Full finite symmetry groups in 3 dimensions

From the result of the last section (and the fact used there that any finite symmetry group fixes a point) we can now calculate all the full symmetry groups. These are the finite subgroups of O(3).

There are three possible cases.

1. If the group contains no opposite symmetries then it is one of the groups of rotations already calculated.

2. If the subgroup contains the central inversion map J given by J(x) = -x then as in Exercises 4 Question 5 it is the direct product G × < J > with G one of the above subgroups.

3. The subgroup G can be what is called a mixed group.

   In this case, G does not contain J. Assuming that there are some opposite symmetries in G, however, the rotations H still form a subgroup { S₁ , S₂ , ... , S_n } of index 2.
   Now we look at the opposite symmetries { R₁ , R₂ , ... , R_n } and consider the set { T₁ , T₂ , ... , T_n } where each T_i = J ∘ R_i. These are a set of rotations since det(J) = det(R_i) = -1 and it is easy to verify that when we multiply two of them together we get one of the S_k's:
   T_i ∘ T_j = J ∘ R_i ∘ J ∘ R_j = J² ∘ R_i ∘ R_j = R_i ∘ R_j (since the map J commutes with every linear map) and this is one of the direct symmetries.
   Also T_i ∘ S_j = J ∘ R_i ∘ S_j = J ∘ R_k for some opposite symmetry R_k and this is T_k.
   It follows that the set K = { S₁ , S₂ , ... , S_n , T₁ , T₂ , ... , T_n } is a group of rotations and so is one of the groups we classified earlier.
   So to describe a mixed group we specify a pair K H of finite rotation groups (one containing the other as a subgroup of half its size). The symmetries are then either elements of the smaller group H (rotations) or opposite symmetries of the form J ∘ T with T ∈ K - H.
   

We do not have too much choice about picking such an H and K since we need to choose rotation groups one of which is a subgroup of index 2 in the other.

1. S₄ A₄
   This is the group of order 24 which is the full symmetry group of the tetrahedron.
   
2. D_2n D_n
   This is a group of order 4n. If n is odd it is the full symmetry group of an n-prism (the Cartesian product of a regular n-gon in the horizontal plane with an interval [-1, 1] in the vertical plane) or indeed of an n-dihedron. (See Exercises 5 Question 3).
   If n is even it is the symmetry group of something called an anti-prism. (See Exercises 5 Question 4).
   
3. D_n C_n
   This is a group of order 2n. This is the symmetry group of a regular n-gon in R² where the elements of order 2 are reflections rather than half-turns. You could get it as the group of a 3-dimensional figure by painting the top of an n-prism a different colour.
   
4. C_2n C_n
   This is a group of order 2n. This can be obtained as a subgroup of the mixed group D_2n D_n of Example b.
   To get a figure with this as symmetry group, start with a prism or antiprism with full symmetry group D_2n D_n and modify it by (say) etching suitable designs on the end faces.
   (See Exercises 6 Question 4)
   

The finite subgroups of I(R³) or of O(3).

Rotation groups			Direct products		Mixed groups
Name	Group	Order	Group	Order	Group	Order
Cyclic	Cn	n	Cn × < J >	2n	C2nCn	2n
Dihedral	Dn	2n	Dn × < J >	4n	DnCn	2n
Tetrahedral	A4	12	A4 × < J >	24	S4A4	24
Cube/octahedral	S4	24	S4 × < J >	48	D2nDn	4n
Dodeca/icosahedral	A5	60	A5 × < J >	120

Remarks

1. The above classification is up to conjugacy in the group O(3) or in the group I(R³).
   In general there will be many subgroups isomorphic to the above groups for every point of R³.

2. Note that in contrast to the situation in O(2) the rotation groups C₂ and D₁ are the same subgroup (up to conjugacy) in I(R³) or in O(3) since both are generated by a half turn.
   It follows that in the above list, C₂ × < J > and D₁ × < J > are also the same. So too are the pairs D₁C₁ , C₂C₁ and D₂C₂ , D₂C₁.