Why is $D_5$ a subgroup of the icosahedral group

Question

According to Wikipedia $D_5$ is a subgroup of the group of rotational symmetries of an icosahedron: http://en.wikipedia.org/wiki/Icosahedral_symmetry. I know this isn't very rigorous, but intuitively this isn't obvious since I couldn't think of any way to map a pentagon ABCDE on an icosahedron (or dodecahedron) to the same pentagon, except with the vertices labeled AEDCB... at least not with just rotations in the icosahedral group (order 60). What would be the rigorous and geometric proofs for these facts?

Answer 1

Put one vertex ($Z$) of the icosahedron on top, and one ($Z'$) on the bottom, touching the table.

The top vertex $Z$ is adjacent to five vertices $A,B,C,D,E$ that form a regular pentagon. $Z'$ is adjacent to the five remaining vertices $A', B', C', D', E'$, which also form a regular pentagon. For each $x$ in $\{A,B,C,D,E,Z\}$, the vertex $x$ is opposite to $x'$.

Let $r$ be a rotation of the icosahedron by $108^\circ$ around the axis $ZZ'$. This rotation permutes the vertices as follows:

$$\begin{pmatrix} Z&Z'&A&A'&B&B'&C&C'&D&D'&E&E' \\ Z&Z'&B&B'&C&C'&D&D'&E&E'&A&A' \end{pmatrix}.$$

Let $f$ be a rotation of the icosahedron that exchanges $Z$ and $Z'$. This turns the entire solid upside down, exchanging the positions of pentagons $ABCDE$ and $A'B'C'D'E'$. It permutes the vertices as follows:

$$\begin{pmatrix} Z&Z'&A&A'&B&B'&C&C'&D&D'&E&E' \\ Z'&Z&A'&A&B'&B&C'&C&D'&D&E'&E \end{pmatrix}.$$

By direct calculation, one can find that $$\begin{align} r^5 & = 1 \\ f^2 & = 1 \\ rf & = fr^{-1} \end{align} $$