I will demonstrate why the group of rotational symmetries of a Dodecahedron is $A_5$. For that, we have to find five nice objects, on which the group of symmetries acts.
One object is "Cubes" inscribed in Dodecahedron. But since my demonstration will be on "Black-Board" rather than "Slide-Show". Can anybody suggest me a simple way to describe five cubes in Dodecahedron, which are the principal objects of permutations in the group of rotations symmetries of Dodecahedron?
A cube has twelve edges, and the cubes you're talking about have one edge along each of the twelve faces of the dodecahedron, on which the edge is a diagonal.
The faces of the dodecahedron are pentagons, with 5 diagonals.
If you start with a given diagonal of a face of the dodecahedron, it uniquely determines one of the five inscribed cubes. You can show what it must look like near a vertex. (See the figure below.)
Once you have this, make a sketch of the dodecahedron near a face of the cube. The dodecahedron over each cube face has a "tent" (double-y, bent-H, what have you) shape:
You should be able to make a convincing argument, based on symmetry, that what you get is in fact a cube, and it does "close up" into a coherent shape when you follow the rules for drawing diagonals.
Practice making sketches with four faces of the dodecahedron showing, and use a different color for the diagonals which form the edges of the cube.
Also, if you look at a main diagonal of the inscribed cube (there are 4 such diagonals), the ends of the diagonal are opposing vertices on the dodecahedron (see above graphic).
For each set of opposing vertices on the dodecahedron there are two inscribed cubes with a main diagonal along the same axis. Rotation along this axis by $2\pi/3$ fixes these two cubes and rotates the other three cyclically among themselves. In other words, this rotation of the dodecahedron is a three cycle among permutations the 5 cubes. All 20 three cycles in $A_5$ are obtained this way, and they generate $A_5$.
