# 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?

• All you need to show is that $gh \ne hg$ for a single pair $g,h \in O(2).$ Try some simple ones! – Anthony Carapetis Mar 22 '18 at 11:24
• Didn't you just say it yourself. You can expand $O(2) = Rot(\phi)+Ref(\theta)$ which you can now apply with the knowledge of how they work. Or do you want to see a proof why rotation don't commute? – Michael Paris Mar 22 '18 at 11:30
• @MichaelParis I was wondering if there was a proof that followed simpily from the fact that $Q^TQ=QQ^T=I$ without any geometric considerations. – berrygreen Mar 22 '18 at 11:31
• In some sense there cannot be a purely algebraic proof that works over any field, because if $\mathbb{F}$ is the field with two elements then $O(2)$ is the cyclic group with two element. – Arnaud D. Mar 22 '18 at 11:32
• One contradiction will do, there are obviously instances where commutativity will hold ($\phi=\theta-\pi n$, $n\in \mathbb{Z}$) but if you show one contradiction, the group isn't Albelian. (You can just pick a $\phi$ & $\theta$ that don't satisfy the above.) – Nobody Mar 22 '18 at 11:34

$$\begin{pmatrix}1&0\\0&\!-1\end{pmatrix}\begin{pmatrix}0&1\\1&0\end{pmatrix}=\begin{pmatrix}0&1\\\!-1&0\end{pmatrix}\neq\begin{pmatrix}0&\!-1\\1&0\end{pmatrix}=\begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}1&0\\0&\!-1\end{pmatrix}$$
...over any field with characteristic$\,\neq2\;$