The Group of Symmetries of the Cube (Rotations) Alan Culligan, Enda O’Connor and James Truman National University of Ireland Galway 1 Outline Orbit-Stabilizier Theorem Rotations Isomorphic to S4 The aspects of the cubes symmetries we will f = face in a cube • Identity = 1. investigate are: |G · f] = |G : StabG(f)| Permutations in S4: (1)(2)(3)(4),R360 through • Verifying the rotations of a cube form a group. (Note: There are 24 rotational symmetries in a any axis. cube i.e. |G| = 24). • Displaying how this group complies with the • Edge Midpoint Rotation = 6 orbit stabilizer theorem. Permutation in S4:(12)(13)(14)(23)(24)(34), Proof. (see figure 3). • Showing the symmetry group of a cube is Take any face on a cube. It is possible to • Diagonal Rotation = 8 isomorphic to S4. move from that face to any face in the cube Permutation in S : • Finding subgroups and proving for Lagrange’s 4 (see figure 1, right). So the orbit of any face is Figure 1:Face Midpoint Rotation. (123), (124), (134), (132), (142)(143), (234), theorem. {1, 2, 3, 4, 5, 6} (where each number is a face in (243), (see figure 2). the cube). The stabilizer of all faces in cube are • Face Midpoint Rotation (horizontal axis) = 6 {id, R ,R ,R } (see figure 1 and apply 90 180 270 Permutation in S4: Rotations of a cube form a appropriate axis). (1234), (1243), (1324), (1342), (1423), (1432), group. |G · f| = 6, |G : StabG(f)| = |24 : 4| = 6 (see figure 1). 6 = 6 • Face Midpoint Rotation (vertical axis) = 3 • Closure. e= edge in a cube Permutation in S4: (12)(34), (13)(24), (14)(23), Let, G be the group of symmetries in a cube and (see figure 1). |G · e] = |G : StabG(e)| let a, b ∈ G. Need to show a ∗ b ∈ G. We Figure 2:Diagonal Rotation. can prove this by showing x ∗ (a ∗ b) = (x ∗ a) ∗ b. Clearly any 3 rotations applied in the Proof. Subgroups same order, on two different occasions, give the same result. Therefore the rotations of a cube Take any edge on a cube and again it’s possible • Stabilizer of each face forms a subgroup of obeys associativity. If a ∈ G then applying a to move from that edge to any edge in the cube. order 4 and stabilizer of each edge forms a to an unmarked cube gives a seemingly identical So the orbit of any edge is {1, 2, 3, ..., 12} (each subgroup of order 2. looking cube. If b ∈ G and applied to this result, number is an edge in the cube. The stabilizer of • For axis shown in figure 1, all rotations we still appear to have an identical looking cube. any edge is {id, R180} (where the axis of rotation through the same axis form a subgroup of Thus a ∗ b ∈ G. goes from the centre of that edge to the centre order 4. of the edge on the face directly opposite, pass- Figure 3:Edge Midpoint Rotation. • Identity. • For axis shown in figure 2, all rotations ing through the centre of the cube. See figure 3, through the same axis form a subgroup of Does an id element exist within the rotations of right). order 3. cube such that: id ∗ x = x ∗ id, x = x, ∀x ∈ G |G · e| = 12, |G : Stab (e)| = |24 : 2| = 12 G • For axis shown in figure 3, all rotations The identity is any rotation that leaves the cube 12 = 12 Abelian through the same axis form a subgroup of completely unchanged. This is R through any 360 order 2. axis. Z(G) = {id} C (x) = {id, any rotation on the same axis as Order of the subgroups are 2,3 and 4 and are • Inverse. G x : ∀x ∈ G} factors of |G| = 24.Verifying Lagrange’s theorem. Each rotation is its own inverse. For example, Hence the group is not abelian. take a rotation x ∈ G. If you apply the rotation The subgroups mentioned on the right are abelian x and then apply the same rotation in the as rotations on the same axis commute with each References opposite direction, you end up where you other. originally started. Thus x−1 ∈ G. [1] Rotation Diagrams. http://garsia.math.yorku.ca/ zabrocki/math4160w03/cubesyms/..