What is the map from nodes of the E8 diagram to conjugacy classes in the binary icosahedral group? Let $G \subset \mathrm{SL}_2(\mathbf{C}^2)$ be a finite subgroup isomorphic to the binary icosahedral group.  Let $Y$ be the minimal resolution of $\mathbf{C}^2/G$.  The irreducible components of the exceptional fiber of $Y$ are naturally in correspondence with nodes of the Dynkin diagram of $\mathrm{E}_8$.
Each of these components has a linking circle in $(\mathbf{C}^2 - \{0\})/G$, whose fundamental group is $G$.  Thus, each component determines a nontrivial conjugacy class in $G$.
There are 8 nontrivial conjugacy classes in $G$, of orders 2,3,4,5,5,6,10,10.  For each node of $\mathrm{E}_8$, which conjugacy class is it labeled by?  In particular, what is the order of this class?
Later: It sounds like Hugh is proposing
$$
\begin{array}{rrrrrrr}
& &     4\\
6 & 3 & 2 & 5 & 10 & 5 & 10
\end{array}$$
but the locations of the (6,3), and of (5,10,5,10) are just guesses.
 A: According to arXiv:math/0503542 by Suter (Fact 5.1) the sizes of the conjugacy classes of $G$ other than the trivial one are 1, 30, 20, 20, 12, 12, 12, 12.  These numbers (plus 1 for the trivial) sum to 120, which is the size of $G$, providing a sanity check.    
In general (also from the same source), if the branch lengths of the finite Dynkin diagram are $p_1$, $p_2$, $p_3$, there are $p_i-1$ conjugacy classes of size $|G|/2p_i$, generated by powers of some element $\gamma_i$ satisfying $\gamma_i^{p_i}$ all equalling the central non-identity element, plus a conjugacy class for it.  
One would therefore expect the $\gamma_i^j$ conjugacy classes to correspond to vertices along the branches, and the central element to correspond to the branch vertex.  Suter actually labels an $E_8$ diagram like this (with powers increasing towards the branch point, which makes sense since that way the branch point is $\gamma_i^{p_i}$ for all $i$) but I don't see any explanation in the paper of what that labelling means for him.  
And, of course, this doesn't begin to answer the question the OP asked...
Later: it seems like we would be most of the way there if one could show that the linking circle for a node, squared, should equal the product of the linking circles for the adjacent nodes.  (Say, assuming the degree of the node is at most two -- this formula doesn't seem to hold for the central node, where the order of the product also potentially enters.)
A: I found a paper of Kirby and Scharlemann which does the computations. 
In figure 2 of the paper, they give the surgery diagram for the Poincar\'e homology sphere:

One obtains a presentation for the fundamental group of the link complement by the usual Wirtinger method. Moreover, 2-handles are attached to each loop with framing 2 to give the 4-manifold, giving 8 more relations, appearing in the 3rd paragraph on p. 116. As indicated there, the group is generated by $a, g$ with presentation $\langle a, g | a^5 = g^3 =(ag)^2 \rangle$. We have $h=(ag)^{-1}$, and the central involution is given by $e= a^5 = g^3 = h^{-2}$ (notice $a,g,h$ are the generators of the outermost vertices of the Dynkin diagram, and $e$ represents the trivalent vertex). From their presentation, one computes $b=a^{-2}, c= a^3, d= a^{-4}, f=g^{-2}$. So indeed, the orders of these elements agree with the orders in your diagram, and essentially agrees with Hugh Thomas' answer. 
