]=x_i^{-1}y_i^{-1}x_iy_i$.

I tried to prove it by induction on $k$ but I failed.

If anyone can prove it,I would be thankful.

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  • $\begingroup$ Who's the author? $\endgroup$
    – Alexander Gruber
    Commented Mar 24, 2014 at 21:23
  • $\begingroup$ @AlexanderGruber: PETER YFF $\endgroup$
    – mesel
    Commented Mar 24, 2014 at 21:29

1 Answer 1

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For the inductive step: $$a_1^{-1}\cdots a_{2k}^{-1}a_1\cdots a_{2k}x^{-1}y^{-1}xy = b_1^{-1} \cdots b_{2k+2}^{-1}b_1\cdots b_{2k+2},$$

where

$b_1=xa_1$, $b_2=a_2x^{-1}$, $\ldots$, $b_{2k-1}=xa_{2k-1}$, $b_{2k}=a_{2k}x^{-1}$, $b_{2k+1}=y^{-1}x$, $b_{2k+2}=y.$

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  • $\begingroup$ Your welcome! I had seen this expression for the product of two commutators before, but it took a while to generalize it to arbitrary $k$, although it's not difficult once you see how to do it! $\endgroup$
    – Derek Holt
    Commented Mar 24, 2014 at 22:01

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