Prove $O(2)$ is non-abelian? $$O(2) = \{Q\in \mathbb{F}^{2\times 2} | Q^TQ= QQ^T=I\}$$
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.

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?

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$\,\neq2\;$
