Past exam of the options II course of the University of Oxford / 2009 / B / B2

A group action assigns to each $g\in G$ and $x\in \Omega$ a point $g\cdot x$ so that $1\cdot x=x$ and $(gh)\cdot x=g\cdot (h\cdot x)$. The orbit of $x$ is $Gx=\{g\cdot x:g\in G\}$. The action is transitive if there is exactly one orbit, equivalently any point can be carried to any other by some group element.

Let $G$ act by left multiplication on its five left cosets of $H$. This gives a group homomorphism $\phi :G\to S_5$. Its kernel is the core of a subgroup, ${\bigcap }_{g\in G}gH{g}^{-1}$: an element fixes every coset precisely when all its conjugates lie in $H$. It is a normal subgroup contained in $H$, so the hypothesis forces the kernel to be trivial. Thus $G$ embeds in $S_5$.

The image has order $60$ and index two, hence is the kernel of a surjection $S_5\to C_2$. All transpositions are conjugate and generate $S_5$, so under any such surjection they must all have the same nonidentity image. This determines the map uniquely as the sign homomorphism. Its kernel is $A_5$, proving $G\cong A_5$.

To prove simplicity directly, the nonidentity conjugacy classes in $A_5$ have sizes $15,20,12,12$. A double transposition has centralizer of order four in $A_5$; a three-cycle has centralizer of order three; a five-cycle has centralizer of order five, yielding the two classes of size twelve. The class sizes together with the identity account for all sixty elements. A normal subgroup must be a union of these classes containing the identity, and its order must divide sixty. None of the proper nontrivial sums of these sizes with one is a divisor of sixty. Therefore $A_5$, and hence $G$, is simple.

Choose an inscribed regular tetrahedron from one of the regular compounds of five tetrahedra in the dodecahedron, and let $H$ be its rotational stabilizer. Its twelve tetrahedral rotations preserve the dodecahedron. Concretely, with $\varphi =(1+\sqrt{5})/2$, use dodecahedron vertices $(\pm 1,\pm 1,\pm 1)$ together with cyclic coordinate placements of $(0,\pm {\varphi }^{-1},\pm \varphi )$. The tetrahedron has the four cube vertices whose coordinate product is $+1$. Cyclic coordinate permutations and even numbers of coordinate sign changes preserve both vertex sets, have determinant $+1$, and give twelve distinct rotations. No other rotation can preserve this tetrahedron, because a regular tetrahedron has only twelve orientation-preserving symmetries.

Their action on its four vertices is faithful and consists exactly of the even permutations: the eight rotations through $\pm 120^{\circ }$ induce three-cycles, the three half-turns induce double transpositions, and there is the identity. Thus $H\cong A_4$, independently of part (ii).

Finally a nonidentity element of $H$ has order two or three. In the dodecahedron's full rotation group, all fifteen half-turns about opposite-edge axes form one conjugacy class, and all twenty third-turns about opposite-vertex axes form one conjugacy class. These geometric conjugacies follow by carrying the corresponding oriented axes to one another by rotations. A normal subgroup containing such an element must contain its whole class, which already has more than twelve elements. Hence no nonidentity normal subgroup of $G$ is contained in $H$, verifying the required extra property.