A question about equivariance to 3D transformations using semi-direct and direct products.

You should read this small section: https://en.wikipedia.org/wiki/Direct_product_of_groups#Semidirect_products. It explains when a group decomposes as a direct product or as a semidirect product. It basically comes down to whether $hk=kh$ for all $h\in H$, $k\in K$ or not. If it helps you, I think of the $\rtimes$ as an inner (or internal) semidirect prouct, and of the $\times$ as an outer (or external) direct product in $\left(\Bbb R^3 \rtimes SO(3)\right) \times S_n$.

$\Bbb R^3 \rtimes SO(3) \cong E^+(3)$ is the group of rigid motions (i.e. orientation-preserving isometries) of $\Bbb R^3$. The semidirect product decomposition expresses that any an isometry can be expressed in a unique way as a rotation followed by a translation. The set of translations is a normal subgroup of $E^+(3)$, but $SO(3)$ isn't, which means we get a semidirect product but not a direct product. In particular, translations don't commute with rotations.

Then $\left(\Bbb R^3 \rtimes SO(3)\right) \times S_n = E^+(3)\times S_n$ means that we consider transformations given by permuting the points followed by moving the whole point cloud with a rigid motion. We use the direct product because:

• Our transformations are uniquely given by a point permutation and a rigid motion.
• Point permutations and rigid motions commute with each other.

Another way to think of the difference between direct and semidirect products might be to ask whether the elements in $H$ and $K$ "interact" with each other. Say we looked at $\Bbb R^3 \times SO(3)$ instead of $\Bbb R^3 \rtimes SO(3)$. This would be like having two separate Euclidean spaces and then translating one of them and rotating the other -> no interaction.