group of rigid motions dodecahedron We have the following theorem 
If |G| = 60 and G has more than one Sylow-5 subgroup, then G is simple.
Since order of the rigid motion of the dodecahedron group is 60, so all we have to do is to show that it has more than one sylow-5 subgroup, but I don't know how to do this as I don't know the elements of this group as I have troubles visualizing it.
 A: If $P$ is one of the pentagons, then the rotations which preserve $P$ (including the identity) are five, and they form a cyclic group of order $5$. Since a non-identity rotation can maps two faces to themselves (and the two faces are opposite on the solid), we see that $G$ has at least as many Sylow $5$-subgroups as there are pairs of opposite faces in the dodecahedron, that is, at least $6$. The usual numerology associated to the Sylow theorems then shows that these are precisely all the Sylow $5$-subgroups.
Interesting, this works for other primes. There are three rotations which fix any given vertex, and they form a cyclic group of order three. Two rotations fix opposite vertices, so this gives us 10 cyclic groups of order $3$, one for each pair of opposite vertices. Sylow numerology then shows these are all the Sylow $3$-subgroups.
The remaining prime is $2$. The Sylow $2$-subgroups have order $4$. To find one, consider two opposite edges in the dodehachedron. There are four rotations which preserve them collectively: you can swap the vertices in both at the same time, or swap the two edges, or do both things. This is a group of order $4$, a direct product of two cyclic groups of order $2$. You should count them correctly! (A previous version of this miscounted them :-) )
A: What is problem in visualization?
 
How many faces it has? Can you pair up them in some nice way? For each pair, consider rotations which take the face of that pair to itself. How many rotations will there be? What it gives?
