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In PlanetMath there is an identity of commutators which I think is wrong. This identity is the fourth identity in Theorem 5 of this page. I think that the correct identity is this one, computed by me:

PlanetMath identity: $[x^z,y]=\left[x,y^{z^{-1}}\right]$, where $x^z=z^{-1}xz$

Identity computed by me: $[x^z,y]=\left[x,y^{z^{-1}}\right]^z$

Am i correct?

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    $\begingroup$ I just looked at OP's profile - age 14. $\endgroup$ – scaaahu May 15 '12 at 12:41
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Yes, you are correct. We have that $[x,y]^z = [x^z,y^z]$, so $$[x^z,y] = \left[ x^z, (y^{z^{-1}})^z\right] = [x,y^{z^{-1}}]^z = [x,y^{z^{-1}}]\left[[x,y^{z^{-1}}],z\right].$$

To see that the claimed identity given by PlanetMath does not hold in general, note that $[x,y^{z^{-1}},z]$ is not trivial in general; or you can just expand (using PlanetMath's convention of $[a,b]=a^{-1}b^{-1}ab$): $$\begin{align*} {}[x^z,y] &= (z^{-1}xz)^{-1}y^{-1}(z^{-1}xz)y\\ &= z^{-1}x^{-1}zy^{-1}z^{-1}xzy\\ {}[x,y^{z^{-1}}] &= x^{-1}(zyz^{-1})^{-1}x(zyz^{-1})\\ &= x^{-1}zy^{-1}z^{-1}xzyz^{-1}, \end{align*}$$ which are different in the free group (both are reduced words but they are not identical).

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