https://hemerotecadigital.uanl.mx/files/original/408/19943/Celerinet_2020_Ano_8_No_1_Enero-Junio.ocr.pdf 350edfc263a412ff1ffefc8a2b3038ac PDF Text Text m :ll ro ' :íl ra z ¡g MATHEMATICS, PHYSICS, COMPUTER SCIENCE, ASTROPHYSICSF C F M . (l)UANL UMVERSlDAD A~7'ÓNO"A DE NUE vo LEÓN FAnHAI) DH1B,r1,\S FISlm M,,TFM,\TlC\S �lng. Rogelio Guillermo Garza Rivera Rector Dr. Santos Guzmán López Secretario General M.A. Emilia Edith Vásquez Farías Academic Secretary Dr. Celso José Garza Acuña Secretary of Cultural Affairs Lic. Antonio Ramos Revillas Publications Director Dr. Atilano Martínez Huerta Director of the Facultad de Ciencias Físico Matemáticas Maximiliano Sánchez Garza Jordi Andrés Martínez Álvarez Collaborators Dr. Álvaro Eduardo Cordero Franco Dr. José Apolinar Loyola Rodríguez Dr. Romeo de Jesús Selvas Aguilar M.C. Azucena Yoloxóchitl Ríos Mercado Dr. Perla Marlene Viera González M.A. Alma Patricia Calderón Martínez Editorial Committee Dr. Álvaro Eduardo Cordero Franco Editor in Chief M.A. Alma Patricia Calderón Martínez Editors Cristina Elizabeth Ballesteros Santos Aide Morales Salazar Editorial Design Celerinet, Year 8, No. 1, January-June. Published on: June 30th, 2020. Gelerinet is a semestral publication edited by the Universidad Autónoma de Nuevo León, through the Facultad de Ciencias Físico Matemáticas. Address: Ave. Universidad S/N. Gd. Universitaria. San Nicolás de los Garza, Nuevo León, México, G.P. 66451. Telephone + 52 81 83294030. Fax:+ 52 81 83522954. celerinet.uanl.mx Editor in Ghief: Dr. Romeo de Jesús Selvas Aguilar. Exclusive Rights Number 04-2014-102111595700-203 licenced by the Instituto Nacional de Derechos de Autor. ISSN 2395-8359. Responsible for last update: Unidad Informática, M.A. Reyna Guadalupe Castro Medellín, Ave. Universidad S/N. Gd. Universitaria. San Nicolás de los Garza, Nuevo León, México, G.P. 66451. Last update: June 30th, 2020. The views expressed in this publication do not necessarily reflect the Editors' views. The partial or total reproduction of the contents and images in this number is forbidden. AII rights reserved © Copyright 2020 celerinet@uanl.mx �1 5 �ACADEMIC PAPERS , ARTICULOS , ACADEMICOS �ACADEMIC / MATHEMATICS Maximiliano Sánchez Garza UANL-FCFM Universidad Autónoma de Nuevo León Facultad de Ciencias Físico l\tlatemáticas San Nicolás de los Garza, Nuevo León, f-.1.[éxico A bstract Ancient Greeks were remarkable mathematicians. They made important contributions for the development of Mathematics (particularly in Geometry), solving problems that arose at the time. Unfortunately, there were three problems they couldn't solve. One of these is directly related with the number 1r and its transcendence. In tlús article, I expose sorne historical data that explains why the transcendence of 1r is important and give a quite simple but tricky proof of this result, which I took from an article written by Steve Mayer. Finally, a brief discussion about this result is exposed in section. K ey Words- Transcendental, algebraic, circle. Resumen Los antiguos griegos fueron matemáticos notables. Hicieron contribuciones importantes para el desarrollo de las matemáticas (particularmente en geometría), resolviendo problemas que surgieron en aquellos tiempos. Desafortunadamente, hubieron tres problemas que no pudieron resolver. Uno de estos está directamente relacionado con el número 1r y su trascendencia. En este artículo, expongo algunos datos históricos para explicar la importancia de la trascendencia de 1r y doy una prueba simple pero astuta de este resultado, la cual tomé de un artículo escrito por Steve fvlayer. Al final, se expone una discusión sobre este resultado. P alabras Clave- Trascendental, algebraico, círculo. Sánchez Garza, M. (2020). 1r is Transcendental. Celerine1. 8 ( 1). 1-4 1 Introduction The mathematicians from the ancient Greece were of great iinportance for the development ofGeometry. They mainly solved problems involving constructions with unmarked straightedge and compass. At that time, there were three problems that they tried to solve (but failed in the process). These problems are the following: l. Cube d:uplication. Given a cube, construct another cube with double of the volume of the original cube. 2. Angle trisection. Given an angle, construct another angle equal to one third of the original angle. 