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Q1. Is it possible to somehow "see" the unsolvability of quintic polynomials in the $A_5$ symmetries of the icosahedron (or dodecahedron)?

Perhaps this is too vague a question.

Q2. Are there consequences of the unsolvability of $A_5$ that can be viewed in terms of the geometry of the icosahedron?

I am seeking some connection for pedagogical reasons, from the geometry to the group theory. One could argue from orbits of vertices of the icosahedron under its various symmetries that $A_5$ has no (nontrivial) normal subgroups, but it is a long way from there to the unsolvability of $A_5$ for students who have not yet studied group theory.

Thanks for any connections you can suggest!

Update 1. Answered by Barry Cipra and Geral Edgar:
No longer $2.25, but still inexpensive through used-booksellers.

Update 2. And here is Jerry Shurman's book (PDF download), as cited by Barry and Sam Hopkins:

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Didn't Felix Klein write a whole book on this? –  Barry Cipra Jun 9 '13 at 13:06
F. Klein, Lectures on the Icosahedron and the Solution of Equations of the Fifth Degree. –  Gerald Edgar Jun 9 '13 at 13:10
Echoing Barry, this is available freely online: people.reed.edu/~jerry/Quintic/quintic.pdf –  Sam Hopkins Jun 9 '13 at 15:37
I can't resist linking to a misnamed but nonetheless lovely icosahedral fountain sculpture: thewilliamandjosephgallery.com/main.php?g2_itemId=5966 -- who wouldn't want one of these bubbling away at the entrance to their math building? –  Barry Cipra Jun 9 '13 at 15:44
Another possible source: "Abel's Theorem in Problems and Solutions" by V.B.Alekseev, based on a course by V.I.Arnold to high school students in Moscow in 1963-64. The approach is somewhat topological, interpreting the Galois group of a polynomial as the monodromy group of a branched cover of the 2-sphere associated to the polynomial. Unsolvability of the quintic is shown by using the geometry of an icosahedron to see that its symmetry group is not solvable. –  Allen Hatcher Jun 10 '13 at 0:20
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1 Answer

up vote 9 down vote accepted

The most direct answer I know for this is the following:

Any maximal subgroup of a finite solvable group has index equal to a power of a prime. The unsolvability of $A_{5}$ is provable, therefore, by exhibiting a maximal subgroup of index 6 or 10 in $A_{5}$. These are both easy to see in terms of the geometry of the icosahedron.
The icosahedron has 12 vertices, so they come in 6 opposite pairs. The stabilizer of one of them has index 6, and it's obvious that this subgroup of index 6 is maximal because (considering a rotation of order 5 about one opposite pair of vertices) the action of $A_{5}$ on these 6 opposite pairs is doubly transitive (and therefore primitive).
Likewise, the icosahedron has 20 faces, which come in 10 opposite pairs. To prove this action is primitive, note that the point stabilizer has elements of order 3 which fix only the pair of opposite faces we have chosen. Using only this, we conclude that the point stabilizer has suborbit lengths equal to 1 and some multiples of 3 (in fact, we get 10=1+3+6 from the suborbit lengths). Since 10 has no proper divisors congruent to 1 mod 3, this must be a primitive action.

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