POLYHEDRON PUZZLES AND GROUPS

JORGE REZENDE

1. Introduction

Consider a polyhedron. For example, a platonic, an archimedean, ora dual of an archimedean polyhedron. Construct flat polygonal plates inthe same number, shape and size as the faces of the referred polyhedron.Adjacent to each side of each plate draw a number like it is shown infigures 1-10. Some of the plates, or all, can have numbers on both faces.We call these plates, two-faced plates. In this article, the two-faced plateshave the same number adjacent to the same side. Figure 4 shows twentyfaces of triangular two-faced plates and their reverse faces are representedin figure 5. Figure 7 shows twelve faces of pentagonal two-faced platesand their reverse faces are represented in figure 8.

Now the game is to put the plates over the polyhedron faces in such away that the two numbers near each polyhedron edge are equal. If thereis at least one solution for this puzzle one says that we have a polyhedronpuzzle with numbers.

In this article we begin in Section 2 by giving some examples of puz-zles. In Section 3 we describe the tetrahedron, the octahedron and theicosahedron symmetry groups. In Section 4 we recall some definitions onpuzzle solutions and their relation with permutation groups. In Section5 we show how to define puzzles using group theory.

This is the first of a series of two articles. In the next one [8], wesuggest some simple mathematical activities using polyhedron puzzles.

2. Definitions using combinatorics

2.1. Platonic and archimedean polyhedra. From now on, assumethat the numbers belong to the set 1, 2, . . . , n, and that all the numbersare used.

If we have plate faces which have the shape of a regular polygon withj sides, one can ask how many possible ways ν are there to draw the

The Mathematical Physics Group is supported by the Portuguese Ministry forScience and Technology.

This paper is in final form and no version of it will be submitted for publicationelsewhere.

1

2 JORGE REZENDE

1 1

32

3 2

Figure 1.

numbers 1, 2, . . . , n, without repeating them on each plate face. Theanswer is

a) For j = 3 (equilateral triangle) and n = 3, then ν = 2 (see figure1).

b) For j = 3 and n = 4, then ν = 8 (see figure 2); 8 is precisely thenumber of the octahedron faces. With these 8 plates we make theoctahedron puzzle (≡ the octahedron (1) puzzle).

44

44

33

33

22

22

11

11

11

43 2

432

Figure 2.

POLYHEDRON PUZZLES AND GROUPS 3

22

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1 1 11

1111

1 1 11

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43 2

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5

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33

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554 33 4

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5 432

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2 34

55 4

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Figure 3.

c) For j = 3 and n = 5, then ν = 20 (see figure 3); 20 is preciselythe number of the icosahedron faces. With these 20 plates we makethe icosahedron (1) puzzle.

d) For j = 3 and n = 6, then ν = 40 (see figures 4 and 5); 40 isprecisely the double of the number of the icosahedron faces.Construct different plates with the numbers written on both faces. Thisgives 20 plates. We call the related puzzle, the icosahedron second puzzle(or icosahedron (2)).

e) Consider again j = 3 and n = 6. Construct different plates with thenumbers written only on one face, but in such a way that the numbersgrow if we read them, beginning with the minimum, counter clock-wise.This gives 20 plates (see figure 4). We call the related puzzle, the icosa-hedron third puzzle (or icosahedron (3)).

4 JORGE REZENDE

66

3 3

222

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1

1

1 1

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1 1

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365

Figure 4.

f) For j = 4 (square) and n = 4, then ν = 6 (see figure 6); 6 isprecisely the number of the cube faces. With these 6 plates wemake the cube puzzle (≡ the cube (1) puzzle).

g) For j = 5 (regular pentagon) and n = 5, then ν = 24 (see fig-ures 7 and 8); 24 is precisely the double of the number of thedodecahedron faces.

Construct different plates with the numbers written on both faces.This gives 12 plates. We call the related puzzle, the dodecahedron firstpuzzle (or dodecahedron (1)).

h) Let again j = 5 and n = 5. Construct different plates with thenumbers written only on one face, but in such a way that the numbersread counter clock-wise, abcd5, are such that abcd form an even permu-tation. This gives 12 plates (see figure 9). We call the related puzzle, thedodecahedron second puzzle (or dodecahedron (2)).

POLYHEDRON PUZZLES AND GROUPS 5

66 64

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Figure 5.

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11

1 1

1

1

44

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223

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Figure 6.

i) Consider n = 4. With j = 3, one has ν = 8 (see figure 2). Withj = 4, one has ν = 6 (see figure 6). Notice that 8 is precisely the number

6 JORGE REZENDE

3

44

555

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555

5 5

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1 1 1

1 1 1

111

1 1 1

Figure 7.

of the cuboctahedron triangular faces and 6 is precisely the number ofits square faces. This is an example of an interesting puzzle using anarchimedean polyhedron. We call it the cuboctahedron puzzle (≡ thecuboctahedron (1) puzzle).

The general formula for ν is

ν = (j − 1)!

(n

j

)=

n!

(n− j)!j.

2.2. More puzzles. Take now a deltoidal icositetrahedron. It has 24deltoidal faces. If we have 24 plates which have the deltoidal shape thenumber of possible different ways to draw the numbers 1, 2, 3, 4, withoutrepeating them on each plate is precisely 24. This an example of an inter-esting puzzle using a dual of an archimedean polyhedron (see reference[6]).

