The five solids as three groups

Five solids, and the temptation is to think of them as five separate objects with five separate symmetries. They are not. There are exactly three groups of rotations among the five, and the same three serve all thirteen Archimedean solids as well, and the reason is short: a solid and its dual occupy the same space, so any turn that leaves one where it was leaves the other where it was too.

Counting the turns without listing them #

There is a way of getting the total that needs no enumeration at all, and it is the argument worth having.

Pick an edge of the solid and give it a direction, so that it points from one of its endpoints to the other. A rotation of the solid carries that directed edge to some other directed edge — and here is the point: for any two directed edges, there is exactly one rotation carrying the first to the second.

That there is at least one is the regularity of the solid: all its edges look alike and all its ends look alike. That there is at most one is because a rotation of space is fixed by what it does to a frame of three directions, and a directed edge supplies them — its own direction, the direction out of the solid at its midpoint, and the perpendicular to both.

So the rotations correspond exactly to the directed edges, and there are two of those per edge. The order of the rotation group is twice the number of edges. Cube and octahedron: twelve edges, twenty-four turns. Tetrahedron: six edges, twelve turns. Dodecahedron and icosahedron: thirty edges, sixty turns.

The figure does not assume this. It builds a rotation for every pair of a vertex and one of its neighbours, checks each one is a genuine rotation rather than a reflection, checks each sends the vertex set to itself, discards duplicates, and then asserts that the count that survives is twice the edge count.

Why duals share a group #

The duality is not a resemblance; it is an identity of symmetry groups, and the reason is that the dual is constructed from the original.

Put a point at the centre of each face of a cube and join points on adjacent faces. What results is an octahedron, and it sits inside the cube in a definite position. Now take any rotation leaving the cube where it was. It permutes the cube’s faces, so it permutes their centres, so it carries the octahedron to itself.

The argument runs both ways and gives an exact correspondence: the rotations of a cube are the rotations of the octahedron inscribed in it, the same twenty-four maps of space, not two groups that happen to be isomorphic.

What the enumeration actually does #

The construction the figures use deserves a description, because it is the difference between a table copied from somewhere and a table produced.

For each vertex $v$ of the solid and each neighbour $u$ of $v$, a frame is built: the direction out to $v$, the direction along the edge to $u$ with the first part removed, and the perpendicular to both. Every such frame is an orthonormal triple, and the rotation taking a fixed reference frame to it is the product of one frame matrix with the transpose of the other.

Each candidate is then subjected to two tests. Its determinant must be $+1$, which rules out reflections — a frame built the wrong way round would produce one, and the check would catch it. And it must send every vertex of the solid to a vertex of the solid, checked against every vertex in turn.

What comes out is a list of matrices, deduplicated, whose length is asserted to be twice the edge count. The classification then groups them by axis, where the axis of a rotation is read off the antisymmetric part of its matrix — except for a half turn, where that part vanishes and the axis has to be read off $R + I$ instead. Each axis’s order is the number of turns on it plus one.

None of that requires knowing the answer in advance, which is the point. A wrong coordinate in the solid would fail the vertex-preservation test; a wrong frame would fail the determinant test; a mistake in the grouping would fail the assertion that the axis counts and the turn counts agree.

The three groups, named #

The three groups have names from the permutations they turn out to be.

The tetrahedral group has twelve elements and is $A_4$, the even permutations of four things. The four things are the tetrahedron’s four vertices, and every rotation permutes them; the odd permutations are the reflections, which are not rotations.

The octahedral group has twenty-four elements and is $S_4$, all permutations of four things. The four things are the cube’s four long diagonals. Every rotation permutes them and every permutation is achieved, which is a small surprise: the cube’s twenty-four rotations and the twenty-four shuffles of four objects are the same group, so everything true of shuffling four cards is true of turning a cube.

The icosahedral group has sixty elements and is $A_5$, the even permutations of five things. The five things are harder to see. A dodecahedron has twenty vertices, and eight of them at a time form the corners of a cube whose edges are diagonals of the pentagonal faces — there are exactly five such cubes, and every vertex belongs to two of them. A rotation of the dodecahedron permutes the five cubes, always evenly, and every even permutation arises.

Finding those five cubes is the hard part of the whole subject and it is worth saying why. The tetrahedron’s four vertices and the cube’s four diagonals are visible objects that a rotation obviously shuffles. The dodecahedron’s five cubes are not visible at all until somebody points them out, and until they are, there is no reason to expect a group of order sixty to be a group of permutations of anything.

Reading a group off a table of axes #

The three tables the figures print are more informative than they look, because the axis counts are the conjugacy classes.

Two rotations are conjugate — the same turn seen from a different position — exactly when they turn by the same amount about axes the solid’s symmetries can carry into each other. So the icosahedron’s sixty split as 1 + 24 + 20 + 15: the identity, twenty-four turns about five-fold axes, twenty about three-fold, fifteen half turns. The cube’s twenty-four split as 1 + 6 + 3 + 8 + 6, with the quarter turns and half turns about the face axes counted separately because a quarter turn is not conjugate to a half turn.

Those splittings are what the next section needs. A normal subgroup is a union of conjugacy classes containing the identity, so the question can this group be taken apart becomes the question which unions of these numbers, including the 1, have a size dividing the group’s order. For sixty split as 1, 15, 20, 24 the answer is that no proper sub-sum works except 1 itself — and that arithmetic is most of the proof.

