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.
$\endgroup$ 72 Answers
$\begingroup$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.
$\endgroup$ $\begingroup$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.
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