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Every element in the symmetry group *G* of **R**^{n} is of the form *T*_{a} ∘ *R* with *R* ∈ *O*(*n*). If one composes two such elements, *T*_{a} ∘ *R*_{1} and *T*_{b} ∘ *R*_{2} then one gets an element whose linear part is the composite *R*_{1} ∘ *R*_{2} .

This is the motivation for the following:

**Definition**

The set of all linear parts of elements of a symmetry group *G* is called the **point group** *P* of *G*.

**Remarks**

- The above remark shows that this is indeed a group.
- It often seems that the point group associated with the symmetry group of a pattern is the group of symmetries of the
*motif*and that the point group is then a subgroup of the group of symmetries.

Alas, this is not always true.

- General lattice

For this pattern, the symmetry group consists only of translations. That is*G*=*L*and so the point group is {*I*}.

[This is ≅ the symmetry group of the motif ≅ the subgroup fixing a point of the lattice.]

- Rectangular lattice

For this pattern, the symmetry group consists only of translations and compositions*T*_{a}∘*V*with*V*a vertical reflection through a lattice point. Hence the point group is generated by*V*and is*D*_{1}

[Again this is ≅ the symmetry group of the motif ≅ the subgroup fixing a point of the lattice.]

- Rhomboid lattice

The symmetry group consists of translations and compositions*T*_{a}∘*H*with*H*a half turn. Hence the point group is generated by*H*and is*C*_{2}

[Again this is ≅ the symmetry group of the motif ≅ the subgroup fixing a point of the lattice.]

- Rectangular lattice

The symmetry group includes a glide reflection which is of the form*T*_{a}∘*R*with*R*a horizontal reflection. Hence the point group is generated by*R*and is*D*_{1}

In this case the symmetry group of the motif and the subgroup fixing a point of the lattice are both trivial and so are*not*the same as the point group. The point group is*not*a subgroup of the symmetry group.

In fact in general, the point group of

**Theorem**

*The point group of a symmetry group G is the factor group G/L where L is the normal subgroup of translations in G.*

**Proof**

Given a symmetry *f*, we may write it as *T*_{a}∘ *R* with *R* ∈ *O*(*n*) and we may define a homomorphism *θ* : *G* *O*(*n*) by *T*_{a}∘ *R* ↦ *R*

It is easy to verify that this is a group homomorphism with kernel the lattice *L* of translations and image the point group. The result then follows from the first isomorphism theorem.

Although the elements of the point group *P* are not elements of the symmetry group *G*, they *do* act on the lattice.

**Theorem**

*If A ∈ P ⊆ O*(2)* and a ∈ L then A*(

**Proof**

Since *A* is in the point group for some *f* ∈ *G* we have *f* = *T*_{v} ∘ *A*.

Then *A* ∘ *T*_{a}(** x**) =

Now calculate

This is

This means that (notwithstanding the fact that

Here are the possible choices for the groups *P* and the lattices on which they can act. We will see later how the final two columns can be filled in.

Point group | Lattice | #groups | Names of groups |

C_{1} | general | 1 | p1 |

C_{2} | general | 1 | p2 |

C_{3} | equilateral | 1 | p3 |

C_{4} | square | 1 | p4 |

C_{6} | equilateral | 1 | p6 |

D_{1} | rectangular rhomboid | 2 1 | pm pg cm |

D_{2} | rectangular rhomboid | 3 1 | pmm pgg pmgcmm |

D_{3} | equilateral | 2 | p3m1 p31m |

D_{4} | square | 2 | p4m p4g |

D_{6} | equilateral | 1 | p6m |

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