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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.

JeongHobin
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  • Are $x$, $y$ and $z$ arbitrary? Then letting $x = e$ in the first commutator yields $y = e$.In a similar fashion we will have $z = e$ from the second and and $x = e$ from the third. – eltonjohn Apr 12 '16 at 06:36
  • Yeah, then if all of them were not identity, how to prove it? – JeongHobin Apr 12 '16 at 06:56
  • Hint: rewrite these as conjugation identities (so y conjugated by x is y^2). Use this to get a finite order for x, then y, then z. Use the finite order of z to get another finite order for x, and use the two orders of x to show its trivial. – Steve D Apr 12 '16 at 07:21
  • Actually, I see an easier way to approach this: the three subgroup lemma shows xyz=1. From here it is easy to show one (and hence all) the generators are trivial. – Steve D Apr 12 '16 at 08:23
  • @Steve D: I doubt whether the OP will be aware of the three subgroup lemma. – Geoff Robinson Apr 12 '16 at 09:43
  • I didn't see how three subgroup lemma will be applicable to this problem. – p Groups Apr 12 '16 at 09:57
  • @pGroups: Sorry, I meant the Hall-Witt identity. I added an answer explaining the approach. – Steve D Apr 15 '16 at 20:44

2 Answers2

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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.

Community
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p Groups
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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.

Steve D
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