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, LR², ... , 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.