Course MT3818 Topics in Geometry

## I: Point group is cyclic

This part of the classification is handled by:

**Theorem**

*If the point group is C*_{n} for n = 1, 2, 3, 4, 6 *then we have one equivalence class for each n.*

**Proof**

Suppose we are given two plane symmetry groups *G* and *H* with the same cyclic point group generated by a rotation *R*.

We construct an isomorphism between their lattices with *f* ∘ *R* = *R* ∘ *F*.

If *n* = 1 or 2 then any isomorphism will do. If *n* = 3, 4 or 6 then choose *u* and *v* to be the shortest vectors in the lattices *L*_{G} and *L*_{H} and take *f*(*u*) = *v*. Since the other generator of the latice may be given by rotating the first, we get the required isomorphism.

The group *G* can be written as a union of cosets *L*, *LR*, *LR*^{2}, ... , *LR*^{n-1} and by mapping these to the corresponnding cosets of *H* we get the required isomorphism.

**Remark**

The equivalence classes of these symmetry groups are called **p1**, **p2**, **p3**, **p4**, **p6**.

One can get patterns with these as symmetry groups by putting an *n*-bladed symbol like at each point of the appropriate lattice.

JOC March 2003