12:42 PM
If we have such group of matrices, there must be matrices $A$, $B$, $C$ corresponding to $(r,0)$, $(s,0)$ and $(id,1)$.
For $A$ we know that $A^4=I$.
This implies that $A$ has to be diagonalizable, and we have $PAP^{-1}=D$ for some regular matrix $P$.
Since $X\mapsto PXP^{-1}$ is an automorphism of $GL(2,\mathbb C)$, we can w.l.o.g. assume that $A$ is diagonal, i.e., $A=\operatorname{diag}(d_1,d_2)$.
We know that $d_i^4=1$, so $d_i\in\{\pm1,\pm i\}$.
We know that $A^2\ne I$, which eliminates some possibilities.
Also a cannot be a scalar multiple of $I$, since then it would commute with each matrix. (And don't want it to commute with $B$.)
So as possibilities for $A$ we get $\operatorname{diag}(1,-1)$, $\operatorname{diag}(1,i)$, $\operatorname{diag}(1,-i)$ and the matrices which are obtained from these three by multiplying by $-1$ or $\pm i$.
Still, I am not sure whether we can get from this that there are no matrices $B$, $C$ with the required properties.
I think we get from $AC=CA$ that $C$ has to be diagonal, too. And again, we have only a few possibilities for $C$, the diagonal elements have to be $\pm1$ (since $C^2=I$).
Also $A^2\ne I$ eliminates some possibilities.
Linear & Abstract algebra
Linear & Abstract algebra