 Solution

A plane crystallographic group is a discrete subgroup of the Euclidean plane's isometry group with compact quotient; equivalently a compact set has translates covering the plane. We prove the crystallographic restriction theorem without assuming its conclusion. The orientation-preserving subgroup $G^{+}$ has index at most two and still has a cocompact group action. Its subgroup $T$ of translations is discrete as an additive subgroup of ${\mathbb{R}}^2$.

Suppose $T$ were trivial. Linear parts of orientation-preserving plane isometries commute, so every group commutator in $G^{+}$ is a translation. Thus $G^{+}$ would be an abelian group. Every member would then commute with the given nontrivial rotation $g$ and preserve its unique fixed point. A discrete subgroup of the compact rotation group about that point is finite. Then $G$ would be finite, contradicting cocompactness on the unbounded plane. Hence $T$ contains a nonzero vector $v$.

If $g$ is a half-turn its order is already two. Otherwise its linear part $R$ rotates $v$ to a linearly independent vector $Rv$, and conjugating translations shows that $Rv\in T$. Thus $T$ spans the plane and is a Euclidean lattice. A discrete additive subgroup meets every bounded closed set in finitely many points: an accumulating sequence would give nonzero differences tending to zero, contradicting isolation of the identity. To see the Euclidean lattice assertion directly, choose a shortest nonzero translation $v$. Translations on its line are integer multiples of $v$. Reducing all parallel components modulo $v$ shows that nonzero perpendicular components are bounded away from zero, since otherwise infinitely many translation vectors would accumulate in a compact rectangle. Choose a vector $w$ with the smallest positive perpendicular component; reducing by $w$ and then $v$ proves $T=\mathbb{Z}v+\mathbb{Z}w$.

Conjugation by $g$ preserves this Euclidean lattice. In its Euclidean lattice basis, $R$ is therefore an invertible integer matrix with determinant one. Its trace is an integer and is also $2\cos \theta$, where $\theta$ is the rotation angle. Since $g$ is nontrivial,These values give angles of order $2,3,4,6$ respectively. Hence