# Rotations around distinct points in the plane cannot be commutative

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.

It's important to also assume that $$f$$ and $$g$$ are nontrivial rotations, or else it is indeed true that $$fg = gf$$.
So, given that $$f$$ is a nontrivial rotation about the point $$p$$ and $$g$$ is a nontrivial rotation about the point $$q$$, and $$p \neq q$$, why don't we check what $$fg$$ and $$gf$$ do to the points $$p$$ and $$q$$ themselves?
Since $$g(q) = q$$, we have $$fg(q) = f(g(q)) = f(q)$$
But since $$f$$ is not centered at $$q$$, we have that $$f(q) \neq q$$. Therefore $$gf(q) = g(f(q)) \neq f(q)$$ using the fact that $$g$$ is a nontrivial rotation, and hence $$g(x) \neq x$$ for all $$x \neq q$$. Thus, we conclude that $$fg(q) = f(q) \neq gf(q)$$ $$\implies fg \neq gf$$