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We now consider some interesting subgroups of *I*(**R**^{2}).

Following the spirit of Klein's *Erlangen program* we look for properties preserved by these subgroups. In particular we look for subsets *X* of *S* = **R**^{2} which are mapped to themselves by all the elements of these subgroups.

The subgroup is then the **symmetry group of the subset** *X*.

**Examples**

**Subgroups generated by a rotation**Rotation by 2

*π*/*n*generates a subgroup isomorphic to*C*_{n}the cyclic group of order*n*.Note that

*C*_{n}is the symmetry group of many figures in**R**^{2}.has symmetry group

*C*_{2}, has symmetry group*C*_{3}, has symmetry group*C*_{4}, etc.**Dihedral groups**The

*Dihedral group**D*_{n}is the group of symmetries of a regular*n*-gon. It has order 2*n*.The group

*D*_{3}consists of rotations by 0, 2π/3 and 4π/3 and three reflections.The group

*D*_{4}consists of four rotations by multiples of π/2 and four reflections -- two through lines joining vertices and two through lines joining mid-points of sides.One may show that in terms of generators and relations,

*D*_{n}= <*a*,*b*|*a*^{n}=*b*^{2}= 1,*bab*=*a*^{-1}>.For

*n*≥ 3 it is a non-abelian group. The group*D*_{3}≅*S*_{3}(the symmetric group on three symbols).The group

*D*_{2}is the symmetry group of a "regular 2-gon" and consists of the identity, rotation by π and reflections in two perpendicular lines.

It is called the**Klein 4-group**. It is sometimes written*V*and is isomorphic to*C*_{2}×*C*_{2}.The group

*D*_{1}is the group consisting of the identity and reflection in a line:

Note that as a group this is a cycle group of order 2 but that it is not the same subgroup as any of the subgroups*C*_{2}of (1) above which are generated by rotations.

Note that for each point ** p** of

**Theorem** (Leonardo da Vinci about 1500)

*The above examples C _{n} and D_{n} for some positive integer n are the only finite subgroups of I*(

**Proof**

Let *G* be a finite subgroup of *I*(**R**^{2}). Then *G* cannot contain a translation or glide reflection since these elements have infinite orders.

It cannot contain rotations about two distinct points otherwise (See Exercises 3 Question 6) it would contain a translation.

If it contains a rotation about a point ** a**, it cannot contain a reflection in any line not through

It can only contain reflections in lines through a common point (Otherwise one could combine two of them to get a rotation about one of the meeting points and a reflection through a different point).

Hence there is a point of

Take this point as the origin and then *G* ⊆ *O*(2).

*H* = *G* ∩ *SO*(2) contains only rotations and hence contains a smallest (non-zero) rotation. All the others must be multiples of this one (otherwise one could combine them to get a rotation whose magnitude was their *hcf*) and so we have *H* ≅ *C*_{n} .

If *G* ≠ *H* then *H* has two cosets in *G* (since *SO*(2) has two cosets in *O*(2)) and the reflections form a coset *rH* and it is easy to verify that if *h* is a generator of *H* then *rhr* = *h* ^{-1} .

That is, *G* ≅ *D*_{n} .

**Remark**

In fact (See Exercises 4 Question 1) any two copies *G*_{1} and *G*_{2} of the groups *C*_{n} of rotations are *conjugate* in *I*(**R**^{2}). That is, for some element *g* ∈ *I*(**R**^{2}) we have *g* *G*_{1} *g*^{-1} = *G*_{2}. Similarly any two copies of the same dihedral group are also conjugate. Thus the classification of the finite subgroups of *I*(**R**^{2}) is *up to conjugacy* in *I*(**R**^{2}).

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