= Solution
Centre the cube at the origin with faces having normals $\pm e_1,\pm e_2,\pm e_3$. Every rotational symmetry fixes the centre and permutes these normals, so its <rotation matrix> is a <signed permutation matrix> of <determinant> one. Conversely any such <matrix> permutes the cube’s coordinates up to signs and preserves the cube. For each of $3!$ coordinate <permutations>, four of the eight choices of signs give <determinant> one. Consequently the <rotational symmetry group of a cube> has
$$
\boxed{|H|=3!\,2^2=24.}
$$
This counts the rotations directly, without an unverified stabilizer calculation. The TeX’s $2^k$ is damaged transcription: the PDF prints 24.
Let $X$ be the $3^6$ face colourings. A colouring fixed by a rotation must be constant on each <permutation cycle> of faces, giving $3^c$ fixed colourings if the rotation has $c$ face cycles. The rotation types are:
|| Rotation
|| Number
|| Face cycle type
|| Fixed colourings
| Identity
| 1
| $1^6$
| $3^6$
| Quarter-turn about opposite face centres
| 6
| $1^2 4$
| $3^3$
| Half-turn about opposite face centres
| 3
| $1^2 2^2$
| $3^4$
| Half-turn about opposite edge midpoints
| 6
| $2^3$
| $3^3$
| Third-turn about opposite vertices
| 8
| $3^2$
| $3^2$
There are three face axes, each with two quarter-turns and one half-turn; six edge axes, each with one half-turn; and four vertex axes, each with two nonidentity third-turns. These rotations are distinct and, with the identity, total 24, so the table is exhaustive. Face-axis rotations fix their two axial faces; the remaining four faces cycle or pair. Edge half-turns pair all faces, while vertex third-turns form two triples.
<Burnside lemma> now gives
$$
\boxed{|X/H|=\frac{3^6+6\cdot3^3+3\cdot3^4+6\cdot3^3+8\cdot3^2}{24}=57.}
$$
The three colours are labelled, <orthogonal reflections> are excluded, and no condition requires every colour to appear.
Back to article page
