= 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 element>[identity], eight <rotations> by $\pm2\pi/3$ 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 $\boxed{H\cong A_4,\quad|H|=12}$ 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.
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