PUZZLES WITH POLYHEDRA AND PERMUTATION GROUPS

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1 PUZZLES WITH POLYHEDRA AND PERMUTATION GROUPS JORGE REZENDE. Introduction Consider a polyhedron. For example, a platonic, an arquemidean, or a dual of an arquemidean polyhedron. Construct flat polygonal plates in the 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 in figure. Some of the plates, or all, can have numbers on both faces. We call these plates, two-faced plates. In this case, they have the same number adjacent to the same side. In figure, c and d are the two faces of a pentagonal two-faced plate example. Now the game is to put the plates over the polyhedron faces in such a way that the two numbers near each polyhedron edge are equal. If there is at least one solution for this puzzle one says that we have a polyhedron puzzle with numbers. These puzzles are a tool in teaching and learning mathematics, and a source of examples and exercises. They may be used in various fields such as elementary group theory and computational geometry. It is obvious that not all puzzles are interesting. But, amazingly or not, there are natural ways of constructing interesting puzzles. From now on, assume that the numbers belong to the set {,,..., n}, and that all the numbers are used. If we have plate faces which have the shape of a regular polygon with j sides, one can ask how many possible ways ν are there to draw the numbers,,..., n, without repeating them on each plate face. The answer is ( ) n ν = (j )! = j n! (n j)!j. For n = and j = (equilateral triangle), this gives ν =. The Mathematical Physics Group is supported by the Portuguese Ministry for Science and Technology. 000 Mathematics Subject Classification. B0, B, 0B0, 0B0.

2 JORGE REZENDE For n = and j =, this gives ν = 8. Note that 8 is precisely the number of the octahedron faces. We naturally call the related puzzle, the octahedron puzzle. a b c d Figure. For n = and j =, this gives ν = 0. Note that 0 is precisely the number of the icosahedron faces. We call the related puzzle, the icosahedron first puzzle (or icosahedron ()). For n = and j =, this gives ν = 0. Note that 0 is the double of the number of the icosahedron faces. Construct different plates with the numbers written on both faces. This gives 0 plates. We call the related puzzle, the icosahedron second puzzle (or icosahedron ()). Consider again n = and j =. Construct different plates with the numbers written only on one face, but in such a way that the numbers grow if we read them, beginning with the minimum, counter clock-wise. See an example in figure b. This gives 0 plates. We call the related puzzle, the icosahedron third puzzle (or icosahedron ()). For n = and j = (square), this gives ν =. Note that is precisely the number of the cube faces. We naturally call the related puzzle, the cube puzzle. For n = and j = (regular pentagon), this gives ν =. Note that is precisely the double of the number of the dodecahedron faces. Construct different plates with the numbers written on both faces. This gives plates. In figure, c and d show the two faces of a plate of this type. We call the related puzzle, the dodecahedron first puzzle (or dodecahedron ()).

3 PUZZLES WITH POLYHEDRA AND PERMUTATION GROUPS Let again n = and j =. Construct different plates with the numbers written only on one face, but in such a way that the numbers read counter clock-wise, abcd, are such that abcd form an even permutation. This gives plates. In figure, c and d show two plates of this type. We call the related puzzle, the dodecahedron second puzzle (or dodecahedron ()). Consider n =. With j =, one has ν = 8. With j =, one has ν =. Note that 8 is precisely the number of the cuboctahedron triangular faces and is precisely the number of its square faces. This is an example of an interesting puzzle using an arquemidean polyhedron. Take now a deltoidal icositetrahedron. It has deltoidal faces. If we have plates which have the deltoidal shape the number of possible different ways to draw the numbers,,,, without repeating them on each plate is precisely. This an example of an interesting puzzle using a dual of an arquemidean polyhedron. These are simple examples of polyhedron puzzles with numbers, which are enough in order to understand the following sections. There are, obviously, others. For more examples see Reference [], which is a development of Reference []. Reference [] is a collection of some of these puzzles paper models.. Polyhedron symmetries and permutation groups Consider a puzzle with numbers,,..., n drawn on the plates. From now on P denotes the set of its plate faces which have numbers drawn, and call it the plate set. S n denotes the group of all permutations of {,,..., n}; σ S n means that σ is a one-to-one function σ : {,,..., n} {,,..., n}. The identity is σ 0 : σ 0 () =, σ 0 () =,..., σ 0 (n) = n. The alternating group, the S n subgroup of the even permutations, is denoted by A n. We shall use also the group {, } S n denoted by S ± n. If δ, δ {, } and σ, σ S n, then (δ, σ ) (δ, σ ) = (δ δ, σ σ ). We denote S + n = {} S n S n. If Λ is a set, then Λ denotes its cardinal. Hence, if G is group, G denotes its order. From now on E denotes the set of the polyhedron edges and F denotes the set of the polyhedron faces. A solution of the puzzle defines a function ε : E {,,..., n}. Denote E the set of these functions. One can say that E is the set of the puzzle solutions. Denote Ω the group of the polyhedron symmetries. Every symmetry ω Ω induces two bijections, that we shall also denote ω, whenever there is no confusion possible: ω : E E and ω : F F. Denote also

