TITLE: A finitely presented group with two simple relations
QUESTION [6 upvotes]: 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.

REPLY [7 votes]: 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 $\tilde{G}'' \cong {\rm PSL}(3,4)$.