Consider again the cube. It has 6 faces that are squares. The number ofpossible different ways to draw the numbers 1, 2, 3, 4, with two repetitions

POLYHEDRON PUZZLES AND GROUPS 7

33

33

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4

22

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2

22

2 3

3

3

3

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55

44

3

3

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32

1 1

1

1

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5

543

25

5

4

1 1

1

1

1

1 1

1

Figure 8.

of the form aabb (the numbers are read counter clock-wise) on each squareplate is precisely 6. This gives the cube (2) puzzle.

Consider again the icosahedron. It has 20 faces that are equilateraltriangles. The number of possible different ways to draw the numbers1, 2, 3, 4, 5, with one repetition on each triangular plate is precisely 20.This gives the icosahedron (4) puzzle (see reference [7]).

These are simple examples of polyhedron puzzles with numbers, whichare enough in order to understand the following sections. There are,obviously, others as we shall see. For more examples see reference [2],which is a development of reference [1]. Reference [3] is a collection ofsome of these puzzles paper models.

3. Polyhedron symmetries

Consider a polyhedron in R3. From now on V denotes the set of thepolyhedron vertices, E denotes the set of the polyhedron edges and Fdenotes the set of the polyhedron faces.

8 JORGE REZENDE

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Figure 9.

The group of the polyhedron symmetries, Ω, called the polyhedrongroup, is the set of all isometries ω of R3, that send vertices to vertices,which implies that they send edges to edges, faces to faces. Every sym-metry ω ∈ Ω induces three bijections, that we shall also denote by ω,whenever there is no confusion possible: ω : V → V , ω : E → E andω : F → F . Denote also by Ω ≡ ω : V → V ≡ ω : E → E ≡ω : F → F, the three sets of these functions. One can say that eachone of these three sets Ω is the set of the polyhedron symmetries. Noticethat not all one-to-one functions F −→ F , E −→ E, V −→ V are in Ω.With the composition of functions each one of these three sets Ω formsa group that is isomorphic to the group of the polyhedron symmetries.If ω1, ω2 ∈ Ω, we shall denote ω1ω2 ≡ ω1 ω2.

When no confusion is possible, ω ∈ Ω represents also the group iso-morphism ω : Ω → Ω, ω(ω1) = ω ω1ω

−1, for every ω1 ∈ Ω. Note thatω1 and ω(ω1) have the same order. Look at the octahedron in figure 18.If ω is a counter clock-wise rotation of 90 around the z-axis, and ω1 is

POLYHEDRON PUZZLES AND GROUPS 9

3

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Figure 10.

a counter clock-wise rotation of 90 around the x-axis, then ω(ω1) is acounter clock-wise rotation of 90 around the y-axis. In simple words,ω transports x over y. Here, a counter clock-wise rotation around thez-axis, for example, means that we look from the positive z-semiaxis.

If Ω1 is a subgroup of Ω, then Ω1 acts naturally on the face set, F : forω ∈ Ω1 and ϕ ∈ F , one defines the action ω ϕ = ω(ϕ).

In the following we only consider polyhedra centered at the origin, Ω+

denotes the subgroup of Ω of the symmetries with determinant 1 and Ω−

denotes the subgroup of Ω of the symmetries with determinant −1.

3.1. The tetrahedron group. Consider the tetrahedron (see figures 11and 12) and its group, Ω.

An element of Ω is, for example, the function ω(x, y, z) = (−x, y, z),that induces the function ω : E −→ E

ω(e1) = e1 ω(e2) = e3 ω(e3) = e2

ω(e4) = e5 ω(e5) = e4 ω(e6) = e6.

10 JORGE REZENDE

x

y

w

z

e6

e2

e3

e5

e4

e1

Figure 11.

This symmetry has determinant−1 and is one of the twelve elements ofΩ with determinant −1. They reverse the orientation. The tetrahedronhas no central symmetry.

e3

e1

e4

e4e2

e5

e6e6

e5

Figure 12.

POLYHEDRON PUZZLES AND GROUPS 11

ωω

e3

e3

e4

e1

e4

e4

e1

e3

e6

e2

e2

e6

e5

e5

e6

e6

e5

e2

Figure 13.

The symmetries with determinant 1 (Ω+), can be seen like this: onetransports a chosen face in such a way that it goes to one of the fourtetrahedron faces; as one has three possibilities of making them coincide(they are equilateral triangles), there are 12 (3 × 4) symmetries withdeterminant 1. Figure 13 shows one of these symmetries. In this casethe function ω : E → E, is the following:

ω(e1) = e4 ω(e2) = e5 ω(e3) = e1

ω(e4) = e3 ω(e5) = e6 ω(e6) = e2.

The advantage of describing in this way the symmetries of Ω+

is that it can be easily adapted to other polyhedra, and used intheir computation in a computer program.

Another way of counting the symmetries de Ω+ is the following: theidentity (1); the rotations of 180 around the three axes defined by thecenters of opposite edges (3); the rotations of 120 and 240 around thefour axes defined by each vertex and the center of the opposite face (8).

