Does SO(3) preserve the cross product? Let $g\in SO(3)$ and $p,q\in\mathbb{R}^3$. I wondered whether it is true that $$g(p\times q)=gp\times gq$$
I am not sure how to prove this. I guess I will use at some point that the last row $g_3$ of $g$ can be obtained by $g_3=g_1\times g_2$.
But I assume there is an easier proof than writing everything out.

If one introduces the cross product in a proper abstract way, this property will be immediate from the definition. First consider any $n$-dimensional real vector space, then each basis has an associated volume form ($n$-linear alternating form), obtained by taking the determinant after expressing the $n$ argument vectors in coordinates with respect to the basis. If the vector space is equipped with an inner product so that it becomes an Euclidean vector space$~E$, then the volume forms of the orthonormal bases for the inner product take two opposite values; choosing one of these volume forms defines the structure of an oriented Euclidean vector space. The automorphism group $O(E)$ of $E$ has a subgroup $SO(E)$ of index$~2$ that fixes the volume forms, and is the automorphism group of the oriented Euclidean vector space.
Now fixing $n-1$ vectors $p_1,\ldots,p_{n-1}$ in an $n$-dimensional oriented Euclidean vector space, there is a linear form $f:v\mapsto\operatorname{Vol}(p_1,\ldots,p_{n-1},v)$, and therefore a unique vector $w$ such that $f(v)=w\cdot v$; by definition, this $w$ is the external product of $p_1,\ldots,p_{n-1}$. For $n=3$, this gives the cross product in an oriented $3$-dimensional Euclidean vector space. Since only the oriented Euclidean structure used in the definition, the external product is automatically compatible with the action of$~SO(E)$; since the (choice of) volume form was actually used in the definition, it is not compatible with the action of$~O(E)$.
