 Solution

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.