text: Q:
  
  What is wrong in my answer relating Todd-Coxeter Algorithm
  
  The question is:
  A triangle group $G^{pqr}$ is a group $<x,y,z|x^p,y^q,z^r,xyz>$, where $p \leq q \leq r$ are positive integers. In each case, prove that the triangle group is isomorphic to the group listed.
  a) The dihedral group $D_n$, where p, q, r = 2, 2, n,
  b) The octahedral group, when p, q, r = 2, 3, 4,
  c) The icosahedral group, when p, q, r = 2, 3, 5.
  I am working with question b), that is $<x,y,z|x^2,y^3,z^4,xyz>$
  have $xyz = 1$ and $x^2 = 1$
  I get $yz = x^{-1} = x$ and $yzyz=1$
  Then $<y,z|y^3,z^4,yzyz>$ is homomorphic to $<x,y,z|x^2,y^3,z^4,xyz>$
  I chose $H=<y>$ and found the $y=(243)$ $z=(1234)$ is the solution of $<y,z|y^3,z^4,yzyz>$, the detailed process is in the followed image:
  
  I found when I set a = 3, I will get b = 2 and c = 4, also all tables are passed.
  Then the order is $|G|=|H|[G:H]=3\times4=12$
  But the order of octahedral group is 24, what in my answer is wrong?
  
  A:
  
  Here is a summary of coset enumeration of the cosets of $H$. We know that $1y=1$.
  Define $1z=2$, $2z=3$, $3z=4$, deduce $4z=1$.
  Deduce $2y=4$ from $1yzyz=1$.
  Define $4y=5$, deduce $5y=2$.
  Define $5z=6$, $6z=7$, $7y=8$, deduce $8z=5$.
  Deduce $6y=3$ from $4yzyz=4$.
  Deduce $3y=8$ from $5yzyz=5$, and hence $8y=6$.
  Finally, deduce $7y=7$ from $8yzyz=8$.
  
  