I am reading Michael Artin's Algebra, encountered paragraphs( 7.3) that I don't quite understand.

so there's the citation

The elementary formula which uses the partition of $S$ into orbits to count its elements. We label the different orbits which make up $S$ in someway, say as $O_1, ..... , O_k$ then $|S| = |O_1|+.....|O_k|$, the formula hs great number of applications. Example, consider the group G of orientation-preserving symmetries of a regular dodecahedron $D$. these symmetries are $11$ rotations. it is tricky to count them without error. Consider the action of $G$ on the set $S$ of the faces $D$. the stablilizer of a face is the group of ortations by multiples of $2\pi/5$ about a perpendicular through the center of $S$. So the order of $G_s$ is $5$. There are $12$ faces, and $G$ acts transitively on them, thus $|G| = 5\times 12 = 60$ or $G$ operates trasnsitively on the vertices $v$ of $D$. There are three rotations including 1 fix a vertex, so $|G_v| = 3$. There are $20$ vertices hence $|G| =3\times 20 = 60$, which checks. There is a similar computation for edges. If $e$ is an edge then $|G_e| = 2$, so since $60 = 2\times 30$, the dodecahedron has $30$ edges.

I don't quite understand when counting using vertex what are the three rotations, and when counting using edge what is exactly $G_e$ meant for ?

let $S$ be the set of $12$ faces of the dodecahedron, and let $H$ be the stablilizer of a particular face $S$, then $H$ also fixes the face opposite to $S$, and so there are two $H$-orbits of order 1, The remaining faces make up two orbits of order $5$. In this case it reads as follows $12 = 1+1+5+5$.

I really don't understand how the order of H-orbits are derived? and infact don't even understand, what are the gemometric meaning of these H-orbits?


1 Answer 1


We used Michael Artin's text last semester, I remember reading this bit. Your first question:

"I don't quite understand when counting using vertex what are the three rotations, and when counting using edge what is exactly $G_e$ meant for ?"

Answering this is going to be difficult because you need a scaled model of the dodecahedron to really see what I'm talking about. At a vertex, there are three faces that meet to give a vertex (each face is a pentagon). Now suppose you rotate the dodecahedron about a vertex. You want to rotate it by an angle such that after the rotation, you get the exact same configuration of the dodecahedron. The clue to what angle you must rotate it by is in the number of faces that meet at a vertex. This number is three so that the angle is $360^{\circ}/3$ which is $120^{\circ}$. So the rotations that fix a vertex are rotation by $120^{\circ}$, $240^{\circ}$ and doing nothing (or $360^{\circ}$), which is three in total. This also means to say that the order of the stabiliser of a vertex is three. When counting the stabiliser of an edge, $G_e$ is just purely notational to mean "stabiliser of an edge", the subscript "e" for "edge". Does this answer your first question?

For the second part of your question, can your provide more context (i.e. copy more of what the text says before and after that paragraph)? This is because I believe we need to look more carefully at what now is acting on the set of 12 faces of the dodecahedron. It can't be the whole group because that acts transitively on the dodecahedron. Perhaps it is being acted on by some subgroup of the whole group? I don't have the book with me now.

$\textbf{Edit:}$ I get your second question now: Fix a face $H$ of your dodecahedron. Now the stabiliser of $H$, $G_H$ has order 5. Now when we let $G_H$ act on the dodecahedron, trivially $G_H$ acting on $H$ will not move it to any other face. So we get one orbit here of order 1. Similarly, $G_H$ acting on the opposite face $H'$ will not move it to any other face so that we get another orbit here of the same order. Now consider the 5 faces of the dodecahedron that are directly adjacent to $H$. When $G_H$ acts on them, they simply rotate about the line perpendicular to $H$ so that $G_H$ acts transitively on these 5 faces. In other words, these 5 faces give us another orbit. Similarly the set of 5 faces adjacent to $H'$ gives us another orbit. So since $5 + 5 + 1 + 1 = 12,$ this illustrates why the sum of the size or orbits must be equal to the size of the whole set (I believe that's what Artin is trying to illustrate). Does this help?

  • $\begingroup$ to clarify, if we have a group action of G on a set X, then any subgroup H of G also acts on that same set X using the very same action. Artin is using H-orbits, to distinguish them from the G-orbits (the orbits of the whole group). given the context, it might have been better to use a different letter besides "H" for the face in question. $\endgroup$ Apr 8, 2012 at 15:57
  • $\begingroup$ @DavidWheeler Thanks for the clarification. I vaguely remember something like that in Artin's text, but as I said I don't have it now. $\endgroup$
    – user38268
    Apr 9, 2012 at 1:28
  • $\begingroup$ @BenjaminLim thanks very much for the explanation. they help a lot, I think I need that mini model of dodecahedron to help understand this. for "what Ge means, I was also asking for the rotations Artion mentioned when rotating around edge". I think the main issue is I don't get the rotation axis when thinking about the rotation. $\endgroup$
    – zinking
    Apr 9, 2012 at 3:12
  • $\begingroup$ @zinking Take an edge, draw a line through that edge until it meets the opposite edge at right angles. Then if you rotate your dodecahedron about that edge, you should see the only rotations that preserve the dodecahedron are $180$ degrees of the full 360. That's why the order of the stabiliser is 2. Hint: I suggest making paper models of these things. If you are here in australia I would gladly pass you mine! Does this answer all your queries now? Please accept my answer above if it does. $\endgroup$
    – user38268
    Apr 9, 2012 at 3:21
  • $\begingroup$ yes, thanks a lot. accepted, not familiar with MathExchange. $\endgroup$
    – zinking
    Apr 9, 2012 at 10:08

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