3. Squaring the circle. Given a circle, construct a square that has the sarne area as the original circle. Nearly two millenniums later, mathen1aticians started working with a new branch of rnathematics, Galois Theory, which Évaiiste Galois developed in the nineteenth century. Vlith this powerful too!, they were able to solve those three problems. They had to introduce notation and concepts to understand this t heory. For example, a complex number et is an algeb raic number over Q if there exists p(x) E Q(:i;] such that p(x) is not identically zero and p(a) = O, where Q(x] is the set of ali polynomials in one variable with rational coefficients. If et is not algebraic over Q, we say it is transcend ental over Q (see [1, pp. 5201) . With this, we have an important theorem that describes the structure of the set that contains these numbers (definitions and proof can be checked in [l, pp. 223-224, 527]). Theorem l. Let A be the set of all algebraic rrnmber-s over Q. Then A is a field. A real number et is constru ctible if, given P e IR2 with at least two points, we can construct two points Q and Q' with straightedge and cornpass from the points in P such that the distance between Q and Q' is o: (for more details, see [l , pp. 531-532]) . We initially select two points A and B from P and set d(A, B) = l. From t his, 1 a.nd O ai·e constructible (since d(A, A) = O) . The important theorem t hat helps us solve these problems is the following (see definitions and proof in [l, pp. 520, 533)) . 1 �Celerinet January - June 2020 T heorem 2. Let a E JR. If a is constructible, then the minimal ir·reducible polynomial of a over Q has degree a power of two. For the first problern, if it was solv-able, then ~ would be constructible. But the minirnurn irreducible polynornial of V2 is x 3 - 2, which clearly does not ha.ve degree a power of two. For the second problern, in general it does not ha.ve a solution. For exarnple, if the angle rneasures 60° then it does not ha.ve a solution (see [5, pp. 132-133] for a proof); but if the angle mea.sures 90° then it has a solution. That lea.ves us with t he last problern. Assume it has a solution. That mea.ns -,/ir is constructible. Frorn theorem 2 we get that .Jir has to be algebraic, and from theorem 1 we conclude that 1r has to be algebraic. So, mathernaticians focused on trying to prove the transcendence of 1r for a long t ime. In 1882, the German mathematician Car! Louis Ferdinand von Lindernann proved t hat 7f is not algebraic, thus solving a problem that t he Greeks couldn't . 2 Proof of the Transcendence of 1r First of ali, !et 1{ be a field and K[x 1 , x 2 , • . . , :i;n] be the set of ali polynornials in n variables with coefficients in K. We say that a polynomial f (x 1 , x 2 , . .• , xn) E K[x 1 , x 2 , .• . , xn] is symmet ric if, for every list ( e1 , E2 , .. . , en) such t hat {e1 , e2 , . . . , e,,} = {l , 2, 3, . . . , n } (tlús is, a permutation of the first n positive integers) , we ha.ve f (x1 , x 2 , ... , Xn) = f (xt 1 , xe2 , • • • , Xtn) . In other words, we can "exchange" variables in the polynornial and we will obtain t he same result . In order to prove the t ranscendence of 1r, we will use the following result (the proof can be seen in [3]) . Theorem 3. (N ewton) Let 1{ be a field and n a po.~itive integer. Let a 1, a2, . . . , in I<(x 1 , x2, . .. , x,.]: L O'j O'n be the f ollowing polynornials Xii Xi2 ... Xi; 1$i1 <í 2 < .. ·<i;$n Then any syrnmetric polynomial in K[x 1 , x 2 , . . • , xn ] can be written as a polynomial in a 1 , a 2 , • . . , O'n with coefficients in 1{ and this polynomials is unique. Remark l. The polynomials v ariables. O'i defined in theorern 3 are called the elemen tal sym m etric polynomials on n Now, we proceed to prove the main result. Theo rem 4. (Lindemann) 1f is transcendental over Q . Proof. To prove this result, we will assume 1r is algebraic and rea.ch a contradiction . By theorern 1, 't7f is also algebraic where i is the imaginary unit (since i is a root of the polynornial equation x 2 + 1 = O). Let P 1 (x) E Q(x] such that Pi (i1r) = O, and !et i 7f = l\'.1 , l\'.2, .. . , Ctn be the roots of Pi , where n = deg( P 1) . Reca.11 t he Euler identity ei?r + 1 = O. From this, (eº 1 + 1) (€02 + 1) '' ' (eºn + 1) Now, consider t he polynomial P2(x) = IJ = 0. (1) [x - (aí + a;)] 1$i<j$n where ea.ch root is of the form l\'.i +a; . Using Viete's Relations, we get that ea.ch coefficient of P2 is symmetric on the nurnbers l\'.i . Since ea.ch elemental symmetric polynon1ial on l\'.i is equal to a rational number (by Viete's Relations 2 �ACADEM IC / MATH EMATJCS on the polynomial P1 ), by theorem 3 we get that P 2 (x) E Q[x]. vVith a simila r idea, we can construct polynomials P3 , P4 , . •. , Pn where the polynornial Pj has as its roots ali the s ums of t he forrn ai, + ai, + · · · + ªi; with 1 < i 1 < i 2 < · · · < Íj < n . Moreover, each of these polynorrúals has rational coefficients. Consider the polynomial P1P2P3 · · · Pn, which can have t he number O as a root . Taking apart all the zero roots of tlús last polynornial and rnultiplying by a convenient integer, we get P (x) E ll1;] whose roots are ali sums of a/s. Suppose P (x) = cx1• + Cr- 1Xr- l + · · · + C¡X + Co where c0 =/= O. Observe that the roots of the polynomial P are the non-zero exponents of e in equation ( 1) when the product is expa nded. Let t hese be f]¡, fh, . . . , fJr - Equation (1) becomes ef'h + efJ• + · · · + ef1r + eº + eº + · · · + eº + 1 = 0, or r (2) k+ L é~' =O i= l where k > O is an integer. Let q be a prime nurnber (we will give another characteristic to it later) and s Consider t he function and define = rq - l. s+q F (x) =L f(il(x) í=O where ¡(i) represents the i-th derivative off and ¡<0 ) = J. Observe that IntegTating on both sides, we get e- "' F (x) - F (O) F(:r;) - e"' F (O) Subst it uting y =- = A:r:, we have F(x) - e"' F (O) Adding equation (3) for ali x fo"' e- y J(y)dy =- =-x foxex-yJ(y)dy. /4 1 e(l- A):x: j(Ax)dA. (3) = {ji and using equation (2), r r L F(fli) + kF(O) = - L i= l i =l ¡ 1 fJí e<I - A)fJ, j(AfJi)dA. (4) O This last equation will help us finish the proof. For this, consider the following lemma ( which will be proved later). Lemma l. The LHS of equation (4) is a non-zero integer for all primes q > ma.,x{k, c,c0 } . Observe that RH S of equation 4 can be written as which tends to zero as q ➔ oo since the factorial funct ion "grows faster" than every exponential function. Then, as q ➔ oo, RHS of equation (4) goes to zero but LHS <loes not goes to zero by lemrna l. This is the desired contradiction.. ■ 3 �Celerinet January • June 2020 To complete the proof of theorem 4, it remains to prove lemma l. Proof of lernma 1. First, note that if0<t<q - 2 if t=q -1 if t > q where mt is an integer. On the other hand, O qlt if t < q - 1 if t > q where lt is an integer. Then, s +q LHS = kc•c;, + q L Clt + kmt) t=q wlúch, for a pri1ne q > rnax{k, e, Co}, it is an integer different from zero (since q would not divide kc• cg) . 3 □ Conclusion As mentioned in the introduction of this article, t he transcendence of 1r is a relevant result in geometry. But this fact also motivated mathematicians to search for other numbers wlúch were not algebraic. Sorne important results have been published since then, an example is the following. Theorem 5. (Gelfond-Schneider) Let O! and fJ be algebraic nurnbers such that O! is difj'erent frorn O and 1 and /3 is irrational. Then O!fJ is transcendental. Moreover, the study of these numbers led to "Transcendental Number Theory", a new branch in rnathematics. For more information about theorem 5 and this branch, consult (4] . In summary, the techniques used by Lindemann (and other mathematicians like Weierstrass, Liouville and Cantor) for proving the transcendence of 1r ande motivated the study of the properties of these kind of numbers and, moreover, opened a new and important branch of number t heory that helps to understand them. 4 About the author Maximiliano Sánchez Garza, a student of thc Bachellor in 11lathematics in the Facultad de Ciencias Físic~ Matemáticas at the UANL. Also member of the organizing committee of the Mexican Nlathematical Olympiad since 2019. Currently doing research in number theory. E-mail: rr1axsanchezl3l2@gmail.com References [l] David S. Durnmit and Richard NI. Foote. Abstrnct Algebro. John vViley and Sons, Inc., 2003. [2] Steve Mayer. The transcendence of 1r . http : //sixthform.info/maths/files/pitrans .pdf, November 2006. [3] Hamza Elhadi S. Daoub. The Fundamental t heorem of Symmetric P olynomials. The Teaching of Mathernatics, XV:55-59, 2012. [4] Kannan Soundararajan. Transcendental Nurnber Theory. http : / /math. stanford. edu/ - ksound/Trans Notes . pdf, September 2011. [5] Felipe Zaldívar. Introducción a la Teoría de Galois. Papirhos, Ciudad de México, México, 2018. 