Past exam of the probability, statistics and options I course of the University of Oxford / 2011 / B (Group Theory)

For a group action, the orbit and stabilizer areThe orbit-stabilizer theorem for a finite group states

For a regular tetrahedron, rotations fixing a vertex rotate about the line from that vertex through the opposite face centre. Their angles are $0,2\pi /3,4\pi /3$, so the stabilizer is a cyclic group of order three. Rotations act transitively on the four vertices, giving $|G_1|=4\cdot 3=12$ by the orbit-stabilizer theorem.

For a cube, the corresponding axis is the body diagonal through the fixed vertex and its opposite. The three incident edges can be cyclically permuted by the same three angles, so again the stabilizer is cyclic of order three. The orbit has eight vertices. Hence

Realize the cube's vertices as $(\pm 1,\pm 1,\pm 1)$ and take the four with coordinate product $+1$. They form a regular tetrahedron, and their negatives form the other one. Every proper rotation of the first tetrahedron also permutes the negatives, hence is a cube rotation. This embeds $G_1$ as a subgroup of $G_2$.

The embedded group preserves each of these two four-vertex sets and acts transitively on each. Thus the resulting action on all eight cube vertices has two orbits, each of size four.