 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-2/8/d/solution

Map each symmetry to the permutation it induces on the three vertices. This is a group homomorphism, because composing geometric motions composes their vertex permutations. In the labels chosen in (b), the map is

g Φ (g) e e r (123) r^{2} (132) s (23) rs (12) r^{2} s (13)

All six images are distinct and exhaust the symmetric group

S_{3}

. Equivalently, the kernel is trivial because a triangle symmetry fixing all three vertices is the identity. Thus the map is bijective and

D_{3} ≅ S_{3} .

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