4 �ACADEMIC / MA TREMATICS Jordi Andrés Mart ínez Álvarez UANL-FCFNI Universidad Autónoma de Nuevo León Facultad de Ciencias Físico Matemáticas San Nicolás de los Garza, Nuevo León, fviéx ico Abstract: Going bac-k to the 80s will probably take us to a very interesting moment, in which, besides ali the emerging music and T .V shows, one of the many things that may come up to our mind is the Rubik 's cube; a puzzle that not only entertained and stressed a lot of people at its time, but it also became a very interesting object for mathematicians to study. In this a.rticle, I will try to make sense of some of the properties of the Rubik's vía group theory, such as patterns, cycles, unreachable cases ancl will give an approach to find an intuitive solution based on mathematical foundations. Key wo rds: Rubik, puzzle, group, permutation. R esumen: Regresar a los años 80 probablemente nos lleve a un momento muy interesante en el que, además de toda la música emergente y los programas de televisión, una de las varias cosas que podría venir a nuestra mente es el cubo de Rubik; un rompecabezas que no sólo entretuvo y estresó a muchas personas en su momento, sino que también se convirtió en un objeto de estudio muy interesante para los matemáticos. En este artículo, intentaremos dar sentido a algunas ele las propiedades d 5_14 bo ele Rubik a través de la teoría de grupos, tales como patrones, ciclos, casos imposibles y ciaremos un tt1,;i::rcamiento para encontrar una solución que resulte intuitiva basada en fundamentos matemáticos. Pala b ras claves: Rubik, Rompecabezas, Grupos, Permutación. Martínez Álvarez, J. A. (2020) . Abstract algebra in the Rubil<s cube. Celerinet. 8 (1) . 5-15 Introduction The first thing that we may notice in a usual Rubik 's cube is its colorful surface. Each face of the cube has a different color, and since there are six of them, we have six clifferent colors, usually white, yellow, green, blue, red and orange. Also, 1,ve have every face divided itself into 9 smaJler squares. 5 �Celerinet January - June 2020 Now, in order to keep track of the pieces of t he cube, let us label each one of t he 54 small squares with the numbers from 1 to 54, as shown in Figure l. Similar ideas can be found in references [l] and [2]. 2 3 4 5 6 7 8 9 10 11 12 19 20 21 13 14 IS 22 16 17 18 25 26 27 46 47 48 49 so 51 52 53 54 23 24 Figure 1 Recall that the action of reordering elements in a set can be thought of as bijectons of a set into itself. T his type of fu nctions are known as permutations, and have a key role in t he study of group theory. Particularly, the symmetric groups will turn out to be useful for this artide. For more about group theory and permutations, see reference [3]. Thus, with this representation, we can basically t hink of all possible arrangements of the cube as a subset of the symmetric group of order 54 (S54 ) where we can swap the numbers in the squares of the same color, since there will be no noticeable difference. As we can see, this is not how the Rubik 's cube works (this representation can be t hought as allowing the stickers in the cubeto change places), so we will need to add sorne more restrictions. After doing a slightly deeper observation, just as Patrick Bossert did in 1981 when he wrote You can do the cube [4] after he disassembled his Rubik ' s cube in order to get a richer understanding of it, we might notice that a regular Rubik 's cube is composed by eight corner pieces, with t hree out of six different colors each one, twelve edge pieces, with two out of six different colors each, and six center pieces, one for each color. In this article, the center pieces ,vill be mostly ignored since they are fixed with screws to an invisible core piece, and instead of being part of the puzzle, we will think of them as the pieces t hat define the color of each face. This adds the restriction that it is only possible for colors to change places if they are in the same kind of piece. This facts allows us to reduce t he possible arrangements of the puzzle to a subset of S 48 (by ignoring center pieces) . Observe that there is no piece, neither corner, edge nor center, with two squares of the san1e color, so we can assign each square from the 54 in total to one and only one position of the solved state (as shown in Figure 2), completing the restrictions for the puzzle. 