Course MT4521 Geometry and topology

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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 , ... , Sn } of index 2.
    Now we look at the opposite symmetries { R1 , R2 , ... , Rn } and consider the set { T1 , T2 , ... , Tn } where each Ti = J comp Ri. These are a set of rotations since det(J) = det(Ri) = -1 and it is easy to verify that when we multiply two of them together we get one of the Sk's:
    Ti comp Tj = J comp Ri comp J comp Rj = J2 comp Ri comp Rj = Ri comp Rj (since the map J commutes with every linear map) and this is one of the direct symmetries.
    Also Ti comp Sj = J comp Ri comp Sj = J comp Rk for some opposite symmetry Rk and this is Tk.
    It follows that the set K = { S1 , S2 , ... , Sn , T1 , T2 , ... , Tn } 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 comp T with TK - 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. D2n Dn
    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 6 Question 4).
    If n is even it is the symmetry group of something called an anti-prism. (See Exercises 6 Question 4).
  3. Dn Cn
    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. C2n Cn
    This is a group of order 2n. This can be obtained as a subgroup of the mixed group D2n Dn of Example b.
    To get a figure with this as symmetry group, start with a prism or antiprism with full symmetry group D2n Dn and modify it by (say) etching suitable designs on the end faces.
    (See Exercises 6 Question 6)


The finite subgroups of I(R3) or of O(3).
Rotation groupsDirect productsMixed groups
CyclicCnnCn × < J >2nC2nCn2n
DihedralDn2nDn × < J >4nDnCn2n
TetrahedralA412A4 × < J >24S4A424
Cube/octahedralS424S4 × < J >48D2nDn4n
Dodeca/icosahedralA560A5 × < J >120


  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.

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JOC February 2010