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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