Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2015 / Paper 2 / 10

The table describes the symmetry group of a regular hexagon. Choose $R$ as a counterclockwise rotation through $\pi /3$ and $m_1$ as reflection in one symmetry axis. With $s=m_1$, the relations areThe rotation entries already are $I,R,\dots ,R^5$, and the table gives $R{m}_j=m_{j+1}$ with indices modulo six. ThereforeThe source writes $D_6$ with the subscript counting polygon vertices; this dihedral group has order 12.

All six reflections square to the identity, and the rotation $R^3$ is the half-turn. These exhaust the nonidentity involutions: $R^k$ has square one only when $2k\equiv 0(\mathrm{mod}6)$, giving $k=3$ apart from the identity. Thus the order-two subgroups areA reflection cannot have order three. The rotations of order three are $R^2$ and $R^4$, and they generate the same subgroup:All these subgroups are cyclic, because a group of prime order is cyclic; the displayed nonidentity elements supply their generators. This gives seven subgroups of order two and one of order three, consistently with involutions in an even dihedral group and the classification of subgroups of a dihedral group.

Act on vectors in the plane of the regular hexagon using its actual rotations and reflections. Choose the first reflection axis as the horizontal axis. A faithful representation is generated byThey are real orthogonal matrices obeying $D_2(R{)}^6=I$, $D_2(m_1{)}^2=I$ and $D_2(m_1)D_2(R)D_2(m_1)=D_2(R{)}^{-1}$. Hence they reproduce the group relations. Since $m_2=R{m}_1$,This is reflection in the line at angle $\pi /6$. More generally $D_2(m_j)=D_2(R{)}^{j-1}{D}_2(m_1)$. The six rotation matrices are distinct, as are the six reflection matrices; their determinant signs distinguish the two sets. Thus no two group elements have the same matrix, proving faithfulness rather than merely the representation property.

Use the left regular representation, defined on the ordered basis $(e_I,e_R,\dots ,e_{R^5},e_{m_1},\dots ,e_{m_6})$ by $D_{\mathrm{reg}}(g)e_h=e_{gh}$. Reading the $m_4$ row of the original PDF table gives the permutationThus an explicit $12\times 12$ permutation matrix, in six-by-six blocks, isEach column has its one in the row labelled by the left product; this specifies the convention completely. Since $P^2=I_6$, the full matrix squares to $I_{12}$ and is its own inverse. Therefore for every integer, including negative integers,This is also immediate from $m_4^2=I$ and the multiplication law for a representation.

The character is the trace. For the planar representation, the rotation trace and reflection trace giveIn the source’s order $(I,R,R^2,R^3,R^4,R^5,m_1,\dots ,m_6)$, these values are $(2,1,-1,-2,-1,1,0,0,0,0,0,0)$.

In the regular representation, the trace counts basis vectors fixed by left multiplication. The equation $gh=h$ forces $g=I$, so a nonidentity element fixes none, while the identity fixes all 12. ConsequentlyIn particular ${\chi }_{\mathrm{reg}}(m_4)=0$, consistently with the zero diagonal in part (d). The two characters have different identity values because the representation dimensions are two and twelve.