Prove $O(2)$ is non-abelian?

Question

$$O(2) = \{Q\in \mathbb{F}^{2\times 2} | Q^TQ= QQ^T=I\}$$ What is the most elegant way to prove that $O(2)$ is non-Abelian?

Here is my thinking: I know that $O(2)$ can be generated by reflections and moreover two reflections result in a rotation. Rotations commute with each other, but reflections do not $$ ROT(\theta/2)= \begin{bmatrix} \cos\theta & \sin\theta \\ \sin\theta & -\cos\theta \end{bmatrix} $$ It suffices to show that $$ROT(\phi/2)ROT(\theta/2)\neq ROT(\theta/2)ROT(\phi/2)$$ for some $\phi,\theta\in (0,2\pi)$. $$ \begin{bmatrix} \cos\phi & \sin\phi \\ \sin\phi & -\cos\phi \end{bmatrix} \begin{bmatrix} \cos\theta & \sin\theta \\ \sin\theta & -\cos\theta \end{bmatrix} =\begin{bmatrix} \cos\phi\cos\theta + \sin\phi\sin\theta & \cos\phi\sin\theta-\sin\phi\cos\theta \\ \sin\phi\cos\theta-\cos\phi\sin\theta & \sin\phi\sin\theta+\cos\phi\cos\theta \end{bmatrix} $$ $$ \begin{bmatrix} \cos\theta & \sin\theta \\ \sin\theta & -\cos\theta \end{bmatrix} \begin{bmatrix} \cos\phi & \sin\phi \\ \sin\phi & -\cos\phi \end{bmatrix} =\begin{bmatrix} \cos\phi\cos\theta + \sin\phi\sin\theta & \sin\phi\cos\theta-\cos\phi\sin\theta \\ \cos\phi\sin\theta-\sin\phi\cos\theta & \sin\phi\sin\theta+\cos\phi\cos\theta \end{bmatrix} $$

We see that indeed the two matrices do not commute since their top-right and bottom-left elements switch signs.

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My question: Is there a purely algebraic proof that does not need geometric consideration? If not, what is the most common proof of this result?

Answer 1

Take for example

$$\begin{pmatrix}1&0\\0&\!-1\end{pmatrix}\begin{pmatrix}0&1\\1&0\end{pmatrix}=\begin{pmatrix}0&1\\\!-1&0\end{pmatrix}\neq\begin{pmatrix}0&\!-1\\1&0\end{pmatrix}=\begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}1&0\\0&\!-1\end{pmatrix}$$

...over any field with characteristic$\,\neq2\;$