Let be $O_{2}$ the orthogonal group, that is, the group of reflections and rotations of $\mathbb{R}^{2}$. His center is $\{ \pm I\} \simeq \mathbb{Z}_{2}$. I'm having problems to study the center of the quotient $\frac{O_{2}}{\{ \pm I\}}$. Someone could clarify?
Thanks in advance.
On
I was ignoring the part "anticommutes with every orthogonal matrix". Indeed, we can look to commutators:
\begin{align} [Rot(\phi), Ref(\psi)] &= Rot(-2\phi)\\ [Rot(\phi), Rot(\psi)] &= I_{2}\\ [Ref(\phi), Ref(\psi)] &= Rot(4(\phi - \psi)) \end{align}
If $AZ(O_{2})$ $\in$ $Z\Big(\frac{O_{2}}{Z(O_{2})}\Big)$:
$$ (AB)Z(O_{2}) = (BA)Z(O_{2}) \Rightarrow [A,B] = \pm I_{2}$$
Looking to the commutators above, because $B$ is arbitrary, $A$ must be a rotation, say $Rot(\phi)$, and $Rot(-2\phi) = \pm I$, which implies $\phi = 0, \pi$ or $\pm \frac{\pi}{2}$. Hence: $$Z\Big(\frac{O_{2}}{Z(O_{2})}\Big) \simeq \mathbb{Z}_{2}$$
So, the center of the quotient isn't trivial.
$O_2$ is generated by rotations and symmetries, which means that all the elements of $O_2$ can be written as $R_\theta S^\epsilon$, where $R_\theta $ is a rotation of an angle $\theta$, $S= \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$ and $\epsilon \in \{0,1\}$. Though it suffices to look for matrices in that form such that $S R_\theta S^\epsilon = \pm R_\theta S^{\epsilon+1}$ and $R_\alpha R_\theta S^\epsilon= \pm R_\theta S^\epsilon R_\alpha$ $\forall \alpha$. This lies to all those matrices such that $S R_\theta=\pm R_\theta S$ and $R_\alpha S^\epsilon= \pm S^\epsilon R_\alpha$ $\forall \alpha$, which are separate conditions for $\theta$ and $\epsilon$. Can you now find out who's the center?