Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 4 / 5 / i / Solution

No prime and no presentation of $G_1$ have p-deficiency at least one. Abelianizing the cyclic squaring relations makes each generator zero: for example $vw{v}^{-1}=w^2$ becomes $w=2w$, so $w=0$, and the other four relations kill $x,y,z,v$. Thus the abelianization of $G_1$ is trivial and $d_p(G_1)=0$ for every prime number $p$. The presentation-independent p-rank of a group bound from the general solution givesfor every group presentation $\mathcal{P}$ of $G_1$. This rules out alternative presentations, not just the one displayed.