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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    $\begingroup$ Didn't Felix Klein write a whole book on this? $\endgroup$ Jun 9, 2013 at 13:06
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    $\begingroup$ F. Klein, Lectures on the Icosahedron and the Solution of Equations of the Fifth Degree. $\endgroup$ Jun 9, 2013 at 13:10
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    $\begingroup$ Echoing Barry, this is available freely online: people.reed.edu/~jerry/Quintic/quintic.pdf $\endgroup$ Jun 9, 2013 at 15:37
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    $\begingroup$ 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? $\endgroup$ Jun 9, 2013 at 15:44
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    $\begingroup$ 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. $\endgroup$ Jun 10, 2013 at 0:20

2 Answers 2


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.


This is an old question, but I wanted to write a proof that $A_5$ is simple via symmetries of the icosahedron, using as little group theory as possible. I don't think that it can lead to a proof of the unsolvability of the quintic without the usual group theory (permutations, normal subgroups, quotients, solvable groups), and, of course, field theory.

Theorem: The group of rotations of the icosahedron is of order 60, simple, and isomorphic to the alternating group $A_5$. Therefore, $A_5$ is simple.

Proof: Consider a regular icosahedron, the convex hull of the 12 points $(0,\pm 1,\pm\phi)$, $(\pm 1,\pm\phi,0)$ and $(\pm\phi,0,\pm 1)$ in $\mathbb R^3$, where $\phi=\frac{1+\sqrt 5}2$ is the golden ratio. It has 12 vertices, 30 edges and 20 faces.

Let $G$ be its group of rotations. It has order $2*30=60$, because if $e$ is an oriented edge, then for any oriented edge $e'$, there is exactly one rotation that carries $e$ to $e'$. Any nontrivial rotation $g$ is of one of three types, according to whether its axis contains a vertex, an interior point of an edge (which must be the midpoint) or an interior point of a face (which must be the center). Its order is 5, 2 or 3, respectively.

To rotate a rotation: Notice that conjugation in $SO(\mathbb R^3)$ is particularly transparent: if $g$ is the rotation of angle $\omega$ about an axis $r$, then any conjugate $hgh^{-1}$ is the rotation of the same angle $\omega$ about the axis $h(r)$.

Proof that $G$ is simple: A normal subgroup $N$ that contains a nontrivial element $g$ must also contain the cyclic group of powers of $g$, and their conjugates, which are all the elements of the same type of $g$. To show that $N=G$, it's enough to see that $G$ can be generated with all the elements of any one of the three types.

  • With vertex rotations: Any vertex $v$ can be moved to a neighbor position $v'$ by rotating about the third vertex $v''$ of any of the two faces $f$ that contain $v$ and $v'$. So you can move any vertex to any place. After that you can rotate about that final place to put the icosahedron in any position.

  • With face rotations: If an edge $e$ joins $v$ with $v'$, then you can move $v$ to $v'$ using a rotation about any of the two faces $f$ and $f'$ that meet at $e$. If you move forth using $f$ and back using $f'$, then you return $v$ to its original position, so the composite is a rotation about vertex $v$. Again, we see that we can move and rotate vertices, so we generate $G$.

  • With edge rotations: Similar to the other cases.

The compound of 5 octahedra: Consider the construction of three orthogonal golden rectangles inscribed in the icosahedron. The six short sides are edges of the icosahedron, and we can paint them with the same color. Observe that, whenever an edge $e$ is chosen, there is a unique way to build the rectangles in such a way that the edge $e$ is one of the short sides, so in this way the 30 edges can be painted with 5 different colors. The midpoints of the edges of each color are the vertices of an octahedron, so we have formed a compound of 5 octahedra.

Proof that $G=A_5$: Each rotation permutes the colors, so we have a morphism $G\to S_5$. The vertex rotations have order 5, so they must give 5-cycles, which are even permutations. Similarly, the face rotations give 3-cycles, which are also even. Finally, consider a half-turn rotation $g$ about the midpoint of an edge $e$ of color $E$. Let $a,b,c,d$ be the consecutive sides of the quadrilateral obtained by joining the two faces that share the edge $e$. The edges $a,b,c,d,e$ have different colors $A,B,C,D,E$. It's now easy to see that $g$ interchanges $A$ with $C$ and $B$ with $D$, leaving $E$ fixed. This permutation is also even. Since any nontrivial rotation of the icosahaedron gives a nontrivial even permutation, we have an injective morphism from the group $G$, of order $60$, to the group $A_5$, of the same order. It must be an isomorphism. $\square$

Exercise: Notice that, at any vertex, the five colors meet, giving a cyclical order of the five colors. Show that the 12 cyclical orders found at the vertices are different.

Regarding the question Q2, there is an interesting (though not quite elementary) application: The fact that the icosahedron group $G$ is simple is a step in the construction of Poincaré's homology sphere, a 3-manifold $M$ that has the same homology of the sphere $S^3$, but is not homeomorphic to $S^3$, because its fundamental group is nontrivial. A construction of this kind allowed Poincaré to reject a preliminary wrong version of his famous conjecture, which was originally stated in terms of homology (a 3-manifold with the homology of the sphere is homeomorphic to a sphere, as happens in dimension 2).

