Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2022 / Paper 2 / 8

The group order of a finite group is its number of elements, $\boxed{|G|=\#G}$. A subgroup $H$ is normal when conjugation by any group element leaves it unchanged:Equivalently, its left and right cosets agree: $gH=Hg$ for every $g$.

Label the triangle's vertices counterclockwise as $1,2,3$. Let $r$ rotate it by $2\pi /3$ and let $s$ reflect it in the axis through vertex 1 and the midpoint of the opposite side. ThenThe six dihedral group elements areThe first three are the identity and the rotations by $120^{\circ },240^{\circ }$; the last three are reflections fixing vertices $1,3,2$, respectively. These are all possible symmetries: the image of one vertex has three choices, and the order of the other two has two choices.

Use rightmost-first composition. From the generator relation,with exponents reduced modulo three and two. The row-times-column multiplication table isThus every symmetry is expressed in the two group generators $r,s$.

Each reflection has order two, giving exactly the three order-two subgroupsConjugation by $r$ sends $r^{j}s$ to $r^{j+2}s$, cycling the three reflections. None of these subgroups is therefore normal.

The two nonidentity rotations generate the same order-three subgroup:It is invariant under conjugation by $r$, and $srs=r^{-1}$ shows invariance under $s$ as well; since $r,s$ generate the group, this proves normality. The complete partition into conjugacy classes isThe rotations are mutually conjugate by $s$, and the reflections by $r$; no rotation is conjugate to a reflection because conjugation preserves the order of a group element.

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 isAll 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