# Cubic Groups

The cubic groups are the groups of symmetries of the platonic solids, T, Td, O, Oh , I and Ih.

• Th - a subgroup of Oh of index two
• Oh - the octahedral group
• Ih - the icosahedral group

#### Cases treated elsewhere:

• T ≅ A4 (T is the group of rotations of the tetrahedron)
• Td ≅ O ≅ S4 (Td is the group of all symmetries of the tetrahedron)
• I ≅ A5 (I is the group of rotations of the icosahedron)

The final column marked "Mull" is the Mulliken symbol used in Physics and Chemistry.

### Th is the group of symmetries of a decorated cube'

The 'decorated cube' is as follows. Take a cube and draw a single new line across each face that divides the square face into two equal rectangles - and this should be done in such a way that each edge of the cube only meets one of these lines (so each edge meets exactly 3 of the rectangles).

Th has order 24 and is isomorphic to $$\mathbf{A}_4\times\mathbb{Z}_2^c$$, where $$\mathbb{Z}_2^c$$ is the centre of O(3), generated by inversion' $$i=-I:\mathbf{x}\mapsto-\mathbf{x}$$, and often denoted Ci. The representations are therefore obtained from those of A4 tensored with those of $$\mathbb{Z}_2$$.

 Th e C2C2 C3 C32 i iC2C2 iC3 iC32 notes # 1 3 4 4 1 3 4 4 |Th|=24 A0 1 1 1 1 1 1 1 1 trivial rep A1 1 1 1 1 -1 -1 -1 -1 alternating rep E1 2 2 -1 -1 2 2 -1 -1 E2 2 2 -1 -1 -2 -2 1 1 A1 ⊗ E1 T1 3 -1 0 0 3 -1 0 0 A1 ⊗ T2 T2 3 -1 0 0 -3 1 0 0 natural rep on cube
• Th is the symmetry group of a volleyball with its markings, and T2 is the representation in question.
• T1 is the representation where i acts trivially, so factors through the rotation group T.

### Oh is the group of all symmetries of the cube

Oh has order 48 and is isomorphic to $$\mathbf{S}_4\times\mathbb{Z}_2^c$$, where $$\mathbb{Z}_2^c$$ is the centre of O(3), generated by `inversion' $$i=-I:\mathbf{x}\mapsto-\mathbf{x}$$, and often denoted Ci

Notation for elements:

• Ck is a rotation of order k (C4 is a rotation by π/2; C2 is a rotation by π around a line through mid points of opposite edges)
• iCk is Ck composed with i.
 Oh e C4 C42 C3 C2 i iC4 iC42 iC3 iC2 notes Mull. # 1 6 3 8 6 1 6 3 8 6 |Oh| = 48 A0 1 1 1 1 1 1 1 1 1 1 trivial rep A1g A1 1 1 1 1 1 -1 -1 -1 -1 -1 alternating rep A1u A2 1 -1 1 1 -1 1 -1 1 1 -1 = A1 ⊗ A3 A2g A3 1 -1 1 1 -1 -1 1 -1 -1 1 = A1 ⊗ A2 A2u E1 2 0 2 -1 0 2 0 2 -1 0 Eg E2 2 0 2 -1 0 -2 0 -2 1 0 = E1 ⊗ A1 Eu T1 3 1 -1 0 -1 -3 -1 1 0 1 symmetry rep of cube T1u T2 3 1 -1 0 -1 3 1 -1 0 -1 = T1 ⊗ A1 T1g T3 3 -1 -1 0 1 -3 1 1 0 -1 = T1 ⊗ A2 T2u T4 3 -1 -1 0 1 3 -1 -1 0 1 = T1 ⊗ A3 T2g
• All reps are absolutely irreducible — even over Q
• The permutation representation on the set of 8 vertices of the cube is A0 + A3 + T1 + T4 = (A0 + A3)⊗(A0 + T1)
• The permutation representation on the set of 6 vertices of the octahedron is A0 + E1 + T1
• The permutation representation on the set of 12 edges of either is A0 + E2 + T1 + T3 + T4
• The permutation representation on the set of 3 diagonals of the octahedron is A0 + E1
• The permutation representation on the set of 4 diagonals of the cube is A0 + T4
• The "orientation permutation" representation on the set of 4 oriented diagonals of the cube is A3 + T1
• The "orientation permutation" representation on the set of 6 faces of the cube is A1 +??
• The "orientation permutation" representation on the set of 8 faces of the octahedron is T1 + ??
• The cube contains 2 inscribed tetrahedra; the permutation rep of this set is A0 + A3

### Ih is the group of all symmetries of the icosahedron

It is isomorphic to A5 x Ci.

 Ih e C5 C52 C3 C2 i iC5 iC52 iC3 iC2 notes Mull. # 1 12 12 20 15 1 12 12 20 15 |Ih|=120 - A0 1 1 1 1 1 1 1 1 1 1 trivial rep A1g A1 1 1 1 1 1 -1 -1 -1 -1 -1 alternating rep A1u T1 3 γ+ γ- 0 -1 -3 -γ+ -γ- 0 1 Symetry of icosahedron T1u T2 3 γ+ γ- 0 -1 3 γ+ γ- 0 -1 = T1 ⊗ A1 T1g T3 3 γ- γ+ 0 -1 -3 -γ- -γ+ 0 1 T2u T4 3 γ- γ+ 0 -1 3 γ- γ+ 0 -1 = T3 ⊗ A1 T2g G1 4 -1 -1 1 0 -4 1 1 -1 0 Gu G2 4 -1 -1 1 0 4 -1 -1 1 0 = G1 ⊗ A1 Gg H1 5 0 0 -1 1 -5 0 0 1 -1 Hu H2 5 0 0 -1 1 5 0 0 -1 1 = H1 ⊗ A1 Hg
• γ+ = 2cos(π/5) = ½(1+√5) (=golden ratio), and γ- = -2cos(2π/5) = ½(1-√5) ( = -(γ+)-1)
• All reps are absolutely irreducible.
• As a subgroup of O(3), Ih contains the central element i:x → -x (inversion in the origin).

So showing that Ih = I x Ci = A5 x Ci.