Use the Todd-Coxeter Algorithm to analyse the group generated by two elements x,y, with the following relations determine the order of the group and identify the group if you can.

$$x^3=y^3=1, xyx = yxy$$

Given function is, $xyx = yxy$ Multiply $x^2$ to both sides we get, $x^3yx = x^2yxy$

Here we know that $x^3 = 1$ so, $yx = yxyx^2$

on applying Right cancellation law $y = yxyx$

On applying Left cancellation Law $xyx = 1$


$yxy = 1$

so, $xyx = yxy = 1$

Hence the order of the Group is 1 and the Types of Group is Cyclic.

  • $\begingroup$ I'm with you until Here we know $x^3 = 1$ so,...I get $ yx = x^2 y x y $ $\endgroup$
    – Will Jagy
    Commented Nov 11, 2013 at 21:21
  • $\begingroup$ why commutative? $\endgroup$
    – Will Jagy
    Commented Nov 11, 2013 at 21:41
  • $\begingroup$ Meanwhile, if you experiment by adding the extra relation $x=y,$ you get the cyclic group of order three. So you have messed up in any case. $\endgroup$
    – Will Jagy
    Commented Nov 11, 2013 at 21:49
  • $\begingroup$ The question says use the Todd-Coxeter algorithm, which you have not attempted to do. I would do it over the subgroup $\langle x \rangle$. Expect 8 cosets! $\endgroup$
    – Derek Holt
    Commented Nov 11, 2013 at 22:01
  • $\begingroup$ @WillJagy I did the manipulations correctly now and found $x^2y^2x^2yxy=1$ $\endgroup$
    – amir
    Commented Nov 11, 2013 at 23:04

1 Answer 1


Note that I did what you were looking for in light of Derek's neat hint

enter image description here

You see that $[G:\langle y\rangle]=8$ so the possibilities for $|G|$ could be $24$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .