Past exam of the mathematics course of the University of Cambridge / 2002 / ia / Paper 1 / 1B (Algebra and Geometry)

For a finite group $G$ acting on a set $X$, write $G_x=\{g\in G:gx=x\}$ for the stabilizer subgroup and $Gx=\{gx:g\in G\}$ for the orbit of $x$. The orbit-stabilizer theorem givesIndeed, $g{G}_x\mapsto gx$ is a bijection from the left cosets of $G_x$ onto $Gx$: two images agree precisely when $h^{-1}g\in G_x$. Each coset has $|G_x|$ elements, proving the formula.

Place the cube vertices at $(\pm 1,\pm 1,\pm 1)$. The four vertices whose coordinate product is $+1$ form $T$; those with product $-1$ form $T^{\prime }$. Within either set, two distinct vertices differ in exactly two coordinates, so every edge has length $2\sqrt{2}$ and each set is a regular tetrahedron.

A member of the cube rotation group is a signed permutation matrix of determinant one. There are $3!$ permutations of the coordinate axes and four choices of signs for each permutation that give determinant one, so $|G|=24$. Such a matrix maps a coordinate-product class to a coordinate-product class. Moreover, the rotation through a quarter-turn $(x,y,z)\mapsto (-y,x,z)$ interchanges the classes. Thus the orbit of $T$ consists of exactly $T,T^{\prime }$. By the orbit-stabilizer theorem, the tetrahedron stabilizer in the cube rotation group has order