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The main result of this section is to classify all such groups.

A finite symmetry group cannot contain translations, glides or screws and so arguing as before, such a group must fix some point which we may as well take as the origin. We are then reduced to considering the linear case.

We start groups of rotations and with the following result about linear groups.

**Theorem**

*Any finite subgroup of SO*(3)* is either a cycle group C _{n} a dihedral group D_{n} or one of the groups of a Platonic solid.*

**Proof**

Let *G* be a finite subgroup of *SO*(3) with order *n*.

Then *G* acts on the unit sphere *S* and each rotation has two fixed points: its **poles**. We will look at the set of all poles on *S* (a finite set).

**Lemma 1**

*Every element of G maps a pole to a pole.*

**Proof of lemma**

If *p* is a fixed point of an element *g* *G* and *h* is any element of *G*, then *hp* is a fixed point of *hgh* ^{-1} *G* and hence is a pole.

We divide the set of poles into **orbits**: subsets of the form *Gp* for some pole *p*.

**Lemma 2**

*If p is a pole of an element of G of maximal order m then the orbit of p contains n/m elements.*

**Proof of lemma**

The subgroup *H* of *G* which leaves *p* fixed has *m* elements. So *G* is a union of cosets *g*_{1}*H*, *g*_{2}*H*, ... , *g*_{r}*H* with *r* = *n*/*m*.

All the elements in a coset *g*_{i}*H* will move *p* to the same point *g*_{i}*p* for *i* = 1, 2, ... , *r*.

Note that if *i* *j* then *g*_{i}*p* *g*_{j}*p* since if *g*_{i}*p* = *g*_{j}*p* then *g*_{i}^{-1}*g*_{j}*p* = *p* and so *g*_{i}^{-1}*g*_{j} *H* and so *g*_{i} and *g*_{j} would represent the same coset.

Hence the orbit of *p* is the set {*g*_{1}*p*, *g*_{2}*p*, ... , *g*_{r}*p*} as required.

The *n* - 1 non-zero rotations in *G* consist of *m* - 1 rotations for each *pair* of poles.

That is ^{1}/_{2} (*m* - 1) *n*/*m* for each orbit.

Hence *n* - 1 = ^{1} /_{2} *n*( (*m*-1)/*m*) where the summation is over the orbits 2 - 2 /*n* = (1 - 1 /*m* ).

Since *m* 2 we have 1 - 1/*m* > ^{1}/_{2} and so we can only have 2 of 3 orbits if *G* is non-trivial.

- The case of two orbits

Suppose these have*n*/*m*_{1}and*n*/*m*_{2}elements.

Then 2/*n*= 1/*m*_{1}+ 1/*m*_{2}2 =*n*/*m*_{1}+*n*/*m*_{2}(since*m*_{1}and*m*_{2}divide*n*)*n*/*m*_{1}=*n*/*m*_{2}= 1 and we have two orbits with one pole in each. This is the case when*G*is a cyclic group*C*_{n}generated by rotation by 2*p*/*n*. - The case of three orbits

1 + 2/*n*= 1/*m*_{1}+ 1/*m*_{2}+ 1/*m*_{3}.

We must have (say)*m*_{3}= 2 1/*m*_{1}+ 1/*m*_{2}= 1/2 + 2/*n*(*m*_{1}- 2)(*m*_{2}- 2) = 4(1 -*m*_{1}*m*_{2}/*n*) < 4

There are only a few possibilities:

*m*_{1}= 2,*m*_{2}=*m*,*n*= 2*m*(This is the dihedral case)

*m*_{1}= 3,*m*_{2}= 3,*n*= 12 (This is the tetrahedral case)

*m*_{1}= 3,*m*_{2}= 4,*n*= 24 (This is the cube case)

*m*_{1}= 3,*m*_{2}= 5,*n*= 60 (This is the dodecahedral case)

Look at the case of *S*_{d}(Cube) for which |*G*| = 24.

Poles are: | Rotation by: | |Orbit| | Geometric feature | |G| |

Red orbit | p/2, p | 6 | face centres | 9 |

Blue orbit | 2p/3 | 8 | corners | 8 |

Green orbit | p | 12 | edge centres | 6 |

**Remark**

Note that we are used to thinking of *D*_{n} as a group acting on **R**^{2} and containing some reflections. As a subgroup of *I*(**R**^{3}) the elements of order two are not reflections but rotations by p.

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