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Place the cube's vertices at $(\pm 1,\pm 1,\pm 1)$. 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 $e=\{(1,1,z):-1\le z\le 1\}$. 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. HenceThere are twelve edges and four elements of the stabilizer subgroup. The orbit-stabiliser theorem therefore gives $|H|=48$, 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 withIt is 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 isIf $H$ 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.