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Hugh Thomas
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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.)

Hugh Thomas
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