Let $ D_6 = \{ e, r, r^2,f,rf,r^2f \} $ be the dihedral group of order $ 6 $ (Recall that we have the following relations: $ r^3=e, f^2=e $ and $ fr^i=f^{3-i}r $ for $ i=1,2 $).
Suppose that $ G=\{ u,v,w,x,y,z \} $ is a group with operation $ * $ where $ z $ has order $ 3 $, $ y * z=u $ and $ v * w =y $. Suppose that we have the isomorphism
$ \phi(f)=v, \phi(rf)=w $. Find $ \phi(r) $ and $ \phi(r^2) $.
My answer:
$ \phi(rf). \phi(f)=\phi(rf^2)=\phi(r). $
So, $ w * v= \phi(r) $. And, $ \phi(r^2) = \phi(r). \phi(r)=(w*v)*(w*v)=w*y*v $.
What am I doing wrong?
Note that $frf=r^2ff=r^2$ so $\phi(r^2)=v\ast w=y$. As $|z|=3$ this forces $\phi(r)=z$.