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There are nine equivalence classes corresponding to the cases where the point group is *D*_{2} (4 of them), *D*_{3} (2 of them), *D*_{4} (2 of them) and *D*_{6} (one only).

**Proof**

In all these cases the point group is generated two reflections *R*_{1} and *R*_{2} in lines *l* and *m* whose product *R*_{1}∘ *R*_{2} is a rotation generating the cyclic subgroup *C*_{n} of *D*_{n}. So the lines *l* and *m* are at an angle of 2*π*/*n* and the reflections *R*_{1} and *R*_{2} have shift vectors ** a** =

We will find that there are 9 possible combinations of point groups and shift vectors.

Let **R**, ** s** be the shortest vectors in

- If
*n*= 2 (so that*l*and*m*are perpendicular) then we can argue as in Stage 2 above:If

^{1}/_{2}**R**+^{1}/_{2}∈*s**L*then both*R*_{1}and*R*_{2}give the**cm**situation (i) above and we get the group**cmm**.If

**R**andform a basis for the lattice*s**L*then we get three possibilities.

- The shift vectors of
*R*_{1}and*R*_{2}are both**0**and we get the group**pmm**

- The shift vectors of
*R*_{1}and*R*_{2}are both non-**0**and we get the group**pgg**

- The shift vectors of
*R*_{1}is 0 and that of*R*_{2}is non-**0**and we get the group**pmg**

- The shift vectors of
- If
*n*= 3 then the axes of reflection*l*and*m*are at 120° or 2π/3.

Choose**R**andas shown.*s*

In this case all the shift vectors will be**0**.

We get one of the following cases

- The point
^{1}/_{3}**R**+^{1}/_{3}∈*s**L*

In this case, the some of the centres of 3-fold rotation are*not*on reflective axes.

The group is called**p31m**

- The vectors
**R**andare a basis for the lattice.*s*

In this case, the some of the centres of 3-fold rotation are*all*on reflective axes.

The group is called**p3m1**

- The point
- If
*n*= 4 then**R**andchosen as above will always be a basis for*s**L*and we get two cases.

- Both shift vectors are
**0**⇒ group is**p4m**(it should really be called**p4mm**)

- One shift vector is
**0**and one is not ⇒ group is**p4g**(it should really be called**p4mg**)

- Both shift vectors are
- If
*n*= 4 then**R**andwill be a basis for*s**L*and all shift vectors are**0**giving one possibility: the group called**p6m**.

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