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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.
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 { S_{1} , S_{2} , ... , S_{n} } of index 2.
Now we look at the opposite symmetries { R_{1} , R_{2} , ... , R_{n} } and consider the set { T_{1} , T_{2} , ... , 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} = J^{2} 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 = { S_{1} , S_{2} , ... , S_{n} , T_{1} , T_{2} , ... , 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.
The finite subgroups of I(R^{3}) or of O(3).
Rotation groups | Direct products | Mixed groups | ||||
Name | Group | Order | Group | Order | Group | Order |
Cyclic | C_{n} | n | C_{n} × < J > | 2n | C_{2n}C_{n} | 2n |
Dihedral | D_{n} | 2n | D_{n} × < J > | 4n | D_{n}C_{n} | 2n |
Tetrahedral | A_{4} | 12 | A_{4} × < J > | 24 | S_{4}A_{4} | 24 |
Cube/octahedral | S_{4} | 24 | S_{4} × < J > | 48 | D_{2n}D_{n} | 4n |
Dodeca/icosahedral | A_{5} | 60 | A_{5} × < J > | 120 |
Remarks
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