Can anybody help me in finding the all possible subgroups of the group
<x,y:x²=y³=(xy)¹²>. Actually i want to find the the number of all subgroups of this group.
$\displaystyle x^2=(xy)^{12}\Longrightarrow x=y(xy)^{11}\,,\,and\,\,also\,\,y^3=(xy)^{12}\Long rightarrow y^2=(xy)^{11}x=(xy)^{11}y(xy)^{11}$ , thus it
follows that $\displaystyle \langle x,y\;;\;x^2=y^3=(xy)^{12}\rangle =\langle {y,(xy)^{11}\rangle \cong F_2=$ the free group in two (free) generators.
Now, did you actually mean to write what you did or you meant to give a presentation of the group in the
form $\displaystyle \langle x,y\;;\;x^2,y^3,(xy)^{12}\rangle$ ? Because this last is NOT a free group as it has non-trivial torsion elements, so
we get two different groups...or I'm wrong, of course. Check this.
Tonio