= Solution
Place the cube's vertices at $(\pm1,\pm1,\pm1)$. Its full <symmetry group of a cube> consists of <signed permutation matrices>. To see completeness, a cube symmetry fixes its <center of a group> and permutes the six face normals, so it sends coordinate axes to signed coordinate axes; conversely every such matrix preserves the cube. Coordinate <permutations> and sign changes take any edge to any other edge, so this is a <transitive group action> on edges.
Choose the edge $e=\{(1,1,z):-1\le z\le1\}$. A symmetry stabilizing it must preserve its axis direction and its midpoint $(1,1,0)$. It can swap the first two coordinates and independently reverse the third coordinate, but cannot change the signs of the two fixed coordinates. Hence
$$
\boxed{\operatorname{Stab}_H(e)=\langle (x,y,z)\mapsto(y,x,z),\ (x,y,z)\mapsto(x,y,-z)\rangle
\cong C_2\times C_2.}
$$
There are twelve edges and four elements of the <stabilizer subgroup>. The <orbit-stabiliser theorem> therefore gives \b[$|H|=48$], including orientation-reversing symmetries, not just the twenty-four rotations.
The action defines a <group homomorphism> $H\to S_{12}$. If an element fixes every edge as a set, it fixes every vertex, since each vertex is the unique common point of its three incident edges. A cube isometry fixing all vertices is the identity. Thus this is a <faithful group action>, and the <cube symmetry action on edges> embeds $H$ as a <subgroup> with
$$
\boxed{[S_{12}:H]=\frac{12!}{48}=9\,979\,200.}
$$
It is \b[not a <normal subgroup>]. Central inversion $\mathbf x\mapsto-\mathbf x$ belongs to $H$ and swaps the twelve edges in six opposite pairs. In $S_{12}$, all <permutations> of cycle type $2^6$ are conjugate, and their number is
$$
\frac{12!}{2^6\,6!}=10\,395>48.
$$
If $H$ were a <normal subgroup>, it would contain this entire <conjugacy class>, impossible for a <group> of order forty-eight. This avoids relying on a classification of normal <subgroups> of the symmetric <group>.
Back to article page