The symmetries with determinant −1 (Ω−) are the compositions of thesymmetries de Ω+ with a symmetry with determinant −1. The cardinalof Ω, the order of Ω, is, therefore, 24.

Let us see two examples:a) Consider, in the figures 11 and 14, the rotations of 0 and 180

around the z-axis. They form a subgroup of Ω+ of order two. The actionof these rotations on the tetrahedron faces, sends every green face to theplace of the other green face and every blue face to the place of the otherblue face. The set of the green faces and the set of the blue faces arethe orbits of this action (see figures 14 and 15). Hence, there are twoorbits, each one of them with two elements. We say that the action of

12 JORGE REZENDE

w

z

Figure 14.

Figure 15.

this subgroup is of the (1, 2× 2) type. The first component, the 1, is thedeterminant of the generator.

b) Consider, in the figures 11 and 16, the rotations of 0, 120 and240 around the w-axis. They form a subgroup of Ω+ of order three.The action of these rotations on the tetrahedron faces keeps the green

POLYHEDRON PUZZLES AND GROUPS 13

w

z

Figure 16.

face fixed and sends every blue face, by order of succession, to the placeof all other blue faces. The set with the green face and the set of the bluefaces are the orbits of this action (see figures 16 and 17). Therefore, thereare two orbits, one with three elements , and one with only one element.We say that the action of this subgroup is of the (1, 1× 1 + 1× 3) type.The first component, the 1, is the determinant of the generator.

3.2. The octahedron (cube) group. Consider the octahedron (seefigures 18 and 19) and its group, Ω. Everything that we say here aboutthe octahedron group can be translated to the cube group interchangingfaces with vertices. In other words the group is the same.

An element of Ω is, for example, the central symmetry ω(x, y, z) =−(x, y, z), that induces the function ω : E −→ E

ω(e1) = e11 ω(e2) = e12 ω(e3) = e9 ω(e4) = e10

ω(e5) = e7 ω(e6) = e8 ω(e7) = e5 ω(e8) = e6

ω(e9) = e3 ω(e10) = e4 ω(e11) = e1 ω(e12) = e2.

The central symmetry has determinant −1. The symmetries with de-terminant 1 (Ω+) can be seen like this: one transports a chosen face insuch a way that it goes to one of the eight faces of the octahedron; asone has three possibilities of making them coincide (they are equilateraltriangles), there are 24 (3 × 8) symmetries with determinant 1. Figure

14 JORGE REZENDE

Figure 17.

e12 e11

e7

xy

z

e9 e10

e6

e3e4

e8e1

e2

e5

Figure 18.

20 shows one of these symmetries. In this case the function ω : E → E,is the following:

POLYHEDRON PUZZLES AND GROUPS 15

e9e12

e2

e5e1

e8

e4

e3

e7

e12

e5

e9

e11e10

e2e3

e6

Figure 19.

ω(e1) = e4 ω(e2) = e7 ω(e3) = e12 ω(e4) = e8

ω(e5) = e3 ω(e6) = e11 ω(e7) = e9 ω(e8) = e1

ω(e9) = e2 ω(e10) = e6 ω(e11) = e10 ω(e12) = e5.

Note once more that the advantage of describing in this way the sym-metries of Ω+ is that it can be easily adapted to other polyhedra, andused in their computation in a computer program.

Another way of counting the symmetries de Ω+ is the following: theidentity (1); the rotations of 90, 180 and 270 around the three axesdefined by opposite vertices (9); the rotations of 180 around the six axesdefined by the centers of opposite edges (6); the rotations of 120 and240 around the four axes defined by the centers of opposite faces (8).

The symmetries with determinant −1 (Ω−) are the compositions ofthe symmetries de Ω+ with the central symmetry. The cardinal of Ω, theorder of Ω, is, therefore, 48.

In reference [5] one can see two examples of such symmetries and adetailed description of the octahedron puzzle case.

3.3. The icosahedron (dodecahedron) group. Consider the icosahe-dron (see figure 21) and its group, Ω. Everything that we say here aboutthe icosahedron group can be translated to the dodecahedron group in-terchanging faces with vertices. In other words the group is the same.

16 JORGE REZENDE

ωω

e3

e7

e2e5

e9e12

e2

e5

e1

e4

e1

e8

e8

e12

e4

e3

e3

e2 e5

e9

e7

e12

e5

e9

e6 e10

e11e10

e12

e11

e7

e2e3

e6

Figure 20.

An element of Ω is, for example, the central symmetry ω(x, y, z) =−(x, y, z), that induces the function ω : E −→ E

ω(e1) = e16 ω(e2) = e17 ω(e3) = e18 ω(e4) = e19 ω(e5) = e20

ω(e6) = e21 ω(e7) = e22 ω(e8) = e23 ω(e9) = e24 ω(e10) = e25

ω(e11) = e26 ω(e12) = e27 ω(e13) = e28 ω(e14) = e29 ω(e15) = e30

ω(e16) = e1 ω(e17) = e2 ω(e18) = e3 ω(e19) = e4 ω(e20) = e5

ω(e21) = e6 ω(e22) = e7 ω(e23) = e8 ω(e24) = e9 ω(e25) = e10

ω(e26) = e11 ω(e27) = e12 ω(e28) = e13 ω(e29) = e14 ω(e30) = e15.

