Does SO(3) preserve the cross product?
You may use the scalar triple product formula $r \cdot (p\times q)=\det(r,p,q)$ to prove that $$ gr \cdot (gp\times gq)=gr \cdot g(p\times q)\tag{1} $$ ($=\det(r,p,q)$) for any vector $r$. Since $g$ is invertible, if $(1)$ holds for every vector $r$, we must have $gp\times gq=g(p\times q)$.
Two ways of seeing it more directly:
- The cross product of two vectors can be expressed in terms of the norms of the vectors and the angle between them, and those properties are preserved by rotations. (Update: As mephistolotl mentioned in the comments, you also need the fact that rotations preserve the orientation. Otherwise, you'll get a vector of correct length, but different direction.)
- Numerically, you would need to show $\epsilon_{ijk} O_{jl} O_{km} =O_{ip}\epsilon_{plm}$, which follows from orthogonality and the fact that $\det O=1$.
Since both expressions $g(p\times q)$ and $g(p)\times g(q)$ are linear in each of the variables $p,q$, it suffices to check the equality for the nine cases $p,q=e_1,e_2,e_3$, i.e. when $p,q$ are basis vectors. This method is quite often employed when dealing with equality of multilinear functions.