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.

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How are you representing your rotations? Matrices? (a bit easy) Complex numbers with unit magnitude? (even easier) – J. M. Nov 28 '11 at 18:58
In $\mathbb{R}^2$ or some plane isomorphic to $\mathbb{R}^2$ yes, matrices. But in this case (notice that both parallel- and continuity-axiom are missing), we have only defined a rotation as a movement with at least one fixed point which is not a reflection. If relevant, we defined a reflection at a line $g$ to be a movement which is the identity for all points on the line and maps all other points of the plane to a point on the other side of the line. A movement is an automorphism which conserves distances. (sorry if some terms are non-standard, I don't know all correct terms in english) – Huy Nov 28 '11 at 19:04
I presume the book you took this from would have a hint somewhere. What book is this? – J. M. Nov 28 '11 at 19:06
There is no book, it is an exercise from the current geometry lecture at my university. A friend of mine attending that lecture told me the only hint his assistant professor gave him was to write $\varphi = \tau_1 \circ \sigma$ and $\psi = \tau_2 \circ \sigma$. – Huy Nov 28 '11 at 19:08
Theorem 2.33 on page 70 of J.M. Aarts Plane and Solid Geometry gives a somewhat informal proof that the rotations about a fixed point are a subgroup of all transformations, which is isomorphic to the abelian group of real numbers in $(-\pi,\pi]$ with the group operation being addition modulo $2\pi$. However, the development of that geometry might be quite different from the development in your class, so I can't say if it's that helpful. – yunone Nov 28 '11 at 19:36

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.