The central symmetry has determinant −1. The symmetries with de-terminant 1 (Ω+) can be seen like this: one transports a chosen face insuch a way that it goes to one of the twenty faces of the icosahedron; asone has three possibilities of making them coincide (they are equilateraltriangles), there are 60 (3× 20) symmetries with determinant 1. Figure

POLYHEDRON PUZZLES AND GROUPS 17

e20

e23

e26

e21

e24

e27

e19

e25

e22

e18

e16

e17

x

z

y

e5

e8

e2

e6

e9

e3

e1

e4

e7

e10

e11

e30

e29

e12

e13

e15

e28

e14

Figure 21.

22 shows one of these symmetries. In this case the function ω : E → E,is the following:

ω(e1) = e5 ω(e2) = e8 ω(e3) = e2 ω(e4) = e10 ω(e5) = e28

ω(e6) = e3 ω(e7) = e14 ω(e8) = e11 ω(e9) = e1 ω(e10) = e27

ω(e11) = e6 ω(e12) = e7 ω(e13) = e30 ω(e14) = e19 ω(e15) = e9

ω(e16) = e20 ω(e17) = e23 ω(e18) = e17 ω(e19) = e25 ω(e20) = e13

ω(e21) = e18 ω(e22) = e29 ω(e23) = e26 ω(e24) = e16 ω(e25) = e12

ω(e26) = e21 ω(e27) = e22 ω(e28) = e15 ω(e29) = e4 ω(e30) = e24.

Another way of counting the symmetries de Ω+ is the following: theidentity (1); the rotations of 72, 144 , 216 and 288 around the sixaxes defined by opposite vertices (24); the rotations of 180 around thefifteen axes defined by the centers of opposite edges (15); the rotations

18 JORGE REZENDE

ωω

e 21

e 18

e 24

e 26

e 23e 17

e 27

e 14

e 21

e 30

e 29

e 20

e 16e 19

e 5

e 10e 30

e 17e 18

e 10

e 7

e 25

e 22

e 28e 8

e 7

e 21

e 22

e 16

e 24e 14

e 1

e 4e 13

e 15e 6

e 11

e 2e 1

e 26

e 23

e 19

e 27

e 5e 2

e 12

e 26

e 9

e 3

e 4e 13

e 28e 11

e 8

e 3e 9

e 23

e 17

e 29

e 12

e 15

e 6

e 29e 20

e 22

e 25

Figure 22.

of 120 and 240 around the ten axes defined by the centers of oppositefaces (20).

The symmetries with determinant −1 (Ω−) are the compositions ofthe symmetries de Ω+ with the central symmetry. The cardinal of Ω, theorder of Ω, is, therefore, 120.

4. Permutation groups and puzzle solutions

Consider a puzzle with numbers 1, 2, . . . , n drawn on the plates. Fromnow on P denotes the set of its plates which have numbers drawn, and callit the plate set. If no confusion is possible, P will also denote the puzzleitself. Sn denotes the group of all permutations of 1, 2, . . . , n; σ ∈ Sn

means that σ is a one-to-one function σ : 1, 2, . . . , n → 1, 2, . . . , n.The identity is σ0: σ0(1) = 1, σ0(2) = 2, . . . , σ0(n) = n. The alternatinggroup, the Sn subgroup of the even permutations, is denoted by An. Ifσ1, σ2 ∈ Sn, we shall denote σ1σ2 ≡ σ1 σ2.

POLYHEDRON PUZZLES AND GROUPS 19

We shall write σ = (α1α2 · · ·αk) · · · (β1β2 · · · βl), if

σ(α1) = α2, σ(α2) = α3, . . . , σ(αk) = α1,

. . . ,

σ(β1) = β2, σ(β2) = β3, . . . , σ(βk) = β1,

where α1, α2, . . . , αk, . . . , β1, β2, . . . , βl ∈ 1, 2, . . . , n.If γ ∈ 1, 2, . . . , n \ α1, α2, . . . , αk, . . . , β1, β2, . . . , βl, then σ(γ) = γ.The permutation (α1α2 · · ·αk) is called a cyclic permutation, or a cycle

(in this case a k-cycle); k is the length of the cyclic permutation.We shall use also the group −1, 1 × Sn denoted by S±

n . If δ1, δ2 ∈−1, 1 and σ1, σ2 ∈ Sn, then (δ1, σ1)(δ2, σ2) = (δ1δ2, σ1σ2). We denoteS+

n = 1 × Sn ≡ Sn, (1, σ) ≡ σ, (−1, σ) ≡ σ−.As before E denotes the set of the polyhedron edges and F denotes the

set of the polyhedron faces. A solution of the puzzle defines a functionε : E → 1, 2, . . . , n. Denote E the set of these functions. One can saythat E is the set of the puzzle solutions.

We shall also consider the group Sn×Ω. If (σ1, ω1), (σ2, ω2) ∈ Sn×Ω,one defines the product (σ1, ω1)(σ2, ω2) = (σ1σ2, ω1ω2). We use here adifferent definition from the one in reference [4].

