Automorphism group of cube is primitive

Question

If S is the set of vertices of a cube, and G is the automorphism group of the cube, then is G primitive?

I'm most having trouble with figuring out how one even approaches this problem.

Answer 1

Geometric approach. Find a property that a collection of points can have that is preserved by all symmetries of the cube. One way to do this is to have the subset of points itself have a geometric interpretation. Two possibilities come to mind: partition the vertex set into antipodal pairs, or partition it into two inscribed tetrahedra. (The latter may not be obvious, I just happen to know that a cube has two inscribed tetrahedra.)

Algebraic approach. Let $G$ be the symmetry group of the cube. For simplicity, I'll assume this is the rotational symmetry group, so orientation-reversing maps are ignored. By writing down all of the types of rotations that are possible (they have axes through either vertices, edges or faces), and then writing down their cycle types as permutations of the four space diagonals, we can verify the standard fact that $G\cong S_4$. (If you wanted orientation-reversing maps too, they can be generated from $G$ and the inversion across the origin, given by the matrix $-I$, and we get $S_4\times\mathbb{Z}_2$.)

If $G$ acts on a set $X$, the orbit-stabilizer says the map $G/\mathrm{Stab}(x)\to\mathrm{Orb}(x)$ given by $g\mathrm{Stab}(x)\mapsto gx$ is well-defined and is an isomorphism of $G$-sets (a $G$-equivariant bijection). In particular, if $G\curvearrowright X$ transitively with a point-stabilizer $H$, then $X$ is isomorphic to the coset space $G/H$ as a $G$-set. Moreover, if $\mathcal{B}\subset\mathcal{P}(X)$ is a block system of $X$, then $G\curvearrowright\mathcal{B}$ transitively too, and if $K$ is a stabilizer of a "point" in $\mathcal{B}$ (containing the same point in $X$ that $H$ stabilizes) then $H\subseteq K$, and the block system $\mathcal{B}$ is isomorphic as a $G$-set to the partition of $G/H$ into fibers of the projection map $G/H\to G/K$.

Then the vertex set is isomorphic to $S_4/\mathbb{Z}_3$ as a $G$-set, where $\mathbb{Z}_3$ is generated by a $3$-cycle (either of the nontrivial rotations around a given axis through a chosen vertex). The $G$-stable partitions (nontrivial block systems) then correspond to intermediate subgroups properly between $S_4$ and $\mathbb{Z}_3$, which can be seen to exist:

• (a) The dihedral group $D_3\cong \mathbb{Z}_3\rtimes\mathbb{Z}_2$ of order $6$, where $\mathbb{Z}_2$ contains any of the three rotations that take the chosen vertex to its antipode (the three rotations are the same up to translates by elements of $\mathbb{Z}_3$). As $\mathbb{Z}_3$ is the point-stabilizer of the chosen vertex, this $D_3$ is the setwise stabilizer of the subset containing this point and its vertex. The fibers of the projection map $S_4/\mathbb{Z}_3\to S_4/D_4$ correspond to the block system of antipodal pairs of vertices.
• (b) The alternating group $A_4$ contains $\mathbb{Z}_3$, and stabilizes the tetrahedron the chosen vertex is on, and the fibers of the projection map $S_4/\mathbb{Z}_3\to S_4/A_4$ form the block system corresponding to subsets of vertices from the two inscribed tetrahedra.