Abstract Algebra Theory and Applications - Computer Science ...

10.2 SYMMETRY 181or reflections of the form( cos θ − sin θT θ =sin θ cos θ).Notice that det(R θ ) = 1, det(T θ ) = −1, and Tθ2 = I. We can divide theproof up into two cases. In the first case, all of the elements in G havedeterminant one. In the second case, there exists at least one element in Gwith determinant −1.Case 1. The determinant of every element in G is one. In this case everyelement in G must be a rotation. Since G is finite, there is a smallest angle,say θ 0 , such that the corresponding element R θ0 is the smallest rotation inthe positive direction. We claim that R θ0 generates G. If not, then for somepositive integer n there is an angle θ 1 between nθ 0 and (n + 1)θ 0 . If so, then(n + 1)θ 0 − θ 1 corresponds to a rotation smaller than θ 0 , which contradictsthe minimality of θ 0 .Case 2. The group G contains a reflection T θ . The kernel of the homomorphismφ : G → {−1, 1} given by A ↦→ det(A) consists of elementswhose determinant is 1. Therefore, |G/ ker φ| = 2. We know that the kernelis cyclic by the first case and is a subgroup of G of, say, order n. Hence,|G| = 2n. The elements of G areThese elements satisfy the relationR θ , . . . , R n−1θ, T R θ , . . . , T R n−1θ.T R θ T = R −1θ .Consequently, G must be isomorphic to D n in this case.□Figure 10.5. A wallpaper pattern in R 2