4.1. The plate group. Some Sn subgroups act naturally on P . Letπ ∈ P and σ ∈ Sn. Assume that a, b, c, . . . are drawn on π, by this order.Then σπ is a plate where the numbers σ(a) = a1, σ(b) = b1, σ(c) = c1, . . .are drawn replacing a, b, c, . . . (see figure 23).

s2b

1c

1

a1

bc

a

s1

a1

b1

c1c b

a

Figure 23.

Let s ∈ S±n and π ∈ P . If s ≡ s1 = (1, σ) ≡ σ, then sπ = σπ. If

s ≡ s2 = (−1, σ) ≡ σ−, then sπ is a reflection of σπ. In this last case, if

20 JORGE REZENDE

the numbers a, b, c, . . . are drawn on π, by this order, then sπ is a platewhere the numbers . . . , σ(c) = c1, σ(b) = b1, σ(a) = a1 are drawn by thisorder (see figure 23).

The plate group, GP , is the greatest subgroup of S±n that acts on P .

If s ∈ S±n and sπ ∈ P , for every π ∈ P , then s ∈ GP .

4.2. The solution group. Let ε : E → 1, 2, . . . , n be a solution ofthe puzzle. The group of this solution, Gε, is a subgroup of Sn × Ω;(σ, ω) ∈ Gε if and only if

σ ε = ε ω.

Denote by Ωε the following subgroup of Ω: ω ∈ Ωε if and only ifthere exists σ ∈ Sn such that (σ, ω) ∈ Gε. Notice that if ω ∈ Ωε thereexists only one σ ∈ Sn such that (σ, ω) ∈ Gε. From this one concludesthat ω 7→ (σ, ω) defines an isomorphism between Ωε and Gε and that(det ω, σ) ∈ GP . This defines gε : Ωε → GP , gε(ω) = (det ω, σ), which isan homomorphism of groups.

For a lot of puzzles (det ω, σ) defines completely ω. It is the caseof all puzzles considered in this article. Hence, when (det ω, σ) definescompletely ω, gε establishes an isomorphism between Ωε and gε(Ωε) ⊂GP . Denote by GPε ≡ gε(Ωε). Finally, Gε and GPε are isomorphic. Wecan identify (σ, ω) with (det ω, σ), and Gε with the subgroup GPε of GP .

4.3. Equivalent solutions. Let ε1, ε2 : E → 1, 2, . . . , n be solutionsof the puzzle. One says that these solutions are equivalent, ε1 ≈ ε2, ifthere are ω ∈ Ω and σ ∈ Sn such that

σ ε1 = ε2 ω.

Notice that (det ω, σ) ∈ GP .If σ = σ0 and det ω = 1, what distinguishes the solutions ε1 and ε2 is

only a rotational symmetry. In this case,

ε1 = ε2 ω

expresses another equivalence relation, ε1 ∼ ε2. When we make a puzzle,in practice, we do not recognize the difference between ε1 and ε2. Weshall say that they represent the same natural solution, an equivalenceclass of the relation ∼.

Figure 24 shows two solutions of the octahedron puzzle that representthe same natural solution.

Let ε, ε1, ε2 ∈ E . As ε1 ∼ ε2 and ε1 ≈ ε imply ε2 ≈ ε, one can say thatthe natural solution represented by ε1 is equivalent to ε.

This equation involving ε1 and ε2 defines an equivalence relation, anda natural solution is an equivalence class of this relation. Notice that ifε1 = ε2, then ω1 is the identity.

POLYHEDRON PUZZLES AND GROUPS 21

4

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1

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1321

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3

Figure 24.

4.4. Maximal solutions. Maximal puzzles. If ε1 and ε2 are twosolutions of a puzzle, we define the relation ε1 ε2 when Gε1 ⊃ Gε2 , orequivalently, when Ωε1 ⊃ Ωε2 . One says that a solution ε of a puzzle ismaximal if it is maximal for this partial order relation. Every polyhedronpuzzle P has a set of Ω subgroups Ω1, Ω2, · · · , Ωk which are the groupsof its maximal solutions. We call it the set of maximal groups of P .

If P1 and P2 are two puzzles of the same polyhedron, we define therelation P1 P2 when every maximal group of P2 is contained in amaximal group of P1. One says that a puzzle P is maximal if it ismaximal for this partial order relation.

4.5. Equivalent puzzles. Consider two puzzles and their plate sets, P1

and P2. One says that they are equivalent if there exists s ∈ S±n such

that the function π 7→ sπ is one-to-one between P1 and P2. We denoteP2 as sP1.

As a first example take the dodecahedron (2) puzzle and its plate setP (figure 9). If s = (1, σ) where σ is a transposition, then sP is the

22 JORGE REZENDE

plate set of an equivalent puzzle (figure 10). Note that P 6= sP . If ε isa solution of the dodecahedron (2) puzzle, then σ ε is a solution of theother puzzle.

Another example concerns the icosahedron (3) puzzle and its plate setP (figure 4). If s = (−1, σ0), where σ0 is the identity, then sP is theplate set of an equivalent puzzle (figure 5). As before P 6= sP and if ε isa solution of the icosahedron (3) puzzle, then εω, where ω is the centralsymmetry, is a solution of the other puzzle.

In practice, equivalent puzzles are viewed as the same puzzle.

