Symmetry group of a cube

The full isometry group of a cube has order $48$. Its transitive actions on the $8$ vertices, $12$ edges, and $4$ main diagonals have stabilizers isomorphic respectively toThe action on main diagonals has kernel $\{I,-I\}$.

For vertices $(\pm 1,\pm 1,\pm 1)$, the symmetry group of a cube consists of signed permutation matrices. A matrix preserving the unoriented line through $(1,1,1)$ must have all three signs equal, so it has the form $\epsilon P$, where $P$ is a permutation matrix and $\epsilon \in \{1,-1\}$. Central inversion commutes with every $P$ and is not itself a permutation matrix. Consequently $(P,\epsilon )\mapsto \epsilon P$ identifies this twelve-element stabilizer subgroup with the direct product of groups $S_3\times C_2$. A body diagonal is different from a face-normal coordinate axis, whose line stabilizer has order sixteen.

The full symmetry group of a cube acts as all signed permutation matrices. The stabilizer subgroups of $(1,0,0)$, $(1,1,0)$ and $(1,1,1)$ have orders eight, four and six, respectively, and their fixed-point subspaces of a group action are the lines through those vectors. The three group orbits have sizes six, twelve and eight. They therefore give distinct axial types for the equivariant branching lemma. More generally the fixed-space dimension of a vector stabilizer subgroup equals the number of distinct nonzero absolute coordinate values, so these are precisely its one-dimensional types.

// Destination: dynamical-systems.bigb

An opposite-vertex line in the full octahedral symmetry group is an unoriented coordinate axis. Its stabilizer subgroup has the block form $\mathrm{diag}(A,\epsilon )$, where $A$ is a signed two-dimensional permutation matrix and $\epsilon =\pm 1$. The eight possibilities for $A$ form the dihedral group of a square, of order eight; the independent axial sign gives a cyclic group of order two. Thus the stabilizer is their direct product, of order sixteen. It is larger than the stabilizer of an individual vertex, because exchanging the two endpoints still fixes the line.

The full symmetry group of a cube acts transitively and faithfully on its twelve edges. For vertices $(\pm 1,\pm 1,\pm 1)$, its signed permutation matrices act transitively on the edges. The stabilizer subgroup of $\{(1,1,z):-1\le z\le 1\}$ consists of independently swapping $x,y$ and reversing $z$, and is a Klein four-group. The orbit-stabilizer theorem gives order $12\cdot 4=48$. Fixing every edge fixes every vertex, as each vertex is the unique intersection of its three incident edges, so the group action is faithful. The resulting subgroup of $S_{12}$ has index $12!/48$ and is not normal: central inversion has cycle type $2^6$, whose conjugacy class in $S_{12}$ has $12!/(2^{6}6!)=10395>48$ elements.

The symmetry group of a cube splits into its central inversion and its rotational symmetry group of a cube. Inversion has determinant $-1$, the rotation subgroup has determinant $1$, and together they generate the full group with trivial intersection. The rotation subgroup acts faithfully on the four body diagonals and is isomorphic to $S_4$. Consequently the full group is this direct product of groups.

The orientation-preserving symmetries of a cube are the $3\times 3$ signed permutation matrices with determinant one. There are $3!2^2=24$ of them. Their group action on the four body diagonals identifies this group with the symmetric group $S_4$. Indeed choose diagonal directions $(1,1,1),(1,-1,-1),(-1,1,-1),(-1,-1,1)$, whose sum is zero and whose only linear relation has all coefficients equal. A rotation fixing every diagonal sends each direction to itself or its negative; the relation forces all four signs equal. All negative would give $-I$, which has determinant $-1$, so the kernel of a group homomorphism is trivial. Both groups have order 24, giving the identification. Its face permutations have cycle types $1^6$, $1^{2}4$, $1^2{2}^2$, $2^3$ and $3^2$, occurring $1,6,3,6,8$ times respectively.

For cube vertices $(\pm 1,\pm 1,\pm 1)$, the four vertices of coordinate product $+1$ and the four of product $-1$ form the two inscribed regular tetrahedra. A cube rotation is a signed permutation matrix of determinant one and maps each product class to one class. A quarter-turn about a coordinate axis exchanges the classes. The orbit of either tetrahedron therefore has size two, while the cube rotation group has order 24. The orbit-stabilizer theorem gives its setwise stabilizer subgroup order 12; this is the tetrahedron's proper rotation group.

Permuting the three pairs of opposite faces gives a group homomorphism from the rotational symmetry group of a cube to $S_3$. Its kernel consists of the identity and the half-turns about the three coordinate axes: these are precisely the diagonal sign matrices with determinant one. Thus it is a normal subgroup of order four, a Klein four-group.

The rotational symmetry group of a cube has three group orbits on ordered pairs of faces: equal faces, opposite faces, and adjacent faces. Their sizes are respectively $6$, $6$ and $24$. Fixing one face leaves its cyclic group of four quarter-turns, which acts transitively on the four adjacent faces; this proves transitivity in the last orbit rather than merely counting its points.