0
$\begingroup$

$$O(2) = \{Q\in \mathbb{F}^{2\times 2} | Q^TQ= QQ^T=I\}$$ 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?

$\endgroup$
  • 2
    $\begingroup$ All you need to show is that $gh \ne hg$ for a single pair $g,h \in O(2).$ Try some simple ones! $\endgroup$ – Anthony Carapetis Mar 22 '18 at 11:24
  • $\begingroup$ 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? $\endgroup$ – Michael Paris Mar 22 '18 at 11:30
  • $\begingroup$ @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. $\endgroup$ – berrygreen Mar 22 '18 at 11:31
  • $\begingroup$ 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. $\endgroup$ – Arnaud D. Mar 22 '18 at 11:32
  • $\begingroup$ 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.) $\endgroup$ – Nobody Mar 22 '18 at 11:34
1
$\begingroup$

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$\,\neq2\;$

$\endgroup$
  • $\begingroup$ @ArnaudD. Thanks. Your comment came while I was adding the info. $\endgroup$ – DonAntonio Mar 22 '18 at 11:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.