4.6. Duality. Dual puzzles. If one has a solution of any puzzle P overa polyhedron, one can construct the dual of the solution over the poly-hedron dual and, therefore, have another puzzle P ′. Equivalent solutionsε1 and ε2 of the puzzle P , correspond to equivalent puzzles P ′

1 and P ′2,

as if σ ε1 = ε2 ω, then P ′2 = (det ω, σ)P ′

1.Consider a puzzle P over a polyhedron and its maximal solutions

ε1, ε2, . . . , εk and construct the puzzles P ′ ≡ P ′1, P

′2, . . . , P

′k, over the

polyhedron dual, that correspond to that solutions. Assume that thesepuzzles are equivalent. Use the same procedure with P ′ in order to pro-duce P ′′

1 , P ′′2 , . . . , P ′′

j and assume that these and P are equivalent puzzles.Then we say that P and P ′ are dual puzzles.

The icosahedron (1) and the dodecahedron (2) puzzles are dual. Thecuboctahedron (1) puzzle and the rhombic dodecahedron puzzle of ref-erence [6] are also dual.

5. From groups to polyhedron puzzles

Let S be a subgroup of S±n and h1, h2 : Ω → S be two isomorphisms.

One says that h1 and h2 represent the same natural isomorphism ifthere exists ω ∈ Ω+ such that

h1 = h2 ω.

Two natural isomorphisms, represented by h1 and h2 are said to beequivalent if there exists s ∈ S and ω ∈ Ω, such that

s h1 = h2 ω.

Consider a puzzle solution ε and its group GPε ≡ Ωε. Take ω1, ω2 ∈ Ωε.If ω1 is a rotation of order k, and ω2 is a rotation of order j, then ω1

transforms ω2 in another rotation of order j, ω3, which is ω1ω2ω−11 ≡

ω1(ω2).The isomorphism between Ωε and GPε suggests that if one wants to

translate the isometries into elements of S±n the function must be such

that if ω1 7→ s1, ω2 7→ s2, then ω3 ≡ ω1(ω2) 7→ s1s2s−11 .

POLYHEDRON PUZZLES AND GROUPS 23

Note that if s1 = (δ1, σ1), s2 = (δ2, σ2), where

σ2 = (α1α2 · · ·αl) · · · ,

then ω3 ≡ ω1(ω2) 7−→ s3 = (δ2, σ3), with

σ3 = (σ1(α1) σ1(α2) · · ·σ1(αl)) · · · .

We assign to every semiaxis of order k (≡ ω) a k-cycle σ = (α1α2 · · ·αk),so that to the counter clock-wise rotation of 2π

k, ω, corresponds the per-

mutation σ. This association must be coherent in the sense that it gen-erates a group isomorphism.

In the cases we are interested in (the tetrahedron, the octahedron andthe icosahedron), it is enough to make the association to two neighborsemiaxes. In these cases the semiaxes are defined by the vertices, theedges (the middle point of each edge), the faces (the center of each face),and have their origin at the polyhedron center.

If we assign to ω and ω1 (two neighbor semiaxes) the cycles σ and σ1,then to the semiaxis ω(ω1) = ωω1ω

−1 we must assign σσ1σ−1. When

σ1 = (α1α2 · · ·αk), then σσ1σ−1 = (σ(α1)σ(α2) · · ·σ(αk)).

In this section we use group theory in order to find puzzles, for a givenpolyhedron, such as, for example, maximal puzzles. These are importantexamples, but others could be given.

To avoid ambiguities, in the puzzles we give in the following, all theedges have numbers, and we use the numbers 1, 2, 3, 4 in the tetrahedronand octahedron (cube) groups cases, and the numbers 1, 2, 3, 4, 5 in theicosahedron (dodecahedron) group case.

5.1. The tetrahedron group. In the tetrahedron a vertex representsa semiaxis of order 3. Associate to a given vertex the permutation (123).It is not difficult to see that the only natural possibility is the one rep-resented in the l.h.s. in figure 25. Something similar happens with per-mutation (132) and the result is in the r.h.s. of the same figure. Hence,there are two natural isomorphisms for the tetrahedron group (see figure25) which are equivalent. This equivalence can be done, for example, byany transposition and an element of Ω+. In this case

S = (1 × A4) ∪ (−1 × (S4 \ A4)).

5.1.1. Cube puzzles. Figure 26 shows the two solutions of the cube (2)puzzle, which represent even better than figure 25 the tetrahedron group.

24 JORGE REZENDE

(13)(24)

(12)(34)

(14)(23)

(14)(23)

(12)(34)

(13)(24)

(14)(23)

(12)(34)

(13)(24)

(13)(24)

(12)(34)

(14)(23)

(132)

(143)

(234)

(123)

(243)

(134)

(124)(142)

Figure 25.

2

3

1

2

4

1

3

1

3

1

2

2

4

1

4

1

2

2

4

4

4

3

3

3

Figure 26.

5.1.2. Octahedron puzzles. Figure 27 represents solutions of two equiv-alent puzzles, that we call the octahedron (2) puzzle. These solutionscorrespond to the natural tetrahedron isomorphisms. Note that

S = (1 × A4) ∪ (−1 × (S4 \ A4))

is precisely the plate group.

