Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2014 / Paper 2 / 9

For a subgroup $H$ of a finite group $G$, Lagrange's theorem statesHere $[G:H]$ is the number of distinct left cosets. Every coset has $|H|$ members, and the cosets partition $G$, proving the identity. In particular the order of a subgroup divides the order of the group. Divisibility is generally necessary, not sufficient, for a subgroup of a given order to exist.

The relation $Rm=m{R}^{-1}$ implies $m{R}^j=R^{-j}m$. Move each reflection generator to the right in any word, cancel pairs using $m^2=I$, and reduce exponents modulo $N$. Every element then has a normal form $R^j$ or $R^{j}m$, $0\le j<N$.

For the actual symmetry group of a regular polygon these are $N$ distinct rotations and $N$ distinct reflections. The determinant distinguishes rotations from reflections in the planar action. Hence the source's dihedral group $D_N$ has order $2N$, with its subscript counting polygon vertices. The calculations below use that convention.

For a regular pentagon, take $R$ as rotation through $2\pi /5$ and $m$ as reflection in the horizontal axis through vertex zero. The distinct elements areEvery nonidentity rotation has order five. The reflections are the five order-two elements, since $(R^{j}m{)}^2=R^j{R}^{-j}=I$.

The reflection $R^{j}m$ has axis at angle $j\pi /5$ to the horizontal, understood modulo $\pi$. If vertices have angles $2k\pi /5$, it sends vertex $k$ to vertex $j-k(\mathrm{mod}5)$. Each axis passes through one vertex and the midpoint of the opposite edge; that vertex stays fixed and the other four vertices exchange in two pairs.

Since $|D_5|=10$, Lagrange's theorem permits proper subgroup orders only $1,2,5$. The complete list isTo show completeness, an order-two subgroup must be generated by its unique nonidentity element, necessarily one of the five reflections. An order-five subgroup is cyclic, since its order is prime; its generator cannot be a reflection, and every nonidentity rotation generates the same rotation subgroup. There is only one order-one subgroup.

Thus there are seven proper subgroups when the trivial subgroup is included, or six nontrivial proper subgroups. These are the prime-sided case of the classification of subgroups of a dihedral group.

The conjugacy class of $g$ is $\{hg{h}^{-1}:h\in G\}$. Rotations commute with one another, while conjugation by a reflection changes $R^j$ to $R^{-j}$. Thus the rotation classes are $\{I\}$, $\{R,R^4\}$ and $\{R^2,R^3\}$.

For reflections,Because multiplication by two permutes the residues modulo five, these conjugates exhaust all five reflections. Consequently the complete class partition isTheir sizes $1+2+2+5=10$ account for every element, as in the conjugacy classes of a finite dihedral group.

A normal subgroup is a union of complete conjugacy classes. The trivial subgroup is normal, and the rotation subgroup is normal because conjugation preserves it; equivalently it has index two. None of the five order-two subgroups is normal, because it contains just one member of the five-element reflection class. Thus the proper normal subgroups are exactly