4 JORGE REZENDE Ω {ω : E E} {ω : F F }, the two sets of these functions. One can say that each one of these two sets Ω is the set of the polyhedron symmetries. With the composition of functions each one of these two sets Ω forms a group that is isomorphic to the group of the polyhedron symmetries. If ω, ω Ω, we shall denote ω ω ω ω. If Ω is a subgroup of Ω, then Ω acts naturally on the face set, F : for ω Ω and ϕ F, one defines the action ωϕ = ω (ϕ). In the following Ω + denotes the subgroup of Ω of the symmetries with determinant. We shall also consider the group S n Ω. If (σ, ω ), (σ, ω ) S n Ω, one defines the product (σ, ω ) (σ, ω ) = (σ σ, ω ω ). On group theory we follow essentially references [] and []... The plate group. Some S n subgroups act naturally on P. Let π P and σ S n. Assume that a, b, c,... are drawn on π, by this order. Then σπ is the plate face where the numbers σ (a), σ (b), σ (c),... are drawn replacing a, b, c,.... Let s S n ± and π P. If s = (, σ) σ, then sπ = σπ. If s = (, σ) σ, then sπ is a reflection of σπ. In this last case, if the numbers a, b, c,... are drawn on π, by this order, then sπ is a plate face where the numbers..., σ (c), σ (b), σ (a) are drawn by this order. The plate group, G P, is the greatest subgroup of S n ± that acts on P. If s S n ± and sπ P, for every π P, then s G P... The solution group. Let ε : E {,,..., n} be a solution of the puzzle. The group of this solution, G ε, is a subgroup of S n Ω; (σ, ω) G ε if and only if σ ε ω = ε. Denote Ω ε the following subgroup of Ω: ω Ω ε if and only if there exists σ S n such that (σ, ω) G ε. Note that if ω Ω ε there exists only one σ S n such that (σ, ω) G ε. From this one concludes that ω (σ, ω) defines an isomorphism between Ω ε and G ε. On the other hand (det ω, σ) G P. This defines g ε : Ω ε G P, g ε (ω) = (det ω, σ), which is an homomorphism of groups. If all the plates are different, as is the case in the given examples and as we shall assume from now on, (det ω, σ) defines completely ω. Hence, g ε establishes an isomorphism between Ω ε and g ε (Ω ε ) G P. Denote G P ε g ε (Ω ε ). Finally, G ε and G P ε are isomorphic. We can identify (σ, ω) with (det ω, σ), and G ε with the subgroup G P ε of G P. One can define a function u ε : F P, that associates to every face ϕ the plate face π that ε puts in ϕ. Define also P ε = u ε (F ). Note that

5 PUZZLES WITH POLYHEDRA AND PERMUTATION GROUPS P ε = F and that G P ε acts on P ε. Then one has for every ϕ F. (det ω, σ) (u ε ω (ϕ)) = u ε (ϕ),.. Equivalent solutions. Let ε, ε : E {,,..., n} be solutions of the puzzle. One says that these solutions are equivalent, ε ε, if there are ω Ω and σ S n, such that Note that (det ω, σ ) G P. Then, σ ε ω = ε. (σ, ω) ( σ σσ, ω ωω ) = (σ, ω ) (σ, ω) (σ, ω ) defines an isomorphism between G ε one has that s σ sσ and G ε. As det ( ω ωω ) = det ω, defines an isomorphism between G P ε and G P ε. If σ = σ 0 and det ω =, what distinguishes the solutions ε and ε is only a rotational symmetry. In this case ε ω = ε expresses another equivalence relation, ε ε. When we make a puzzle, in practice, we do not recognize the difference between ε and ε. We shall say that they represent the same natural solution, an equivalence class of the relation. Let ε, ε, ε E. As ε ε and ε ε, implies ε ε, one can say that the natural solution represented by ε is equivalent to ε. For ε E, represent by [ε] the set of natural solutions equivalent to ε. Choose now ω Ω, such that det ω =. For ε E and s = (δ, σ) G P, denote ε s = σ ε ω, where ω is the identity if δ = and ω = ω if δ =. The set {ε s : s G P } includes representatives of all natural solutions equivalent to ε. Then [ε] = G P G P ε. The cardinal of all the natural solutions is then given by [ε] G P G P ε, where the sum is extended to all different equivalence classes [ε].

