How to show that rotations around the same point are commutative? This seems to be a very easy exercise and yet I fail at the last part:
I want to show that the rotations around a point $O$ are an Abelian sub-group of the group of all movements in a plane which fulfils Hilbert's combination-, order- and congruence-axioms. I have already shown that for rotations $\varphi$ and $\psi$, $\varphi \circ \psi$ is a rotation and that the inverse of a rotation is a rotation as well. So what I have left to do is to show that $\varphi \circ \psi = \psi \circ \varphi$.
I tried writing $\varphi = \tau_1 \circ \sigma$ and $\psi = \tau_2 \circ \sigma$ for three reflections $\tau_1, \tau_2$ and $\sigma$ but this did not yet lead me to any success. What is the trick to show commutativity?
Thanks a lot for any answers.
 A: Let $\varphi$, $\psi$ be rotations around $O$. We have to show $\varphi \circ \psi = \psi \circ \varphi$.
Let $\sigma$ be a reflection at a line through $O$. Then, we know that  $\tau_1 :=\varphi \circ \sigma$ is also such a reflection and then $\varphi = \tau_1 \circ \sigma$. Similarly, we define $\tau_2 := \psi \circ \sigma$, another reflection at a line through $O$, and have $\psi = \tau_2 \circ \sigma$. Hence, we have to show
$$\tau_1 \circ \sigma \circ \tau_2 \circ \sigma = \tau_2 \circ \sigma \circ \tau_1 \circ \sigma \iff \tau_1 \circ \sigma \circ \tau_2 = \tau_2 \circ \sigma \circ \tau_1.$$ 
As $(\tau_1 \circ \sigma \circ \tau_2)^{-1} = \tau_2 \circ \sigma \circ \tau_1$, this is equivalent to $(\tau_1 \circ \sigma \circ \tau_2)^{-1} = \tau_1 \circ \sigma \circ \tau_2$. Furthermore, we know that $\sigma \circ \tau_2$ is a rotation around $O$, thus $\tau_1 \circ \sigma \circ \tau_2$ is a reflection at a line through $O$ and therefore $(\tau_1 \circ \sigma \circ \tau_2)^{-1} = \tau_1 \circ \sigma \circ \tau_2$ holds.