5.1.3. Cuboctahedron puzzles. Figure 28 represents solutions of two equiv-alent puzzles, that we call the cuboctahedron (3) puzzle. These solutionscorrespond to the natural tetrahedron isomorphisms. Note that, as inthe octahedron case, S is precisely the plate group.

5.2. The octahedron (cube) group. There is only one natural iso-morphism for the octahedron (cube) group (see figure 29). In this case

S = −1, 1 × S4.

POLYHEDRON PUZZLES AND GROUPS 25

3 1

2

3 1

1

22

34 4

44 4

4

2 2

3

31

11 3

2

Figure 27.

2

2

2

1

3

4

3

4

3

3

2

1

44

4

2

1

1

3

2

3

1

1

43

3

1

3

4

1

2

1

4

4

1

4

23

1

3

2

3

2

1

4

2

4

2

Figure 28.

5.2.1. Cube and octahedron puzzles. If one looks for puzzles with solu-tions that have as their group Ω+ ≡ S4, we find the cube (1) and theoctahedron (1) puzzles. Those solutions are maximal. This means thatthere are no puzzles where S = −1, 1 × S4 is a solution group.

5.2.2. Cuboctahedron puzzles. The cuboctahedron (1) puzzle has only amaximal natural solution which is shown in the l.h.s. of figure 30. As itsgroup is precisely S = −1, 1 × S4, this solution is excellent in order torepresent the octahedron (cube) group. This puzzle and this solution iscompletely rediscovered using S.

In the r.h.s. of figure 30 is represented the solution of the cubocta-hedron (2) puzzle which has also S as group. These two are the onlypossibilities for cuboctahedron puzzles with S as maximal group.

26 JORGE REZENDE

(13)

(34)(23)(24)

(12) (14)

(24)

(12)(14)(234)

(132)(143)

(1234)

(1432)

(1423)

(1243)

(1342)

(1324)

(142)

Figure 29.

4

4

3

3

1

4

3

1

3

22

2

1

2

1

2

4

3

4

3

1

34

1

2

3

1

2

1

1

3

4

44

1

4

2

3

3

2

2

1

32

2

1

4

4

Figure 30.

Note that the rhombic dodecahedron puzzle of Reference [6] has onlyone natural solution which is dual of the l.h.s. solution in figure 30. Thisproperty was crucial in choosing the puzzle.

POLYHEDRON PUZZLES AND GROUPS 27

5.2.3. Rhombicuboctahedron puzzles. Consider a rhombicuboctahedron squareface which has no common edge with a triangular face. Associate to thisface the permutation (1234). To a neighbor triangular face one mustassign the permutation (132), as it is shown in figure 31, if one wantsto have a puzzle with a maximal group containing S4. Figure 31 showssixteen possibilities and none of them has a central symmetry. All ofthem are maximal solutions of puzzles with S4 as a group.

There are six puzzles. All the puzzles have the six plates of figure 6.Four of them have the eight plates represented in figure 2 (see the firstthree rows in figure 31) and the other two have eight triangular platesof the type aaa (see the last row in figure 31). The first puzzle has theremaining square plates of the type abac (6 solutions). The second puzzlehas the remaining square plates of the type aabb (2 solutions). The thirdpuzzle has the remaining square plates of the type aaaa (2 solutions).The forth puzzle has the remaining square plates of the type abab (2solutions). The fifth puzzle has the remaining square plates of the typeabac (2 solutions). The sixth puzzle has the remaining square plates ofthe type aabb (2 solutions).

The first puzzle is the one presented in Reference [6]. It was chosenbecause two of its six solutions with S4 as a maximal group are dual oftwo maximal solutions of the deltoidal icositetrahedron puzzle presentedin the same Reference [6].

5.2.4. Snub cube puzzles. As in the rhombicuboctahedron situation thereare sixteen cases, that are shown in figure 32, if one wants to have puzzleswith S4 as a maximal group (the largest one possible, as the snub cubehas no central symmetry).

In the figure the edge marked with x must have a 2 or a 3 in each ofthe sixteen cases. Hence, there are, in fact, thirty two possibilities.

There are five puzzles. All the puzzles have the six plates of figure 6.Three have the eight plates represented in figure 2 and two have eighttriangular plates of the type aaa.

The first puzzle has the remaining triangular plates of the type abc (8solutions). The second puzzle has the remaining triangular plates of thetype aab (14 solutions). The third puzzle has the remaining triangularplates of the type aaa (2 solutions). These are the first three puzzles (thefirst three columns in figure 32).

The last two puzzles (the last column in figure 32) are as follows.The fourth puzzle has the remaining triangular plates of the type abc (4solutions). The fifth puzzle has the remaining triangular plates of thetype aab (4 solutions).

28 JORGE REZENDE

44

13

42

44

32

13

22

3

34

44

44

21

32

21

13

42

43

2

214

3

1

34

44

21

23

32

41

43

1

13

32

14

22

1

32

3

12

14

21

43

43

4

32

31

14

12

1

12

21

23

2

43

3

23

21

33

14

42

1

11

21

13

2

32

24

3

14

13

21

43

2

32

14

3

14

32

44

3

14

14

(143)

(142)

(243)

(132)

(23)

(12)

(14)(1234)

(34)

Figure 31.

