Problem: We know the symmetry group of the cube defined by the points $(x,y,z) = (\pm 1,\pm 1,\pm 1)$ consists of 48 matrices, which are $ \begin{pmatrix} \pm 1 & 0 & 0 \\ 0 & \pm 1 & 0\\ 0 & 0 & \pm 1 \\ \end{pmatrix} $, $ \begin{pmatrix} \pm 1 & 0 & 0 \\ 0 & 0 & \pm 1\\ 0 & \pm 1 & 0 \\ \end{pmatrix} $, $ \begin{pmatrix} 0 & \pm 1 & 0 \\ 0 & 0 & \pm 1\\ \pm 1 & 0 & 0 \\ \end{pmatrix} $, $ \begin{pmatrix} 0 & 0 & \pm 1 \\ \pm 1 & 0 & 0\\ 0 & \pm 1 & 0 \\ \end{pmatrix} $, $ \begin{pmatrix} 0 & 0 & \pm 1 \\ 0 & \pm 1 & 0\\ \pm 1 & 0 & 0 \\ \end{pmatrix} $, $ \begin{pmatrix} 0 & \pm 1 & 0 \\ \pm 1 & 0 & 0\\ 0 & 0 & \pm 1 \\ \end{pmatrix} $.
Now define function $f$ mapping from this group to matrices where $f((a)_{ij}) = (a)_{ij}^2$, i.e. it squares every entry of the matrix. Prove that $f$ is a homomorphism, and find its kernel and image.
My attempt:
To show $f$ is a homomorphism I need to show that $f(AB) = f(A)f(B)$. It is impossible to verify this case by case for this symmetry group, so I need to find a simpler approach. I think the number of verifications can be reduced by considering the fact that product of two rotations is simply another rotation or reflection, and since repeated multiplication of, for example, the $R_{\pi/4}$ around the $z$-axis with itself generates the other 3 rotation matrices around the $z$-axis, I just need to verify $f(R_{\pi/4} R_{\pi/4}) = f(R_{\pi/4}) f(R_{\pi/4})$ and omit the rest of the 3 cases. So am I on the right track? Or am I missing a much simpler and cleaner way to verify $f$ being homomorphic?
In terms of the second part of the problem, my attempt is that the kernel is $ \begin{pmatrix} \pm 1 & 0 & 0 \\ 0 & \pm 1 & 0\\ 0 & 0 & \pm 1 \\ \end{pmatrix} $, and the image is just the six matrices described above without the $\pm$ signs. Am I correct on this part?
I really appreciate any help on this problem!
Ignoring the signs for a moment, all of those matrices are permutation matrices. So their action on basis vectors is just to swap them with some other basis vector and, including the signs again, perhaps also negate them. From this the first result is relatively intuitive. If you want to be more formal, you can express the statement in the following manner. Let $e_i$ be a basis vector, i.e. $e_1=(1,0,0)$ and so forth. Then for any matrix $M$, $Me_i = \pm e_{\pi(i)}$ for some permutation, $\pi$, on $\{1,2,3\}$ depending on $M$. It's relatively clear that $f(M)e_i = e_{\pi(i)}$ regardless of the sign of $Me_i$. From here proving $f(AB)=f(A)f(B)$ is relatively straightforward using basic facts about permutation matrices. Rotations and reflections are a geometrical perspective on permutations, but they add unnecessary complication in this case.
The kernel and image of $f$ is clear, and you are correct about them.