= 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 $T=\mathbb Zv+\mathbb Zw$.
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,
$$
2\cos\theta\in\{-2,-1,0,1\}.
$$
These values give angles of order $2,3,4,6$ respectively. Hence
$$
\boxed{\operatorname{ord}(g)\in\{2,3,4,6\}}.
$$
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