TITLE: commutator identity on planetmath
QUESTION [8 upvotes]: 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?

REPLY [8 votes]: 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).