I am trying to prove that given two isometry on the plane that consist of rotations of the plane around distinct points contain a translation in their group (please see Groups containing two rotations.)
I have showed that they need to be in the kernel and everything but cannot show that $fg$ is not equal to $gf$.
I know that $a$ and $p(a)$ have the same length as orthogonal operators preserve lengths. It makes sense that they don’t equal but how do we know for sure? So basically is: $a+b’=b+a’$ where $a$ and $a’$ have the same length same goes for $b$. How to show this is not possible for non trivial vectors.