6 JORGE REZENDE.. Equivalent actions. Consider two groups H and H that act on two sets K and K. For ρ j H j and x j K j, ρ j x j ( K j ) represents the action of ρ j on x j. One says that the action of H on K is equivalent to the action of H on K if there exists an isomorphism ξ : H H and a one-to-one function ζ : K K, such that for every ρ H and every x K one has ξ (ρ) ζ (x) = ζ (ρx). Note that if the action of H on K is equivalent to the action of H on K, and H is a subgroup of H, then the action of H on K is equivalent to the action of H = ξ (H ) on K. Assume that the action of H on K is equivalent to the action of H on K. Let K and K be the sets of the equivalence classes of such actions. Then there exists a bijection Z : K K, such that for every ρ H and every X K one has and ξ (ρ) Z (X) = Z (ρx) X = Z (X). These two conditions are necessary in order to have equivalent actions. Example.. If ε and ε are equivalent solutions of the polyhedron puzzle, the actions of G P ε and G P ε on P are equivalent. Let ξ (s) = σ sσ and ζ (π) = (det ω, σ ) π. Then ξ (s) ζ (π) = σ sσ (det ω, σ ) π = (det ω, σ ) sπ = ζ (sπ). Example.. If ε and ε are equivalent solutions of the polyhedron puzzle, the actions of Ω ε and Ω ε on F are equivalent. Let ξ (ω) = ω ωω and ζ (ϕ) = ω ϕ. Then ξ (ω) ζ (ϕ) = ω ωω ω ϕ = ω ωϕ = ζ (ωϕ). Example.. If ε is a solution of the polyhedron puzzle, the actions of Ω ε on F and of G P ε on P ε = u ε (P ) are equivalent. Let ξ (ω) = g ε (ω ) = (det ω, σ ) and ζ (ϕ) = u ε (ϕ). Then ξ (ω) ζ (ϕ) = ( det ω, σ ) u ε (ϕ) = (u ε ω (ϕ)) = ζ (ωϕ). Denote F = {F, F,..., F k } and P = {P, P,..., P k } be the sets of the equivalence classes of such actions. Then u ε induces a one-to-one function U : F P, U (F j ) = P j, F j = P j, and ( det ω, σ ) U (F j ) = U (ωf j ), for every j =,,..., k, ω Ω ε.

7 PUZZLES WITH POLYHEDRA AND PERMUTATION GROUPS 7 From now on we shall deal only with the action of a subgroup Ω of Ω on F, or the action of a subgroup G of G P on P. Hence, when we say that two groups H and H are equivalent (H H ), we are talking about these actions.. Examples Some of the examples we give in this section can easily be studied directly. It is the case of the cube, the octahedron and the dodecahedron () puzzles. We give also results for the dodecahedron (), icosahedron () and () puzzles. These results were obtained mostly with a computer. Similar results for the icosahedron () puzzle are left to the reader. There are good reasons for presenting results for these two icosahedron puzzles. The icosahedron () puzzle is mathematically rich, has a lot of natural solutions (over one million!). On the other hand, the icosahedron () puzzle is also very instructive because it has few natural solutions ( only ), which means that it is difficult to do it without the help of a computer... The cube puzzle. In this case there is only one equivalence class. G ε Ω + S. G ε =. G P = 8. If one puts a plate on a face then one has different possibilities for the other plates ( natural solutions): 8/. These two different possibilities and the equivalences G ε Ω + S can be used to translate the cube symmetries into permutations, in the same way as we shall do later with the icosahedron () puzzle and its canonical solution... The octahedron puzzle. There three equivalence classes. One can distinguish them in the following way. Take a solution. On every vertex of the octahedron, note the numbers that correspond to its four edges. There are two possibilities for a vertex: a) four distinct numbers; b) three distinct numbers, with one of them repeated. In the solution, count the number of vertices where the situation a) happens. They can be, or 0, that distinguish the three equivalence classes. The first class group is of order. The second class group is of order 8. The third class group is of order. As G P = 8, one has that if one puts a plate on a face then one has different possibilities for the other plates ( natural solutions): Although the cube and the octahedron are dual polyhedra, puzzles are not. Solutions can be dual. The first class of the octahedron puzzle is dual of the cube puzzle class.

