The commutator identity in $2$-engel groups Let $G$ be a group such that $[x,y,y]=1$ for all $x,y\in G$. How to show that $[x,y,z]=[y,z,x]$?
I remember that this was true but I could not obtain it. I can see that
$$[x,y,z][y,z,x][z,x,y]=1$$ but I could not proceed from this equation.
 A: I shall use these definitions: $[x,y]=xyx^{-1}y^{-1}$ and $x^y=yxy^{-1}$. So, in general:


*

*$[x,y]=x(x^{-1})^y= y^xy^{-1}$;

*$[x,y]^{-1}=[y,x]$;

*$[x,yz]= [x,y][x,z]^y$;

*$[xy,z]= [y,z]^x[x,z]$.


Assume now $[x,y,y]=e$.


*$x^G\subseteq C(x)$ as $e=[y,x,x]= [[y,x],x]= [x^yx^{-1},x]= [x^{-1}x]^{x^y}[x^y,x]= [x^y,x]$.

*Moreover, elements of $x^G$ mutually commute because $x^yx^z=zx^{z^{-1}y}xz^{-1}= zxx^{z^{-1}y}z^{-1}= x^zx^y$.

*$[x,G]\subseteq C(x)$ as $[x,y]x=x(x^{-1})^yx=xx(x^{-1})^y=x[x,y]$. 

*Moreover, elements of $[x,G]$ mutually commute because $[x,y][x,z]=x(x^{-1})^yx(x^{-1})^z= x(x^{-1})^z x(x^{-1})^y=[x,z][x,y]$. 

*$[x,y]^{-1}= [y,x]= x^yx^{-1}= x^{-1}x^y=[x^{-1},y]$. 

*$[x,y]^{-1}= [y,x]= [y^{-1},x]^{-1}= [x,y^{-1}]$. 

*$[x,y^{-1},z^{-1}]=[x,y,z]$. 

*$[x,y,z]= [[x,y],z]= [x,y]z[x,y]^{-1}z^{-1}=$ $ [x,y][x,z][x,z]^{-1}z[x,y]^{-1}z^{-1}=$ $ [x,y][x,z][x^{-1},z]z[x^{-1},y]z^{-1}= $ $[x,y][x,z]x^{-1}zxx^{-1}yxy^{-1}z^{-1}= $ $ [x,y][x,z]x^{-1}zyxy^{-1}z^{-1}= $ $[x,y][x,z][x^{-1},zy]= $ $[x,y][x,z][x,zy]^{-1}$. 

*$[x,y,z]^{-1}= $ $([x,y][x,z][x,zy]^{-1})^{-1}=$ $ [x,y]^{-1}[x,z]^{-1}[x,zy]= $ $[x,y^{-1}][x,z^{-1}][x,y^{-1}z^{-1}]^{-1}= $ $[x,z^{-1}][x,y^{-1}][x,y^{-1}z^{-1}]^{-1}= $ $[x,z^{-1},y^{-1}]=$ $[x,z,y]$. 

*$[z,x,y]= [[x,z]^{-1},y]= [x,z,y]^{-1}=[x,y,z]$. 

