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.
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 :-) )