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Prove that the group $G$ with generators $x,y,z$ and relations $z^y=z^2$, $x^z=x^2$, $y^x=y^2$ has order $1$.

This is a problem on Page 56 of Derek J.S. Robinson's A Course in the Theory of Groups (GTM 80). I think $z^y$ means the result of $y$ acting on $z$, and may be defined as $y^{-1}zy$.

Suppose that $F$ is a free group generated by $x,y,z$. The epimorphism $\pi: F \rightarrow G$ has its kernel $K$ generated by $z^yz^{-2}$, $x^zx^{-2}$, $y^xy^{-2}$. How to prove $K=F$? Or, how to prove $x,y,z \in K$? I've tried but didn't find the right way.

Thank you very much.

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Yes: if $a$ and $b$ are elements of the group, then $a^b = b^{-1}ab$. For this kind of problem, you usually want to manipulate the relations given to conclude that $x=y=z=1$, rather than trying to work with $K$ and attempting to show that $K=F$ (it really amounts to the same thing, since showing $x=1$ amounts to showing $x\in K$, etc; it's just easier to manipulate the identities). –  Arturo Magidin Aug 9 '11 at 4:02
    
@Arturo Magidin: I see. I'll make attempts in this direction. Thank you very much. –  ShinyaSakai Aug 10 '11 at 14:47
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1 Answer

up vote 3 down vote accepted

EDIT: I changed the steps to be somewhat simpler.

Since this is tagged homework, I will only give hints (I hope!).

  1. Look at $z^{yx}$, and write it in two fairly different ways; use the equality between these two words to write $x$ as a word in $y$ and $z$.

  2. Now look at $y^x$, and write it in two different ways; use the equality between these two words to show $x$ is a power of $z$.

  3. Use the fact $x$ is a power of $z$ to write $z^x$ in two different ways, and show $x$ is trivial; conclude your group is trivial.

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Thank you very much for the hint. I tagged it homework because I am still a beginner in group theory. But I am sorry I still have difficulty solving this. From $z^{yx}=(z^y)^x=(z^2)^x=x^{-1}z^2x$, and $z^{yx}=(y^{-1}zy)^x=(y^{-1})^xz^xy^x=(x^{-1}y^{-1}x)(x^{-1}zx)y^2=x^{-1}y^{-1}z‌​xy^2$, I see that $z^2x=y^{-1}zxy^2$. As $z^2=y^{-1}zy$, $x=y^{-1}xy^2$. Then how can I write $x$ in $y$ and $z$? Many Thanks. –  ShinyaSakai Aug 10 '11 at 14:41
    
From the relations given for the group, derive the further relations $z^x=zx^{-1}$, $x^y=xy^{-1}$, $y^z=yz^{-1}$. –  user641 Aug 10 '11 at 16:16
    
Thus $z^{yx}=z^{2x}=z(x^{-1}z)x^{-1}=z(zx^{-2})x^{-1}=z^2x^{-3}$. –  user641 Aug 10 '11 at 16:17
    
We also have $z^{yx}=(x^{-1}y^{-1})z(yx)=(y^{-2}x^{-1})z(xy^2)=y^{-2}(zx^{-1})y^2=z^4(y^{-2}x‌​^{-1}y^2)=z^4y^2x^{-1}$. –  user641 Aug 10 '11 at 16:28
    
So $z^2x^{-3}=z^4y^2x^{-1}$, or $x^2=y^{-2}z^{-2}$, and conjugating both by $z^{-1}$ gives $x=zy^{-2}z^{-3}=z^{-11}y^{-2}$. –  user641 Aug 10 '11 at 16:30
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