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 has index at most two and still has a cocompact group action. Its subgroup of translations is discrete as an additive subgroup of .
Suppose were trivial. Linear parts of orientation-preserving plane isometries commute, so every group commutator in is a translation. Thus would be an abelian group. Every member would then commute with the given nontrivial rotation and preserve its unique fixed point. A discrete subgroup of the compact rotation group about that point is finite. Then would be finite, contradicting cocompactness on the unbounded plane. Hence contains a nonzero vector .
If is a half-turn its order is already two. Otherwise its linear part rotates to a linearly independent vector , and conjugating translations shows that . Thus 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 . Translations on its line are integer multiples of . Reducing all parallel components modulo 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 with the smallest positive perpendicular component; reducing by and then proves .
Conjugation by preserves this Euclidean lattice. In its Euclidean lattice basis, is therefore an invertible integer matrix with determinant one. Its trace is an integer and is also , where is the rotation angle. Since is nontrivial,
These values give angles of order respectively. Hence

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