I am working on a family of finitely presented groups and I asking me the following question. Let $\phi$ be an application from $\left\{1,\dots,n\right\}$ to $\left\{1,\dots,n\right\}$ (not necessary bijective), we consider the group defined by: $$G_{n,\phi} = \langle x_1 , \dots, x_n \mid x_i^2, (x_i x_j)^4, x_i x_{\phi(i)} x_{i+1} x_{\phi(i)} \rangle. $$ Is $G_{n,\phi}$ isomorphic to $\mathbb{Z}/2\mathbb{Z}$?
It is clear that the abelianisation of $G_{n,\phi}$ is $\mathbb{Z}/2\mathbb{Z}$, so the question is equivalent to prove that $G_{n,\phi}$ a monogenous group or an abelian group. Futhermore, we have that $x_i$ and $x_{i+1}$ commute.
I have tested on GAP the first cases, and it seems to be true when $n\leq 7$. But I don't found a method to simplify this presentation for all $n$ and $\phi$. If you have any idea, it is welcome.
Edit I: The question is equivalent to prove that the group $K_{n,\phi}$ is trivial, where: $$ K_{n,\phi} = \langle y_1,\dots,y_n \mid y_1, (y_i y_j^{-1})^4, y_i y_{\phi(i)}^{-1} y_{i+1} y_{\phi(i)}^{-1} \rangle. $$ In this group we have the properties $(y_i y_{i+1}^{-1})^2=1$, which implies that the order of $y_2$ and $y_n$ is 2.
Edit II: I have tried to simplify the set of relations of $K_{n,\phi}$, and it seems that the groups: $$ S_{n,\phi}=\langle y_1,\dots,y_n \mid y_1, y_i^4, y_i y_{\phi(i)}^{-1} y_{i+1} y_{\phi(i)}^{-1} \rangle, $$ is always trivial. I have tested it on GAP for all $\phi$, and all $n\leq7$. In fact, I have also remarked that the groups: $$ T_{n,m,\phi}=\langle y_1,\dots,y_n \mid y_1, y_i^{2^m}, y_i y_{\phi(i)}^{-1} y_{i+1} y_{\phi(i)}^{-1} \rangle, $$ seems to be trivial. Once again, I have checked on GAP for all $\phi$ and all $n\neq 6$ and $m\leq 6$.
It is clear that $T_{n,1,\phi}$ is always trivial and that we have a natural morphism $\psi_m:T_{n,m,\phi}\to T_{n,m-1,\phi}$. For the moment, I do not see how use these last properties.