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Problem: Show that the group given by the presentation $$\langle x,y,z \mid xyx^{-1}y^{-2}\, , \, yzy^{-1}z^{-2}\, , \, zxz^{-1}x^{-2} \rangle $$ is equivalent to the trivial group.

I have tried all sorts of manners to try to show that the relations given by the presentation above imply that $x=y=z=e$. However, I am stuck and would appreciate any hints as to how I should move forward.

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  • $\begingroup$ Do you know any reference where you found this question. I saw it in previous year question paper of my institution. $\endgroup$ – Bhaskar Vashishth Feb 5 '15 at 21:53
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This is a very well-known presentation of the trivial group, to be compared with the presentation of Higman's infinite group with no finite quotient. I do not know of any easy proof.

The proof I'm going to give is due to Bernhard Neumann in An Essay on Free Products of Groups with Amalgamations (Philosophical Transactions of the Royal Society of London, Series A, 246, 919 (1954), pp. 503-554.)

So we have to prove that in a group satisfying $$ xyx^{-1} = y^2 \qquad (R_1)$$ $$ yzy^{-1} = z^2 \qquad (R_2)$$ $$ zxz^{-1} = x^2 \qquad (R_3)$$ the elements $x$, $y$ and $z$ are trivial.

By inverting $(R_1)$, multiplying on the left by $y$ and on the right by $x$, we get $$yxy^{-1} = y^{-1}x.$$ This easily gives $$y^i x y^{-i} = y^{-i} x \qquad(R_1^{[i]}),$$ for every integer $i$, by induction. The same argument on the second relation gives $$z^i y z^{-i} = z^{-i} y. \qquad (R_2^{[i]})$$

If we now conjugate $(R_3)$ by $y$, the left-hand side becomes $$\begin{align}yzxz^{-1}y^{-1} &= z^2y\cdot x \cdot y^{-1}z^{-2}\\ & = z^2y^{-1}xz^{-2}\\ & = z^2 y^{-1}z^{-2}\cdot z^2xz^{-2}\\ & = y^{-1}z^2\cdot x^4 \end{align}$$ (the first equality is a double use of the relation $yz = z^2y$, a reformulation of $(R_2)$ ; the second uses $(R_1^{[1]})$ and the last uses the inverse of $(R_2^{[2]})$ and $R_3$ twice).

On the other hand, the left side becomes $$\begin{align} yx^2y^{-1} &= (y^{-1}x)^2 \\ &= y^{-1}xy^{-1}x \\ &= y^{-3}x^2 \end{align}$$ (the first equality uses the $(R_1^{[1]})$ twice, the last uses the inverse of $(R_1)$).

Put together, we have proven $y^{-1}z^2x^4 = y^{-3}x^2$, which gives $$z^2 = y^{-2}x^{-2}.\qquad (R^*)$$

If we conjugate $y$ by $z^{-2}$, we now get on the one hand $$\begin{align}z^{-2}y z^2 &= x^2 y^2 \cdot y \cdot y^{-2} x^{-2} \\ &= x^2 y x^{-2} \\ &= y^4 \end{align}$$ (the first equality uses $(R^*)$ twice, the last uses $(R_1)$ twice.) But, on the other hand, $z^{-2}yz^2 = z^2 y$ because of $(R_2^{[-2]}$). So we finally get $z^2 y = y^4$, which translates to $$z^2 = y^3.$$

This proves that $y$ and $z^2$ commute. The relation $(R_2)$ then boils down to $z = z^2$, which gives $z = 1$. Because of the symmetries in the presentation, this proves that the group is trivial.

Not very enlightening, but the fact that the corresponding group with 4 generators is highly nontrivial somehow reduces my hopes of ever finding a "good reason" for this group to be trivial.

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Here is a proof-

$G=\langle x,y,z \mid xyx^{-1}=y^{2}\, , \, yzy^{-1}=z^{2}\, , \, zxz^{-1}=x^{2} \rangle$. See that $xyx^{-1}y^{-1}=y$ tells you that $y$ belongs in the Commutator subgroup generated by $x$ and $y$ and similarly $x$ belongs in Commutator subgroup generated by $x$ and $z$ and $z$ belongs in Commutator subgroup generated by $z$ and $y$ and Now this gives you that $G$ is perfect i.e. $G=G'$. Now If you can prove $G$ is solvable , you are done as only perfect solvable group is trivial group.

So consider the subgroup generated by $H=\langle x,y \rangle$ and show that $H$ is solvable. For that consider $H_1=\langle y \rangle < H$ and it is easily seen that $H_1 \unlhd\ H$ so what is factor group $H/H_1$? Yeah correct, $H/H_1\ \cong\ \langle x \rangle$ which is abelian and hence $H$ is solvable.

Now only thing remains is to check that $H=G$, i.e. $z \in \langle x,y \rangle$. Now is the tedious calculation work you will have to do, in order to prove this, use the relators given and express $z$ in terms of $x$ and $y$.

I hope this helps!

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