A construction of the Poincaré sphere: Consider the 2-to-1 universal cover $\pi:S^3\to SO(\mathbb R^3)$, where $S^3=S_{\mathbb H}$ is the sphere of unit quaternions. (Explicitely, if $q$ is a unit quaterion, then $\pi(q)$ acts on $\mathbb R^3=\{v=x\mathrm i+y\mathrm j+z\mathrm k:\ x,y,z\in\mathbb R\}$ by conjugation, sending $v\mapsto qvq^{-1}$.) Its kernel is $\{\pm 1\}$. Define the binary icosahedral group $2G:=\pi^{-1}(G)$. It's nonabelian and of order 120. Let $M=SO(\mathbb R^3)/G=S^3/2G$ be the Poincaré homology sphere.

Fundamental group and homology of the Poincaré sphere: Notice that, since the quotient map $q:S^3\to M$ is a universal cover, we have $\Pi_1(M)=2G$. Since the Poincaré sphere is an orientable compact 3-manifold with one connected component, its bottom and top homology groups $H_0(M)$ and $H_3(M)$ are isomorphic to $\mathbb Z$. It remains to be proven that $H_1(M)$ and $H_2(M)$ are trivial. By Poincaré duality, it is enough to prove that $H_1(M)=0$, since $H_2(M)\cong H^1(M)\simeq H_1(M)$.

The homology group $H_1(M)$ is obtained by abelianization of the fundamental group $\Pi_1(M)=2G$, so we must prove that the commutator subgroup $[2G,2G]$ equals $2G$. Observe that $\pi([2G,2G])=[G,G]=G$, where the first equality holds because $\pi$ maps $2G$ onto $G$, and the second one holds because $G$ is nonabelian and simple, so the nontrivial normal subgroup $[G,G]$ must be equal to the whole group. So $\pi$ restricts to a surjective morphism $[2G,2G]\to G$ and there are two possibilities:

  • The morphism is 2-to-1, so $[2G,2G]=2G$ and we are done.

  • The morphism is an isomorphism.

To rule out the second possibility, we prove that $-1$, which is in the kernel of $\pi$, is an element of $[2G,2G]$. We do so by expressing it as a product of commutators of elements of $2G$. For that, observe that $i\in 2G$ because the rotation $\pi(i)$, which maps $i\mapsto i$, $j\mapsto -j$ and $k\mapsto -k$, preserves the set of vertices of the icosahedron. In the same way we see that $j\in 2G$, and then we can compute $[i,j]=iji^{-1}j^{-1}=ij(-i)(-j)=ijij=kk=-1.$

I think that there is also a more conceptual proof that the restriction $\pi|_{[2G,2G]}^G:[2G,2G]\to G$ is not 1-to-1 along these lines: If it was 1-to-1, then $\pi$ would admit a section along $G$. Since the discrete subgroup $G$ is ``pretty dense'' in $SO(\mathbb R^3)$, this section can (hopefully) be extended to a section defined on the whole rotation group $SO(\mathbb R^3)$, which is absurd.

Off-topic comment about the density of $G$ in $SO(\mathbb R^3)$: I think that any rotation $h$ is of the form $h=gk$, with $g\in G$ and $k$ a rotation of angle at most $\frac{2\pi}6$. The situation I have in mind is the following. Let $h$ be a rotation of angle $\frac{2\pi}6$ about the direction $(1,1,1)$. It sends the convex hull of

$\{(0,\pm 1,\pm\phi), (\pm 1,\pm\phi,0), (\pm\phi,0,\pm 1)\}$ to the convex hull of

$\{(0,\pm\phi,\pm 1), (\pm\phi,\pm 1,0), (\pm 1,0,\pm\phi)\}$.

The picture shows the superposition of the two icosahedra, known as compound of two icosahedra. The rotation $h$ can be decomposed as $g=hk$ in 8 ways. More visually, there are 8 rotations of angle $\frac{2\pi}6$ that move the red icosahedron to the yellow icosahedron.

Since the angle-of-rotation distance in $SO(\mathbb R^3)$ is just the double of the geodesic distance in $S^3$ (which is the angle of vectors), to verify what I claim we just have to prove that $2G$ is $\frac{2\pi}{12}$-dense in $S^3$. This, if true, can be checked numerically with error of less than 1 degree. But it would be more interesting to have a picture of the Voronoi-cell decomposition of $S^3$ with respect to the discrete set $2G$. I wonder if this leads to the other usual presentation of the Poincaré sphere (as a quotient of the dodecahedron, identifying opposite faces by a $\frac{2\pi}{10}$ right-handed twist).

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    $\begingroup$ Wow! Beautiful exposition! $\endgroup$ May 27, 2014 at 17:13
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    $\begingroup$ As I scrolled down through this answer I thought to myself "O'Rourke". When I got to the bottom I said Aha! I knew it! $\endgroup$ Jun 25, 2014 at 3:03

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