The one that will not come apart #

The last of the three is the important one, and the reason is a property that has nothing to do with geometry.

A group can sometimes be broken into a smaller group and a quotient, repeatedly, until only the simplest pieces are left — and a group for which every stage of that breaking has an abelian quotient is called solvable. The tetrahedral and octahedral groups are solvable. The icosahedral group is not: $A_5$ has no normal subgroup at all except the identity and the whole thing, so there is nowhere to start.

$A_5$ is the smallest non-abelian group with that property, and its order is sixty — the same sixty as the icosahedron’s turns. That is not a coincidence of numbers; the group is the icosahedron’s rotations, and the fact that it cannot be taken apart is the fact that the general equation of the fifth degree has no formula in radicals.

The connection runs through the correspondence between field extensions and groups. Solving an equation by radicals corresponds to a chain of subgroups with abelian quotients; a polynomial whose group is $A_5$ admits no such chain; and the icosahedron is where $A_5$ lives most visibly. Klein wrote an entire book about it, and the title — Lectures on the Icosahedron and the Solution of Equations of the Fifth Degree — puts a solid and a theorem about algebra in one sentence deliberately.

Where the three groups turn up elsewhere #

Having only three groups for five solids and thirteen Archimedean ones and thirteen duals is an economy worth pausing on, and the groups reappear well outside geometry.

$S_4$, the cube’s rotations, is the group of permutations of four things, so every question about arranging four objects is a question about the cube. The solvability of $S_4$ is why the quartic equation has a formula, and the chain of subgroups that solves it corresponds to a chain of geometric conditions on the cube: the three pairs of opposite faces, the two ways of splitting four diagonals into two pairs of two.

$A_5$ appears as the rotations of the icosahedron, as the smallest non-abelian simple group, and as the group of the general quintic. It is also the symmetry group of the Petersen graph’s complement and shows up in the theory of modular curves — a group that small being that widely useful is unusual and is a consequence of being simple.

$A_4$ is the smallest group in which the converse of Lagrange’s theorem fails: it has order twelve and no subgroup of order six, so a divisor of the order need not be the order of a subgroup. That fact is normally taught as an abstract counterexample, and it is a statement about the tetrahedron.

What the group does not include #

Everything above is about rotations, and each solid has twice as many symmetries as that, because reflections are symmetries too. The full symmetry group of the cube has forty-eight elements, of the tetrahedron twenty-four, of the icosahedron 120.

Restricting to rotations is not an arbitrary simplification. A rotation is a motion that can be carried out physically — a solid model can be turned into any of its rotated positions — whereas a reflection cannot, and a mirror-image object is a different object to anyone who has to build it. The rotations are the symmetries of the thing as an object in space; the full group is the symmetries of the thing as a set of points.

The distinction has a consequence worth noting: the full group is not always the rotation group doubled in the obvious way. For the tetrahedron the full group is $S_4$ — the same group as the cube’s rotations — while the tetrahedron’s own rotations are $A_4$. Two different solids, two different senses of symmetry, one group of twenty-four appearing in both.

What the picture cannot show #

An axis is drawn as a line and a rotation is not a line; it is a motion, and a still picture of a motion shows only what is left fixed by it. The tables beside each solid say what the drawing cannot: how many turns each axis carries and what their orders are.

Worse, the axes themselves are not drawn — only the solid and the counts are. A figure with fifteen two-fold axes drawn through an icosahedron is a figure with fifteen lines through a shape with thirty edges, which is illegible, and the honest response was to compute the classification and print it rather than to draw a thicket. The same choice was made about the eight symmetries of a square, where eight is few enough to draw and sixty is not.

And nothing here shows the group structure. The counts of axes are a classification by conjugacy class, which is a genuine invariant, and two different groups can have the same class sizes. That $A_5$ has no normal subgroup is proved by examining every union of conjugacy classes and testing it for closure — an argument made elsewhere and not visible in any drawing of a solid.

The ladder from here #

Rungs above: the full symmetry groups with reflections, and the three-way relation between a rotation group, its double and the group with a central inversion. The finite subgroups of the rotations of space, which are exactly these three plus two infinite families — a classification as clean as the solids’ own. The binary polyhedral groups, which live in the quaternions and double each of these. Klein’s icosahedral function, which solves the quintic with something other than radicals. And the crystallographic restriction, which explains why five-fold symmetry, the icosahedron’s own, cannot tile space.

The count that is really a theorem #

The arithmetic of this rung is worth separating from the geometry, because it is what carries.

Twice the edge count is the order of the group, and the derivation used nothing about the solid except that all its directed edges look alike. That is the orbit–stabiliser relation in disguise: the group acts on the directed edges, the action has one orbit, and nothing but the identity fixes a directed edge, so the group’s size is the number of directed edges.

Written that way it applies far beyond solids. The size of a group acting on a set is the size of one orbit times the size of the thing that fixes a point of it, and almost every count of symmetries in this collection is that sentence with different nouns — the colourings nobody can tell apart, the blocks a subgroup cuts out, the number of ways a necklace can be arranged. The five solids are a good place to meet it because the orbit is visible and the stabiliser is trivial, which is the easiest case there is.