6 �ACADEMIC / MATH EMATICS 1 2 3 4 u s 6 7 8 9 10 11 17 18 19 12 L 13 20 F 21 14 IS 16 22 23 24 41 42 43 44 D 45 46 47 48 Figure 2 This representation of the solved state will be used as a reference for virtually every property studied in this article. From here, the next thing we have to analyze is t he "legal " movements of the cube. Other useful ideas a nd approaches and can be found in 11]. Figure 3: Core piece Figure 4: Edge piece 7 �Celerinet January - J une 2020 F igure 5: Corner piece Construction In a Rubik's cube, there are 6 possible 1novements, one for each face, so we will caH then1 U, D, F, B, R and L accordjng to Singmaster notation, introduced by David Singmaster [5], which a re permutat ions in S48. Hence, there is a n operation defined for them. A special kind of permutations are cycles. In [6], a cycle is defined as a permutation o in S,,, where o(a1) = a2 , o(a,2) =a3 , ... , o(aj) = llj +i for j < k, o(ak) = a 1 and o(s) = s for any s in S if s is d ifferent from ai, a,2 , ... , ~ - As a result, every permutation can be written as a product of ilisjoint cycles. From here, every movement is defined in disjoint cycle dec:omposition as follows: U := (1386)(2 5 7 4)(9 33 25 17)(10 34 26 18)(11 35 2719) D: = (41 43 48 46)(42 45 47 44)(14 22 30 38)(15 23 31 39)(16 24 32 40) , F : = (17 19 24 22)(18 21 23 20)(6 25 43 16)(7 28 42 13)(8 30 4111) B : = (33 35 40 38)(34 37 39 36)(3 9 46 32)(2 12 47 29)(1 14 48 27) R : = (25 27 32 30)(26 29 31 28)(3 38 43 19) (5 36 45 21)(8 33 48 24) L : = (9 11 16 14)(10 13 15 12)(1 17 41 40)( 4 20 44 37)(6 22 46 35). Thus, we can define the Rubik's cube group Gas: G: = (U, D, F, B, R, L} (1) Note: Here, if .tvl, N are permutations in G , MN represents doing .tvI, t hen N. In t he standard notation of composition of permutations, t his is represented by Nivl, but we will not use the standard notation. From here, we can easiJy see t hat G is a non-abelian group, since, for example, we can verify that FU~UF. In the Jangua.ge of the Ru bik 's cube, t he identity permutation represents doing no movements at ali. 8 �ACADEMJC/ MATHEMATI CS Properties The information we have just given about the 1novements of the cube will make sense of son1e of its properties. Theorem 1 The subgroups of G: (U, D), (R, L), (F , B), are abelian. Proof This can easily be observed by noticing that the permutations that represent both elements that generate each subgroup are disjoint. ■ T his fact is, physically, a consequence of opposite faces not having any pieces in common. Another important result, is that: R4=D4=f4=L4=B4=U4 (2) This is a direct consequence of the fact that every cycle in the disjoint cycle decomposition of the movements has length 4. Geometrically, tbis makes sense, since every face is squar&-shaped, and making four quarter turns to a square lea.ves it intact. Theorem 2 The order of G is 43,252,003,274,489,856,000. Proof. In orcler to count all the permutations in the cube, we shall consider a.11 possible arrangements of the eight corner pieces and the twelve edge pieces. As a first approach, we will temporarily ignore the permutations that define t he puzzle, and will only consider the restrictions given before t he Construction section. Let's first study the edge pieces. Since there are 12 of them, there are 12! ways to relocate the pieces. Since every edge piece has two orientations, there are i 2 orientations for the twelve edge pieces in total. Thus, by the multiplication principie, there are at most, 12!x212 ways to arrange ali twelve edge pieces. An analogous procedure can be done for the corner pieces. Since there are 8 of them, there are 8! ways to relocate the pieces. Since every c:orner piece has 3 orientations, t here are 38 orientations for the eight corner pieces in total. Thus, there are at n1ost 8!xs8. Again, by the multiplication principie, we may conclude that there is a total of: (3) 9 �Celerinet January - June 2020 possible arrangements for the cube. This number actually represents ali the possible states of the cube that can be reached by taking the cube apart by pieces (like we did in the Introduction section) and reassembling its pieces into different positions and orientations. In order to complete the constraints, we must once again consider t he definition given for the group G, but before we do this, we will require to define an equivalence relation as an auxiliary. According to F igure 2, let x, ye:{l, 2, ... , 48}. vVe will say that: x~y~ x and y belong to the same piece in the cube. For example, 2 and 34 belong to the same edge piece, but 42 doesn't, while 1 and 9 belong to the same corner piece, but 41 doesn 't . Since it is clearly an equivalence relation, this induces the equivalence clasS:Js: 1, 3, 8, 6,41,43, 47,49, namely, c1 , c2, c3, c4 , es andes (for corners), respectively, and: 2, 5,7,4,20, 21, 29,37, 42, 45,47, 4~ namely, e 1, e2 , ... , e12 (for edges) respectively. Let X be the set of ali the equivalence dasses defined above. vVe could define an action of the group G on the set X, but instead, for conven ience, we will define a function on X. Let NI be an element of G, and Jet: be a map such that: (4) The reader is encouraged to remeinber that permutations in G are bijections the need to define this new function. 011 {l, 2, ... , 48}, hence \\IIoreover, the funct ion f),1 itself is a bijection on X, since we can define f),.1-1 to be f).,r t for aU M in G. From here, for any NI in G, we will think of f),1 as a permutation in SxNow, let's consider the permutation fu in disjoint cycle decomposition. This is : Since each cycle has length 4, their parity is odcl (this cycles can be written as a product of thxee transpositions), and the product oftwo odd permutations is even. Thus, the permutation fu is even. T his property actually holds also for f0, fL, fR, fF and f8. NJoreover, since every l\11 in G is a product of powers of U, R, L, B, D and F, f,.1 must be an even permutation for ali lvI in G. Since exactly half of the permutations are even, this reduces the possible states of t he cube to: 12!x212 x8! x 38/ 2= 12!xi 1 x8!x38 (6) The property that has been proved is often presentecl as saying that it is impossible to have ali the cube solved except by two swapped edge pieces or two swapped corner pieces. The remaining two constraints state that only a third of the possible corner piece orientations are valid, and only half of the possible edge piece orientations ate va.lid. This leaves us with a total of: 10 �ACADEMJC / MATH.EMATI CS 12!x i 1 x8!x38/ 6=12!x210 x8!x37 (7) possible arrangements for the cube, which is equal to 43,252,003,274,489,856,000. ■ T he Jast two constraints that were mentioned have formal proofs written in reference [1], and along with the one that was proved in this article, are sometimes referred as the Fundamental Theorem of Cubing. However, in reference [7] a rather visual and int uit ive proof is provided by Robert Snapp, and both constraints a re a consequence of rotation inva riants of corner and edge pieces by using congruencies mod 3 for corners, and mod 2 for edges. This two properties physically imply t hat it is iJn possible for a cube to be completely solvecl except by a twisted corner piece, or a twisted edge piece. Theorern 3 The order of the subgroup of G, (UR), is 105. Proof. In order to prove this, we can rewrite 105 into its prime factor decomposit ion. This is, 7x5x3. Let's analyze what happens after doing (UR)7. \ ·Ve will keep track of, for example, the edge piece corresponding to 2 in Figure 2. Let: (8) vVhere n goes from 1 to 7. T hen, from t he definitions given for U a nd R, (9) E 2= {36,45, 21, 5, 7, 4, 2} So, after 7 UR moves, 2 goes back to its origiJ1al posit ion. T his means that not only the piece corresponding to 2 goes back to its original place, but it also rema ins well oriented. Since 34 belongs to the same edge piece as 2, then 34 is also back to its original posit ion. vVe can, again, check this by using t he definition, but t he consistency of the definit ion will save us the effort. Observe that, because of the cyclic nature of (UR)", t he sets E3G, E4s, E2 1, E5 , E7 , E4 and E2 a re identical, except for the order of appearance of t heiJ- elements, ancl there will be no repeated elements, so their cardinality is also 7. Thus, doiJ1g (UR)7 will leave ali edge pieces from the U face intact. \.Ve can check t hat t his prope1ty holds for the rest of the edge pieces. In fact, it is only nece.ssary to now check the ones in the R face, since the rest will not be affected by the permutation UR. Now, !et' s a.nalyze the behavior of (UR)5. Proceeding ana logously to t he edge pieces, we will define the set: Cs={(UR)"(8)} (10) \.Yhere n goes from l to 5. Again, from the defini tions, it occurs t hat C8 ={6, l. 38, 43, 19} (11) 11 �Celerinet January - June 2020 So 8 goes to the position of 19. It is