*November 21, 2004*

**Introduction**

**Regular polyhedra**

**Inscribing one regular polyhedron in another**

The cube can be inscribed in this way in the octahedron, but also reciprocally. The same reciprocity holds between the icosahedron and the dodecahedron. Such polyhedra are said to be dual. The tetrahedron is self-dual.

You can permute the set {1, 2} in two ways: (1, 2) and (2, 1). What about the set {1, 2, 3}? It is easy to see that we have six possibilities: (1, 2, 3), (2, 3, 1), (3, 1, 2), (2, 1, 3), (3, 2, 1), (1, 3, 2). The result is the same, for any sets of three objects, not only numbers. The number of permutations depends only on the number of elements of the set, and not on their nature. For a set with n elements, this number is 1·2·...·n, and is named the factorial n! of the number n.

An ordered set of the first n positive integers can also be used to express the permutation, as a specific operation to be applied to another ordered set. For example, (2, 1) shows that the elements of an ordered set of two elements are reverted. (2, 3, 1) shows that an ordered set of three, for example (1, 2, 3), changes its order to (2, 3, 1). Such transforms are named permutations of the set {1, 2, ... , n}.

We can multiply two permutations, which means that we apply them successively to the ordered set. All the permutations of a set with n elements forms a grup, because the operation of multiplication of permutations is associative, has a neutral element (which is the identical permutation (1, 2, ... , n), leaving the order unchanged), and every permutation has an inverse which cancels its effect. The group of permutations of the set {1, 2, ... , n} is named the symmetric group of degree n, denoted by $S_n$, and having n! elements.

A transposition is a permutation which interchanges only two elements. For example, (2, 1, 3), (3, 2, 1), (1, 3, 2) are transpositions of the group $S_3$. Any permutation can be decomposed as a product of transpositions in a non unique way. Yet, there is something independent of the way we decompose the permutation as a product of transpositions. If the number of transpositions in such a decomposition is even, it will be even for any other decomposition of the same permutation, and we name such a permutation even. Otherwise, we call it odd. The set of even permutations of a set with n elements forms a subgroup of $S_n$, named the alternating group, denoted by $A_n$, and having n!/2 elements.

**The symmetries of a regular polyhedron
**

By rotating the regular tetrahedron around one of its heights with 120º or 240º, this one remains unchanged. We say that the regular tetrahedron is unchanged by these transformations. We can rotate a regular polyhedron so that, after this transformation, it occupies exactly the same position, but having the faces not necessarily in the same positions. Also, they admit symmetry planes. In fact, this is the idea about the regular polyhedra – their rich symmetry.

The transformations leaving unchanged a polyhedron are named symmetry transformations. One can multiply the transformations. Because each symmetry transformation interchanges the faces of the polyhedron, we can associate to the transformation a permutation from $S_n$, n being the number of faces of the polyhedron. The symmetry transformations leaving invariant a polyhedron form a subgroup of $S_n$, named the automorphism group of the polyhedron.

How many symmetry transformations have each of the regular polyhedra? One easy method to count them is the following. Choose a face a. After a transformation it will take the place of another face a’. Since the polyhedron has n faces, we have n possibilities. Let’s choose a face b, neighbor to the first face before transformation, a. After the transformation, b goes into a face b’, neighbor to the face a’. Each face has the same number of edges k. The face b’ can be one of the k faces neighbor to a’. Therefore, we have nk possibilities. These transforms are the rotations of the polyhedron. But the face a is a polygon, therefore it has two sides. Consequently, when we move the face a in a’, this can flip. In this case, the transformation is no longer a rotation. It can no longer be obtain simply by moving the polyhedron, but by taking its mirror image.

Thus, among all transformations of the polyhedron, a special subgroup is formed by the nk rotations, but the total number of transformations is 2nk. Both these groups are subgroups of $S_n$. For the regular tetrahedron we obtain 2·4·3 = 24 automorphisms, from which the rotations are 4·3 = 12. The cube’s automorphisms group contains 2·6·4 = 48 automorfisms, from which 6·4 = 24 are rotations. The regular octahedron has 2·8·3 = 48 automorfisms, from which 8·3 = 24 rotations, like the cube. Both the icosahedron and the dodecahedron have 2·4·3 = 24 automorfisms, 4·3 = 12 rotations.

We see that the dual polyhedra have the same symmetries.

