Past exam of the Probability, statistics and options II course of the University of Oxford / 2009 / B / B2

A left action is a map $(g,\omega )\mapsto g\omega$ satisfying $1\omega =\omega$ and $(gh)\omega =g(h\omega )$. The orbit of $\omega$ is $G\omega =\{g\omega :g\in G\}$. The action is transitive when there is exactly one orbit on the nonempty acted-on set, equivalently when any point can be moved to any other by a group element. An action is equivalently a homomorphism from $G$ to the permutation group of the set.

Act by left multiplication on the five cosets of $H$. The kernel is the core ${\bigcap }_{g\in G}gH{g}^{-1}$, a normal subgroup contained in $H$, so the hypothesis makes it trivial. Thus $G$ embeds in $S_5$, and its image has order $60$ and index two. An index-two subgroup is normal, so the quotient is of order two. Every three-cycle maps trivially to that quotient, since its order divides three. Three-cycles generate $A_5$, so the image contains $A_5$; their equal order gives $\boxed{G\cong A_5}$.

For completeness, prove simplicity rather than merely invoking it. The nonidentity even cycle types in $S_5$ are three-cycles, products of two disjoint transpositions and five-cycles. In $A_5$, a three-cycle has centralizer of order three, so its conjugacy class has size $20$. A double transposition has centralizer of order four in $A_5$, giving class size $15$. A five-cycle has centralizer its cyclic subgroup of order five, giving class size $12$; the $24$ such elements therefore split into two classes of size $12$. These account for all $60$ elements with the identity.

A normal subgroup is a union of whole conjugacy classes, containing the identity, and its order must divide $60$. No nonempty proper sum of the class sizes $15,20,12,12$, plus one, is a divisor of $60$ greater than one: the possible proper nontrivial sums are $13,16,21,25,28,33,36,40,45,48$. Only the full union has order $60$. Hence $\boxed{A_5\text{ is simple}}$, and so is $G$.

Choose one regular tetrahedron inscribed in the dodecahedron and let $H$ be its stabilizer among rotations of the dodecahedron. The twelve rotations of that tetrahedron preserve the dodecahedron; geometrically these are its identity, eight rotations by $\pm 2\pi /3$ about tetrahedral vertex axes and three half-turns about opposite tetrahedral edge axes. They act faithfully on its four vertices and give the twelve even permutations. Thus $\boxed{H\cong A_4,|H|=12}$ without using part (ii).

To check the core condition geometrically, the fifteen half-turns of the dodecahedron, about opposite-edge-midpoint axes, are conjugate under its rotations. A normal subgroup contained in $H$ cannot contain even one of these half-turns, because normality would then force all fifteen into a group of order twelve. On the other hand every nontrivial normal subgroup of $A_4$ contains a half-turn: if it contains a three-cycle, its conjugates and their inverses generate $A_4$; otherwise its nonidentity elements are among the three double transpositions, which are conjugate in $A_4$. Apply this observation to any normal subgroup of the full rotation group lying inside $H$. It must be trivial, proving exactly the required condition.