Simplicity of rotational symmetry group of dodecahedron without proving it is $A_5$ I am trying to prove that the rotational symmetry group $G$ of a regular dodecahedron is $A_5$.
I am trying to do this in an unusual way: I want to show that $G$ is simple. Since I can show $G$ is order 60 and can prove that any simple group of order 60 is $A_5$  I just need to argue it is simple.
I am aware of the standard proof that $G$ is $A_5$ by embedding cubes/ tetrahedrons: I want to prove that $G$ is simple without first showing it is $A_5$
 A: Here is a combinatorial way, by simply counting elements and using the regularity of your dodecahedron. In the isometry group, you have


*

*the identity element

*$6\times 4=24$ elements of order $5$ (rotations with axe face-face),

*$15\times 1=15$ elements of order $2$ (rotations with axe edge-edge),

*$10\times 2=20$ elements of order $3$ (rotations with axe vertex-vertex). 
Remark furthermore that by regularity of the dodecahedron, you act transitively on the set of axes face-face, on the set of axes edge-edge and the set of axes vertex-vertex. Since the rotations associated to each axe are of prime order, if a normal subgroup contain one rotation of given order, it contains all rotations of this order.
As a result we only have the following possibility (identity + some conjugacy classes) for the order of the normal subgroup:
$$1\text{, }16\text{, }21\text{, }25\text{, }36\text{, }40\text{, }45\text{, }60\text{.}$$
and using Lagrange's theorem the only possibility are $1$ and $60$. 
