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We now look at which groups can be the symmetry groups of lattices.

Note that if *L* is the lattice of translations of a symmetry group *G* then *L* is a normal subgroup of *G* and the quotient group *G*/*L* acts on *L* by conjugation.

The main result is:

**The Crystallographic restriction**

Any rotation in the symmetry group of a lattice can only have order 2, 3, 4, or 6.

**Proof**

We will give the proof for **R**^{2}. The proof for **R**^{3} is similar. It is harder for higher dimensions!

Let *L* be the lattice and let *M* be the set of all centres of rotations in *S*_{d}(*L*). This will include *L* since rotation by π about any lattice point is in *S*_{d}, but will in fact be bigger. It will, however, still be discrete.

Now let *p* ∈ *M* be the centre of a rotation *R* by 2*π*/*n*.

Let *p*_{1} ∈ *M* be a closest point of *M* which is the centre of a rotation *R*_{1} by 2*π*/*n*.

Let *p*_{2} = *R*_{1}(*p*).

Now if *T* is *any* transformation mapping a point *x* to *T*(*x*) then conjugating a rotation about *x* by *T* gives a rotation (by the same angle) about *T*(*x*).

Thus conjugating *R* by *R*_{1} gives a rotation *R*_{2} by 2*π*/*n* about the point *p*_{2} and the diagram shows that if *n* > 6 the point *p*_{2} would be closer to *p* than *p*_{1} contradicting the definition of *p*_{1} .

A similar proof using this diagram:

with *p*_{3} = *R*_{2}(*p*_{1}), rules out the possibility that *n* = 5.

**Remark**

Note that the cases *n* = 2 , 3 , 4 and 6 are all possible.

A *general lattice* can have half-turns as symmetries, a *square lattice* can be left fixed by rotations by 2π/4 while an *equilateral lattice* can have rotations by 2π/3 or 2π/6 as symmetries.

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