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.
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.
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!
I found a few different methods in efforts to finding possible simpler/intuitive solutions. They are still complicated, but all centered around the idea of finding a simple identity involving 3 elements and applying commutator iteratively.
Hopefully someone can double check these to make sure that I'm not producing some junk out of nowhere. I probably have burnt my brain after staring at all these anyways.
List of identities for easy dictionary: \begin{align*} &R_x: zxz^{-1} = x^2, && R_x': xz^{-1}x^{-1} = z^{-1}x, \\ &R_y: xyx^{-1} = y^2, && R_y': yx^{-1}y^{-1} = x^{-1}y, \\ &R_z: yzy^{-1} = z^2, && R_z': zy^{-1}z^{-1} = y^{-1}z. \end{align*}
First Solution: Starting from $R_x$, our goal is to eliminate $x$ on RHS: \begin{align*} x^2 &= \underline{z}xz^{-1} & (R_z) \\ &= yzy^{-1}\underline{z^{-1}x}z^{-1} & (R_x') \\ &= yz\underline{y^{-1}x}z^{-1}x^{-1}z^{-1} & (R_y'^{-1}) \\ &= yzyx\underline{y^{-1}z^{-1}}x^{-1}z^{-1} & (R_z) \\ &= yzyxz^{-1}y^{-1}\underline{zx^{-1}z^{-1}} & (R_x) \\ &= yzy\underline{xz^{-1}}y^{-1}x^{-2} & (R_x) \\ &= yzyz^{-1}\underline{x^2y^{-1}x^{-2}} & (R_y^{-1}) \\ &= yz\underline{yz^{-1}y^{-4}} & (R_z) \\ &= \underline{yz^3y^{-3}} & (R_z) \\ &= z^6y^{-2}, \end{align*} Here I underlined the part that we are operating with, with the relation in the parenthesis. Then by symmetry, we have the following $3$ identities: $$ x^2y^2 = z^6, \quad y^2z^2 = x^6, \quad z^2x^2 = y^6. $$ Next, one can easily check that, by applying $R_x$, we have $ z^2x^2 = x^8z^2 $. Hence, by this and the last two identities above, we can infer that \begin{align*} y^6 &= z^2x^2 = x^8z^2 = y^2z^2x^2z^2 \\ \Rightarrow y^4 &= z^2x^2z^2 = x^8z^4 = y^2z^2x^2z^4 \\ \Rightarrow y^2 &= z^2x^2z^4 = x^8z^6 = y^2z^2x^2z^6 \\ \Rightarrow x^2 &= z^{-8}. \end{align*} This and $R_x$ then implies that $ x = x^2 $ and we can conclude the proof by symmetry.
Second Solution: In Serge Lang's algebra, right before this result, he included the cocycle relation $$ [x, yz] = [x,y] \ {^y[x,z]}, $$ where $ {^a b} = aba^{-1} $ is the exponent notation for conjugation, making me suspicious of using this for the proof. I did find an easy solution, well, conceptually, through this approach:
Same dictionary as before: \begin{align*} &R_x: zxz^{-1} = x^2, && R_x': xz^{-1}x^{-1} = z^{-1}x, \\ &R_y: xyx^{-1} = y^2, && R_y': yx^{-1}y^{-1} = x^{-1}y, \\ &R_z: yzy^{-1} = z^2, && R_z': zy^{-1}z^{-1} = y^{-1}z. \end{align*}
Interchanging the role of $x,y$, we have $$ [y,xz] = [y,x] \ {^x [y,z]}. $$ Then applying all the commutators (so that we don't end up with a trivial useless equation), we have \begin{align*} &LHS \stackrel{R_x}{=} [y,x^{-1}zx] = yx^{-1}z\underline{xy^{-1}x^{-1}}z^{-1}x \stackrel{R_y}{=} yx^{-1}\underline{zy^{-2}z^{-1}}x \stackrel{R_z'}{=} yx^{-1}y^{-1}zy^{-1}zx \\ = &RHS \stackrel{R_y, R_z}{=} y^{-1}xzx^{-1} \stackrel{R_x^{-1}}{=} y^{-1}x^{-1}z. \end{align*} Then main observation is the following: Since we have exactly one $z$ in the above equation ignoring $x,y$, therefore, if we just routinely apply $R_y$, we should be able to write $z$ in terms of $x,y$ in a simple way. With this in mind, we can manipulate as follows:
\begin{align*} &y^{-2}x^{-1} = x^{-1}y^{-1}zy^{-1}\underline{zxz^{-1}} \stackrel{R_x}{=} x^{-1}y^{-1}zy^{-1}x^2 \\ \Rightarrow \ &z = y\underline{xy^{-2}x^{-1}}x^{-2}y = y^{-3}x^{-2}y = x^{-2}y^{-11}. \end{align*} Here we applied $R_y$ repetitively in the last line (conjugation of $y$ by $x$ doubles its exponent for quick mental calculation). Rewrite above result as $ zx = y^{-11} $ using $R_x$. Then the rest is easy: $$ z = y^{-11}x^{-1} = x^{-1}y^{-22} = x^{-1}zxzx = x^{-1}zx^3z = x^5 z^2, $$ which implies that $z$ commutes with $x$ and completes that proof.
Third Solution: I probably have spent an unnecessary amount of time on this question, but I thought that exchanging $x,y$ does not make any sense, so I cooked up another solution without doing so.
Dictionary: \begin{align*} &R_x: zxz^{-1} = x^2, && R_x': xz^{-1}x^{-1} = z^{-1}x, \\ &R_y: xyx^{-1} = y^2, && R_y': yx^{-1}y^{-1} = x^{-1}y, \\ &R_z: yzy^{-1} = z^2, && R_z': zy^{-1}z^{-1} = y^{-1}z. \end{align*}
Again, starting from the cocycle relation $$ [x, yz] = [x,y] \ {^y[x,z]}, $$ applying all $3$ commutators gives \begin{align*} RHS &= y\underline{yx^{-1}y^{-1}} \stackrel{R_y'}{=} yx^{-1}y \\ = LHS &= [ x, z^2y ] = xz^2\underline{yx^{-1}y^{-1}}z^{-2} \\ &\stackrel{R_y'}{=} \underline{xz^2x^{-1}}yz^{-2} \\ &\stackrel{R_x'^{-1}}{=} x^{-1}zx^{-1}\underline{zyz^{-2}} \\ &\stackrel{R_z'^{-1}}{=} x^{-1}\underline{zx^{-1}z^{-1}}yz^{-1} \\ &\stackrel{R_x}{=} x^{-3}yz^{-1}. \end{align*} Isolating $z$ from the equation above then gives $$ z = y^{-1}\underline{xy^{-1}x^{-3}}y = \underline{y^{-3}x^{-2}}y = x^{-2}y^{-11}, $$ which is the same identity we got in the second solution.
