The class equation of the icosahedral group, I, is $60 = 1 + 20+12+12+15$. The conjugacy class of order 15 contains elements of order 2, and their centralizers are of order 4. The centralizer contains elements of order 2.

The question from Artin is, what elements are in the centralizer?

  • $\begingroup$ Remember that the center of a (finite, to be sure) direct product is the direct product of the centers. $\endgroup$ – DonAntonio Feb 11 at 20:45
  • $\begingroup$ Which group is this? Are you asking about the group $A_5 \times C_2$? $\endgroup$ – the_fox Feb 12 at 6:30
  • $\begingroup$ @the_fox From the class equation, the group must be $A_5$, the group of rotations of the icosahedron. The elements of order $2$ are $180^\circ$ rotations fixing a pair of opposite edges. I guess each one of these must commute with two others, but I was never much good at 3D geometry! $\endgroup$ – Derek Holt Feb 12 at 8:19
  • $\begingroup$ @DerekHolt Yes, I suspected that was the intended group, but Wolfram says that the icosahedral group is the one I mentioned: mathworld.wolfram.com/IcosahedralGroup.html Hence my question. (I'm not good at geometry either.) $\endgroup$ – the_fox Feb 12 at 8:31
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    $\begingroup$ You get $A_5 \times C_2$ if you include reflections. That is the full automorphism group of the associated graph. I think the $15$ pairs of opposite edges must decompose in some natural way into five triples, where the involutions within a triple commute with each other. $\endgroup$ – Derek Holt Feb 12 at 8:37

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