8 8 JORGE REZENDE.. The dodecahedron () puzzle. This puzzle has three equivalence classes that we can distinguish in the following way. Consider a solution. On every dodecahedron vertex note the numbers that are on the edges around the vertex. There are 0 possibilities, but not all of them belong to the solution. There are some repetitions: 7 or. The solutions that have 7 repetitions on the vertices are equivalent. The solutions that have repetitions on the vertices belong to different equivalence classes. One of these classes has the repetitions on opposite vertices. The other has the repetitions on vertices that belong to the same edges. The first class group is of order 8. The second class group is of order. The third class group is of order. As G P = 0, one has that if one puts a plate on a face then one has ( 0 different possibilities for the other plates (0 natural solutions): ) The dodecahedron () puzzle. This puzzle has equivalence classes that can be distinguished in the following form. Consider two opposite dodecahedron edges. There are other four that are orthogonal to these two. The six edges are over the faces of a virtual cube where the dodecahedron is inscribed. There are five such cubes. The first equivalence class has the same number associated to the edges that belong to the faces of each cube. The second equivalence class has the same number associated to the edges that belong to the faces of one of the fives cubes. The first class group is of order 0 and the second class group is of order. As G P = 0, one has that if one puts a plate on a face then one has different possibilities for the other plates ( natural solutions): The icosahedron () puzzle. The icosahedron () puzzle has 9 equivalence classes: have groups of order, have groups of order, have groups of order, has a group of order, have groups of order, 0 have groups of order, has a group of order 8, have groups of order 0, have groups of order, have groups of order, has a group of order 0. As G P = 0, one has that once one puts a plate over a face there are 0 different possibilities (0 natural solutions): 0 ( There are two possible natural solutions with a group of order 0. One of them, that we call canonical solution, is dual of the dodecahedron () solution with a group of order 0. The group of these two solutions ).

9 PUZZLES WITH POLYHEDRA AND PERMUTATION GROUPS 9 z x y Figure. is equivalent to the icosahedron group: G ε Ω {, } A. The canonical solution is represented in figure... The icosahedron () puzzle. This puzzle has 97 equivalence classes: 90 have groups of order, and 7 have groups of order. As G P =, one has that, once one puts a plate over a face there are different possibilities ( natural solutions): Note that the actions of all these 7 groups of order are equivalent. Figure shows representatives of all these 7 equivalence classes.. More on icosahedron puzzles As we have already seen there is a natural solution of the icosahedron () puzzle which is dual of the dodecahedron () natural solution with a group of order 0 (see figure ). This group is equivalent to the icosahedron group Ω: G ε Ω {, } A. As this solution is very easy to construct, one can use it in order to translate in terms of

10 0 JORGE REZENDE Figure. {, } A everything that happens in Ω. For example, all subgroups of Ω are equivalent to the corresponding subgroups of {, } A... Subgroups of the icosahedron group. The icosahedron group Ω has different equivalence classes of subgroups: of order ({σ 0 }); of order ; of order ; of order ; of order ; of order ; of order 8; of order 0; of order ; of order 0; of order ; of order 0; of order 0 (Ω {, } A ). In the following a, b, c, d, e {,,,, } are different numbers, and in angle brackets we give the number of equivalence classes of solutions in the icosahedron () puzzle.... Groups of order. The equivalence classes are generated by (ab) (cd) 8, (, (ab) (cd)) and (, σ 0 ). Look at figure. The rotation around the z-axis by an angle of 80, corresponds to the permutation () (). This rotation belongs to the first equivalence class. The reflection using the xy-plane (orthogonal to the z-axis) as a mirror, corresponds to (, () ()). This reflection belongs to the second equivalence class. The third equivalence class corresponds to the central symmetry (x, y, z) (x, y, z).

