Does this group presentation define a nontrivial group?

Given a presentation $$ \langle x,y,z : x^y=x^2, y^z=y^2, z^x = z^2 \rangle, $$ where $x^y$ is just the usual conjugation (that is, $x^y$ is defined to be $y^{-1} xy$). Can we say for sure, whether this presentation defines a nontrivial group?

Solution 1:

This is actually a "well-known" presentation of the trivial group. I think it might have been originally proved trivial by Graham Higman, but I am not not certain. It can be done by hand, but it is IIRC a challenging exercise to do so!

It had the interesting feature that it can be used to construct a sequence of more complicated presentations of the trivial group that defeat coset enumeration programs. Write the presentation as

$G_1 = \langle x,y,z \mid y^{-1}xyx^{-2} = z^{-1}yxy^{-2} = x^{-1}zxz^{-2}=1 \rangle$.

Now define a new group $G_2$ with generators $a,b,c$, where the three relations are derived by substituting $x=b^{-1}aba^{-2}$, $y=c^{-1}bcb^{-2}$, $z=a^{-1}cac^{-2}$ in the presentation of $G_1$. So $G_2$ has three generators and three relations each of length $25$.

Since $G_1$ is trivial, the elements $b^{-1}aba^{-2}$, $c^{-1}bcb^{-2}$ and $a^{-1}cac^{-2}$ of $G_2$ must be trivial, but then, using the triviality of $G_1$ again, we deduce that $G_2$ is trivial. But I think GAP will struggle to prove that given the presentation of $G_2$.

You can repeat this construction to give groups $G_n=1$ with three generators and three relations, each of length $5^n$.