Cyclic Group Presentation Show that the the group with presentation $$\langle x, y\ \mid\ x^2=y^2x^2y,\ (xy^2)^2=yx^2, \ yx^{-1}y^2=x^7\rangle $$ is cyclic of order 24.
This presentation was obtained using the Todd-Coxeter process for a subgroup of index 2 in the group presented in problem 476854.

$$x^2=y^2x^2y$$  
$$x=x^{-1}y^2x^2y=y^{-1}(yx^{-1}y^2)x^2y$$
$$x=y^{-1}x^9y$$
then we can say that;
$$y^2x^2y=x^2=y^{-1}x^{18}y$$
$$y^2x^2=y^{-1}x^{18}$$
$$y^3=x^{16}$$
Now, $$(xy^2)^2=yx^2$$
$$xy^2xy^2=yx^2$$
$$xy^2x(y^3)=yx^2y=y^3x^2y^2=x^{18}y^2$$
$$y^2x^{17}=x^{17}y^2$$
$$y^5x=xy^{5}$$
As a last step;
$$x^2=y^2x^2y$$
$$x^{18}=x^2y^3=y^3x^2=(y^5x^2)y=x^2y^6$$
$$x^2y^3=x^2y^6$$
$$e=y^3$$
Thus, we can say that $x,y^2$ will commute with each other so will $x,y^4=y$. And $x=y^{-1}x^9y=x^9\implies x^8=e$. From that point you can easily conclude that $G$ is an cyclic group of order $24$.

N.Q.E.D