5.3. The icosahedron (dodecahedron) group. In the icosahedronthe center of a face represents a semiaxis of order 3. Associate to a givenface the permutation (123) as it is shown in figure 33. Then there are 8possibilities for the vertices of this face, but only two of them, the greenones in the figure, are coherent in the sense that they generate a groupisomorphism.

There are two natural isomorphisms for the icosahedron (dodecahe-dron) group (see figure 34) which are equivalent. This equivalence ismade, for example, by any transposition and an element of Ω+. In thiscase

S = −1, 1 × A5.

5.3.1. Icosahedron and dodecahedron puzzles. Using this isomorphism onecan recover the icosahedron (1) and the dodecahedron (2) puzzles, whichhave precisely S as maximal group (see Reference [4]).

5.3.2. Icosidodecahedron puzzles. For the icosidodecahedron there is onlyone puzzle with S as maximal group (in fact, there are two equivalent

POLYHEDRON PUZZLES AND GROUPS 29

4242

23

3

23

42

3

31

1

1

31

23

1

42

31

14

4

4

1 1

14

31

4

23

1 4

3 1

34

2

2

1 1

44

1

14

2

31

4 2

14

3 4

1

2 3 32

2

3

11

44

1

3

44

14

2 3

4 2

3 4

4

23 3

2

2

42 2

1

4

1

34

1 1

44

1

4

41

21

23

34

2

2 3 32

2

42 2

1

4

1

41

13

3

3

34

12

44

1

1

42

32

3

2 3 32

2

4 2 21

4

1

12

13

3

3

41

23

2

34

12

2

43

43

42 2

1

4

1

23

13

3

3

12

34

2

41

23

12

14

13

3

3

23

41

2

12

34

23

2x

23

41

34

41

Figure 32.

puzzles). Figure 35 shows the possibilities if one wants to have A5 asa group (two puzzles). Only one of them, the green one, has a centralsymmetry.

References

[1] Jorge Rezende: Puzzles numericos poliedricos. Diario da Republica, III Serie,no 300, p. 22647 (1988).

[2] Jorge Rezende: Jogos com poliedros e permutacoes. Bol. Soc. Port. Mat. 43, 105–124 (2000).http://gfm.cii.fc.ul.pt/Members/JR.pt PT.html

[3] Jorge Rezende: Puzzles com poliedros e numeros. SPM: Lisboa 2001.http://www.spm.pt/SPM/lojaSPM.html

[4] Jorge Rezende: Puzzles with polyhedra and permutation groups.http://gfm.cii.fc.ul.pt/Members/jr puzzles-poly-perm.pdf

30 JORGE REZENDE

(123)(123)

(123)

(123)

(123)

(123)

(123)(123)

(12534)

(14235)

(15243)

(12435)

(15234)

(14253)

(13245)

(13452)

(14532)

(13254)

(13542)

(15432)

(12345)

(14523)

(12453)

(12354)

(15423)

(12543)

(14325)

(13524)

(15342)

(15324)

(13425)

(14352)

Figure 33.

(12)(54)

(14)(35)

(25)(34)

(15)(23)

(13)(54)

(14)(23)

(12)(35)

(15)(24)

(12)(34)

(24)(35)

(13)(25)

(13)(25)(14)(23)

(15)(34)

(14)(25)

(12)(45)

(15)(34)

(24)(35)

(14)(23)

(13)(45)

(15)(23)

(12)(34)

(14)(25)

(12)(35)

(25)(34)

(13)(24)

(13)(24)(15)(23)

(14)(35)

(15)(24)

(14)(35)(15)(34)

(14)(25)(15)(24)

(14)(23)(15)(23)

(145)(154)

(142)

(234)

(135)

(152)

(235)

(134)

(143)

(245)

(123)

(153)

(254)

(123)

(354)(345)

(235)

(152)

(234)

(142)

(13245)

(15342)

(13524)

(12543)

(14325)

(13254)

(14352)

(13425)

(12453)

(15324)

(12354)(12345)(14253)(15243)

(15234)(14235)

(14532)(15432)

Figure 34.

[5] Jorge Rezende: On the Puzzles with polyhedra and numbers (2001).http://gfm.cii.fc.ul.pt/Members/jr poliedros-puzzles en.pdf

[6] Jorge Rezende: Three puzzles with polyhedra and numbers.http://gfm.cii.fc.ul.pt/Members/jr three-puzzles en.pdf

[7] http://gfm.cii.fc.ul.pt/Members/jr icosaedr4.pdf[8] Jorge Rezende: Some mathematical activities with polyhedron puzzles.

POLYHEDRON PUZZLES AND GROUPS 31

2

53

14

44

2

43

15

55

3

14

25

55

3

15

24

44

1

42

12

53

4

25

1

52

23

31

5

31

12

43

5

24

31

42

23

31

4

31

31

52

(123)

(13524)

(123)

(13425)

Figure 35.

Grupo de Fısica-Matematica da Universidade de Lisboa, Av. Prof.Gama Pinto 2, P-1649-003 Lisboa, Portugal, e Departamento de Matematica,Faculdade de Ciencias da Universidade de Lisboa

E-mail address: rezende@cii.fc.ul.ptURL: http://gfm.cii.fc.ul.pt/Members/JR.pt PT.html