Some commutator identities On the way to study Lang's algebra, I cannot solve this problem. See page 69. 
Let G be a group and denote the commutator of x and y by & $[x,y]=xyx^{-1}y^{-1}$. 
I wanna prove that if $[x,y]=y, [y,z]=z, [z,x]=x$ then $x=y=z=e$. 
I tried before posting it, but I don't have a clue. Please give me some hints or solution. 
Thanks in advance. 
 A: Writing commutator identities explicitly gives 
$$xyx^{-1}=y^2, yzy^{-1}=z^2, zxz^{-1}=z^2.$$
It appeared here, in similar way, that a group $\langle x,y,z\rangle$ with these relations is trivial.
A: Here is the approach I outlined in the comments above.
From the identity $[x,y]=y$, one sees that $xyx^{-1} = y^2$.  I will write $y^x$ from now on, for $xyx^{-1}$; similarly, I write ${}^xy$ for $x^{-1}yx$.
We also see from this commutator relation that ${}^yx=yx$ and $x^y=y^{-1}x$. Similar identities can be deduced from the other two commutator relations.
So a quick check shows $[x,y^{-1}]=y^{-1}$, and so
\begin{align}
[[x,y^{-1}],z] &= y^{-1}(zyz^{-1}) \\
&= y^{-1}z^{-1}y
\end{align}
(The second equality comes from $y^z=z^{-1}y$). This means that
$$ [x,y^{-1},z]^y=z^{-1} $$
Similarly, we get
$$ [y,z^{-1},x]^z = x^{-1} $$
and
$$ [z,x^{-1},y]^x = y^{-1} $$
The Hall-Witt identity then shows $z^{-1}x^{-1}y^{-1}=1$, or
$$ z = x^{-1}y^{-1} $$
Let's put that value for $z$ into the final commutator relation; we get
\begin{align}
x &= [z,x] \\
&= [x^{-1}y^{-1},x] \\
&= x^{-1}y^{-1}xyxx^{-1} \\
&= x^{-1}(y^{-1}xy) \\
&= x^{-1}yx
\end{align}
(The last equality comes from ${}^yx=yx$).  Conjugating both sides by $x$ shows $x=y$.  Thus $y=[x,y]=1$, and hence $x=y=1$, and so $z=1$ as well; the group is trivial.
