# A finitely presented group with two simple relations

Is the group $G$ with the presentation $\langle x,y \;|\; x^7=1, y^2 x y=x^4\rangle$ solvable? infinite? I have computed by GAP the following fators of the derived series of $G$: $G/G'\cong C_3 \times C_7$, $G'/G'' \cong C_2 \times C_2 \times C_2 \times C_7$, and $G''/G^{(3)}$ and $G^{(3)}/G^{(4)}$ are elementary abelian $2$-groups of rank $8$ and $16$, respectively. I couldn't go further, it apparently needs more time and .... Maybe a simple trick needs here.

• I am confident that the group is infinite, but I don't have a formal proof. If you calculate the lower $2$-central series of the derived group of $G$ (which has index $21$ in $G$), then you will find that series (apparently) continues indefinitely with a periodic sequence of layers of sizes $3,2,3,3,2,3,3,2,3,3,2,\ldots$. I calculated it to class $50$, and the sequence continues with this pattern. I have seen this behaviour before and it typically indicates a representation of small degree over an infinite local field. I believe there are some techniques for calculating such representations. – Derek Holt Jun 7 '15 at 21:07

The group $G$ is not solvable, since its quotient $$\tilde{G} := \langle x, y \ | \ x^7 = 1, y^2xy = x^4, y^{15} = 1\rangle$$ is a group of order $423360$ such that $\tilde{G}'' \cong {\rm PSL}(3,4)$.
• @AlirezaAbdollahi you could try to compute with GAP a presentation of the kernel of $G\to\tilde{G}$ and see if its abelianization is infinite? I don't remember if the index $\sim 10^5$ is within reasonable computing time. – YCor Jun 7 '15 at 19:40
• @YCor: I did not (couldn't maybe better to say) compute the kernel of $G\rightarrow \tilde{G}$. Instead as I mentioned in the question, I was unable to find the abelianization of $G^{(4)}$, where its index in $G$ is $19730006016$. – Alireza Abdollahi Jun 7 '15 at 20:27