**The symmetries of regular polyhedra and the permutations**

Let’s label the vertices of a regular tetrahedron with the numbers {1, 2, 3, 4}:

Let us now label the cube’s vertices such that the ends of each large diagonal have identical labels from the set {1, 2, 3, 4}, like below:

Let’s choose one of the two regular tetrahedra inscribed in the cube. Its vertices are labeled with the numbers {1, 2, 3, 4}. A rotation of the cube rotates also the tetrahedron. Each cube rotation which let the tetrahedron in place corresponds to a rotation of the tetrahedron, and it is an even permutation. The even permutations of the cube’s labels correspond to transformations which interchange the two tetrahedra. Thus, the cube’s rotations correspond to the symmetric group $S_4$.

By labeling the faces of the regular octahedron, we obtain that it has the same symmetry groups as the cube (being its dual).

To obtain the symmetry group of the regular dodecahedron, let’s label its edges with the numbers {1, 2, 3, 4, 5}, like in the image:

**Group operations with polygons**

**"Calculator" for the Klein group**

The Klein’s group has four elements {E, A, B, C}, and its multiplication is given by the label:

**Calculator for the groups $A_3$ and $S_3$**

We start with the group $A_3$ of the even permutations of a set with 3 elements. Its elements are the permutations (1, 2, 3), (2, 3, 1), (3, 1, 2). We construct a card shaped as an equilateral triangle and we label its vertices with the numbers 1, 2 and 3:

The rule is: we label the vertices with the numbers {1, 2, 3} clockwise, and the edges with the permutations given by the order in which we met the vertices by starting from that edge and go clockwise.

The group $S_3$ contains, in addition to the even permutations (1, 2, 3), (2, 3, 1), (3, 1, 2) from the group $A_3$, the odd permutations (2, 1, 3), (3, 2, 1), (1, 3, 2). This is why we will allow, besides the rotations maintaining the triangle in plane, transformations obtainable by lifting the triangle from the table and flipping it. On its back face we will mark the odd permutations:

**Calculator for the groups $C_n$ and $D_n$**

A finite group such that all its elements can be obtained by multiplying one element (named generator of the group) with itself, is named cyclic group. We denote the cyclic group with n elements by $C_n$. The cyclic group $C_n$ is isomorphic with the group of integers modulo n, $\mathbb{Z}_n$. The cyclic group C3 is isomorphic with the alternating group $A_3$, but this doesn’t hold for n > 3. We can consider the group $C_n$ as representing the plane rotations of a regular polygon with n edges. If we allow mirror symmetries, obtained by taking the polygon outside the plane and flipping, the number of the possible symmetries doubles, and their group is named the dihedral group, $D_n$. We observe from the definition that for n = 3 the dihedral group is isomorphic to the permutation group of a set of three elements: $D_3$ ~ $S_3$, but, as for the cyclic group, we can’t generalize for n > 3.

**Polyhedral calculators of permutations**

**Calculator for the group $S_4$**

Once we position the cubes, we can simply read all the results of the multiplications with the chosen permutation (4132) with any element of $S_4$ on the result cube.

The reader is invited to study the symmetries of the permutations written on the cube.

Another interesting property of this cube is that, applying a rotation, we obtain on the initial position of the identical permutation a permutation showing how the four large diagonals of the cube were permuted.

**Calculator for the group $A_5$**

## 6 comments:

hi, very good exposition. can you please tell me where you gathered the material from ? a bibliography ? also, did you actually create the polyhedra yourself out of cartonboard or paper to visualize this ? thanks

Thank you.

The way I used polyhedra to represent group multiplication was the result of my own work. The ideas come to me when I studied group representations and the symmetry groups of regular polyhedra. I created the models several years ago on the computer and out of carton.

hi,im bernard and currently making a thesis paper regarding the permutation of the platonic solids,did the formed cycles from the permutation is disjoint?and how is it related to the ruffini's theorem?

Hi Bernard,

you may find helpful

this online book.

Hello, and thank you for the interesting article. The vertices of the dodecahedron can be assigned single digits and the edges assigned the permutation. This is consistent with your cube example and neater. For 6 element permutations the polyhedron becomes instead a plane tiled with hexagons.

cheers,

Doug Barrett

Edmonton

Doug,

> The vertices of the dodecahedron can be assigned single digits and the edges assigned the permutation.

I preferred to label the edges to be consistent with Klein's coloring, to use it also in that context.

One can label each vertex of the dodecahedron by starting from the solution presented in the article. One can use for example the following rule: each vertex is labeled with the second digit which appears in all three permutations which meet in that vertex. This rule exploits the fact that all three permutations at the same vertex have the same number on the second position. The resulting group is $A_5$.

> For 6 element permutations the polyhedron becomes instead a plane tiled with hexagons.

Actually, the group $S_6$ is not a subgroup of the group of isometries of the plane. But you can obtain its subgroups of the form $Z_n$ and $D_n$, where $n$ divides $6$.

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