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Let $a_1,\ldots,a_{2n+1}$ be elements of a group $G$. Then $$a_1\cdots a_{2n+1}a_1^{-1}\cdots a_{2n+1}^{-1}$$ is a product of $n$ commutators.

The case $n=1$ was proven in this question and the proof of the above statement should use that and induction. I'm ashamed to admit that again I have no clue. It's probably quite analogous, but I really can't think of anything but blind guessing.

So it would be nice if someone could help me out.

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Actually, what was proven in that question was the $n=1$ case, not the $n=3$ case. – Arturo Magidin Feb 11 '12 at 0:44
up vote 4 down vote accepted

Let $g = a_3a_4 \cdots a_{2n+1}$. Then

$a_1a_2 \cdots a_{2n+1} a_1^{-1} a_2^{-1} \cdots a_{2n+1}^{-1} = (a_1a_2g a_1^{-1}a_2^{-1} g^{-1})(g a_3^{-1}a_4^{-1} \cdots a_{2n+1}^{-1})$

which is a product of two expressions of the same type, the first with $n=1$ and the second with $n$ one less than the original $n$.

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Here's a proof that it's the product of $4n-3$ commutators, but that's not what you asked for!

Let $g=a_2a_3\cdots a_{2n}$ and $h=a_2^{-1}a_3^{-1}\cdots a_{2n}^{-1}$. Now rewrite your product $$ a_1^{-1}ha_{2n+1}^{-1}a_1ga_{2n+1} $$ as $$ g^{-1}(ga_1^{-1}ha_{2n+1}^{-1}a_1ga_{2n+1}g^{-1})g.$$

Conjugation doesn't effect being the product of $n$ commutators, so let's focus on $$ ga_1^{-1}ha_{2n+1}^{-1}a_1ga_{2n+1}g^{-1}.$$

Now note that $$ [a_1g^{-1},a_{2n+1}h^{-1}] = ga_1^{-1}ha_{2n+1}^{-1}a_1g^{-1}a_{2n+1}h^{-1}.$$

And so we're reduced to writing the difference between the above two products as some combination of commutators. But the difference between the two is $$ ha_{2n+1}^{-1}ga_{2n+1}\cdot a_{2n+1}^{-1}ga_{2n+1}g^{-1}.$$

By induction, we're done.

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This may not be what the OP asked for, but since it shows that the element in question lies in the commutator subgroup, it may well be good enough... – Pete L. Clark Feb 11 '12 at 3:01
But it is clear that the image of the element in question is zero in the abelianization, so for that there is no need of Steve's argument. – Mariano Suárez-Alvarez Feb 11 '12 at 3:53

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