Past exam of the mathematics course of the University of Cambridge / 2007 / ii / Paper 3 / 3G / Solution

For a plane crystallographic group, the translation subgroup consists of translations by vectors in a rank-two lattice $\Lambda$; the lattice is this additive group of translation vectors. The point group is the image under the linear-part homomorphism $G\to O(2)$, equivalently $G/\Lambda$. Conjugating a translation by an isometry with linear part $A$ gives translation by $Av$, so every point-group element preserves $\Lambda$.

Choose a lattice basis. A point-group rotation is represented in that basis by an integral matrix, because it maps lattice vectors to lattice vectors, and its inverse does likewise. The trace is therefore an integer. In an orthonormal basis the same rotation has trace $2\cos \theta$, which lies between minus two and two. Hence $2\cos \theta \in \{-2,-1,0,1,2\}$. The value two gives the identity, while the other values give angles $\pi$, $\pm 2\pi /3$, $\pm \pi /2$, or $\pm \pi /3$. Thus every nonidentity rotation has orderThis proves the crystallographic restriction theorem from the lattice action rather than assuming the classification of wallpaper groups.