11 PUZZLES WITH POLYHEDRA AND PERMUTATION GROUPS... Groups of order. The unique equivalence class is generated by (abc). In figure, the rotations around the axis x = y = z by angles multiple of 0 correspond to group generated by ().... Groups of order. Let σ = (ab) (cd) and σ = (ac) (bd). Note that σ σ = σ σ = (ad) (bc). The equivalence classes are generated by σ and σ 0, σ and (, σ 0 ), σ and (, σ ) 0, respectively.... Groups of order. The unique equivalence class is generated by (abcde). In figure, the rotations around the axis x = 0, z = + y (this is the z -axis in figure ), by angles multiple of 7, c orrespond to group generated by ().... Groups of order. Let σ = (ab) (cd) and σ = (ae) (cd). The equivalence classes are generated by σ and σ ( S ), (, σ ) and (, σ ) 0, (abc) and (, σ 0 ) 7, respectively.... Groups of order 8. Let σ = (ab) (cd) and σ = (ac) (bd). As before, note that σ σ = σ σ = (ad) (bc). The unique equivalence class is generated by σ and σ and (, σ 0 )...7. Groups of order 0. Let σ = (ab) (cd) and σ = (ac) (be). The equivalence classes are generated by σ and σ ( D, the dihedral group of order 0), (, σ ) and (, σ ) 0, (abcde) and (, σ 0 ), respectively...8. Groups of order. Let σ = (ab) (cd) and σ = (acd). The equivalence classes are generated by σ and σ ( A ), S and (, σ 0 ) ( {, } S ) 0, respectively...9. Groups of order 0. The unique equivalence class is generated by D and (, σ 0 ) ( {, } D ) Groups of order. The unique equivalence class is generated by A and (, σ 0 ) ( {, } A ).... Groups of order 0. The unique equivalence class is A Groups of order 0. The unique equivalence class is {, } A Ω.

12 JORGE REZENDE.. Subgroups of plate groups. All the subgroups of the icosahedron group have cyclic groups of order, or as generators. Let us look at the actions on the faces of such cyclic groups. The first equivalence class of the order groups has 0 orbits with elements and the determinant of the generator is : (, 0 ). The second equivalence class has 8 orbits with elements, orbits with element and the determinant of the generator is : (, 8 + ). The third equivalence class has 0 orbits with elements and the determinant of the generator is : (, 0 ). The equivalence class of the order groups has orbits with elements, orbits with element and the determinant of the generator is : (, + ). The equivalence class of the order groups has orbits with elements and the determinant of the generator is : (, ).... The icosahedron () puzzle. The plate group is G P = {, } S. The only cyclic groups of G P whose actions on the plates have orbits of the type (, 0 ), (, 8 + ), (, 0 ), (, + ) and (, ) are those generated, precisely, by (, (ab) (cd)), (, (ab) (cd)), (, σ 0 ), (, (abc)) and (, (abcde)).... The icosahedron () puzzle. Let σ = () and σ = () () (). The plate group, G P, is generated by (, σ ) and (, σ ). The only cyclic groups of G P whose actions on the plates have orbits of the type (, 0 ), (, 8 + ), (, 0 ) and (, + ) are generated by (, σ), ( ) ( ), σσ j,, σ k σ, (, σ ), j =,,, k = 0,,. There are no subgroups of order. It is easy to see directly that there are no solutions corresponding to the subgroups of order. The solutions corresponding to subgroups of order, are only those related to the generators ( ), σσ k, k = 0,,... Finding solutions of a puzzle with a prescribed group. A way to find a solution for a puzzle is to use the group relations between the polyhedron and the plate groups. Take, for example, the icosahedron () puzzle and the third equivalence class for groups of order 0. This class is generated by an element σ of order and (, σ 0 ). The numbers on the edges must respect the central symmetry and a rotational symmetry group of order. Let σ = (). In figure, that illustrates this example, a, b {,,,, }, and a j = σ j (a), b j = σ j (b), j =,,,. The numbers a and b must be chosen so that a b, a σ (b) = b, b σ (a) = a. These three conditions give 9 possibilities, but only of them correspond to solutions of the puzzle: (a, b) = (, ), (, ), (, ), (, ), (, ). The third one is the canonical

13 PUZZLES WITH POLYHEDRA AND PERMUTATION GROUPS z a a b b a b b b b a a Figure. solution. The first and the fourth ones belong to the same equivalence class. The same happens with the second and the fifth ones. These last two classes are the ones already listed. References [] Jorge Rezende: Puzzles numéricos poliédricos. Diário da República, III Série, n o 00, p. 7 (988). [] Jorge Rezende: Jogos com poliedros e permutações. Bol. Soc. Port. Mat., 0- (000). PT.html [] Jorge Rezende: Puzzles com poliedros e números. SPM: Lisboa 00. [] M. A. Armstrong: Groups and Symmetry. Berlin: Springer-Verlag 988. [] R. A. Wilson, R. A. Parker and J. N. Bray: ATLAS of Finite Group Representations. Grupo de Física-Matemática da Universidade de Lisboa, Av. Prof. Gama Pinto, 9-00 Lisboa, Portugal, e Departamento de Matemática, Faculdade de Ciências da Universidade de Lisboa address: URL: PT.html

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