= Solution
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 $\phi=(1+\sqrt5)/2$, use <dodecahedron> vertices $(\pm1,\pm1,\pm1)$ together with cyclic coordinate placements of $(0,\pm\phi^{-1},\pm\phi)$. 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 $\pm120^\circ$ induce three-cycles, the three half-turns induce double transpositions, and there is the identity. Thus \b[$H\cong A_4$], 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.
Back to article page
