 Solution

Use the matrices of solution. Their projective classes satisfy $[-M]=[M]$, since $-I_2\in N$. The group homomorphism is therefore explicitlyThese four classes are distinct: the two diagonal representatives are not scalar multiples of one another, the two off-diagonal representatives are not scalar multiples of one another, and a diagonal representative cannot be a nonzero scalar multiple of an off-diagonal one.

Although $A$ and $B$ anticommute as matrices, their projective classes commute:Furthermore $[A{]}^2=[B{]}^2=[I_2]$ and $[C{]}^2=[-I_2]=[I_2]$. Thus every nonidentity element of the image has order two. An explicit group isomorphism to the Klein four-group isIndeed reordering products introduces only a sign, which disappears in the projective linear group, so this map turns multiplication into addition modulo two in each coordinate. Equivalently, $\ker \tilde{R}=G\cap N=\{\pm I_2\}$ and the first isomorphism theorem givesThis projective image of the square symmetry group illustrates how a noncommutative group can have an Abelian projective image: its scalar commutator becomes the identity class.