now not that obvious what happens, but notice that 8, 19 a nd 25 belong to the same corner piece. In consequence, 19 goes to the position of 25 and 25 takes the place of 8. This can be thought as a third of a counterclockwise turn of a corner piece, and the consistency of the definitions of the movements will allow us to sldp the analysis for (UR)5(19) and (UR)5 (25) . If we repeat t his process for the rest of the remaining five corner pieces affected by UR, we will notice that this behavior holds for ali of them, so (UR) 5 has (on the corner pieces) the effect of only rotating them. Finally, three of this counterclockwise turns of a corner piece will take them back to their original orientation. Hence, the order of the subgroup (UR) is t he least common multiple of 7, 3 and 5, which is 105. ■ Theorern 4 The order of the subgroup (R2 U2) is 6. The proof for this theorem is analogous to the one made for Theorem 3. Theorern 5 Let S=RUR-1u·1 . Then, the order of the su bgroup (S) is 6. Again, the proof for this theorem is anaiogous the one made for Theorem 3, and this theorem is often used to solve the Rubik' s cube, as we will see in the next section. Solution Before directly providing a solut ion, !et us begin by t hinking of possible approaches to solve the puzzle. Our first attempt might be to fine! the sequeuce of moves that lec! to our scrambled position. This means, identifying the single element of G, out of the 43,252,003,274,489,856,000 that scraJnbled our cube. ,vhile this is the optima! solution, it is quite complicated to identify this elen1ent. A situation that would simplify finding a solution for the puzzle ,vould be the existence of a sequence of movements such that, no matter how it is scrambled, will solve the cube after doing a finite number of repetitions of the sequence. This way, we could just repeat this sequence until the cube gets solved. Despite not being efficient because of the number of permutations that was calculated in the Properties section, it would be effective. However, the existence of such a sequence is equivalent to say that the group G is cyclic, but cyclic groups are abelian, and, as pointed out in the Construction section, this is not the case. A n1ore realistic approach would be to reduce the group G into smaller subgroups that are easier to think of. To begin the solution for the puzzle, a very standard procedure would be to first solve one face of the cube. This is, to make one face of the cube, for example, U, identical to its representation in Figure 2. Since every other piece is i1-relevant for this step, this gives the solver plenty of freedom to reach this goal. In fact, with a little bit more reasoning, t he solver can, with sorne practice, very intuitively solve the second !ayer without undoing the fac:e that has just been solved. This is, assuming we solved the U face, to 12 �ACADEMJC/ ~1ATHEMATICS place the four edge pieces corresponding to the numbers 20, 21, 36 and 37 into t heir right positions and orientations. T he reader is encouraged to try this on their own. The complications may appear in the third !ayer (the face corresponding to D, assum ing we sta1ted by solving U), since our movements are now very restricted, because whenever we want to solve something from the third layer, we will be forced to undo some of the parts that we ha.ve just solved, and then solve everything back again, but the advantage here is that there are only four corner pieces and four edge pieces left to solve. Experienced readers may associate thjs behavior to commutators and conjugates. Remember that, in groups, for two elements x and y, the commutator of x and y is defined as (12) Thls definition may remember us to Theorem 5 from the Properties section, since, as it was defined, S=[R,U] . Now, for t he last layer, there are two things that we gradually have to solve. First, we have to ta ke the edge and corner pieces to their correct places, and then make sure they are placed with their correct otientations. The reader will be a.ble to design their own sequences of moves using the tools presented in here. For example, in order to relocate the corner pieces, we will take advantage of 3-cycles a nd commutators. First, identify the three corner pieces that you want to relocate as shown in Figure 6. As a consequence of what was proven in Theorem 2, we can reach any permutation by doing this. A '71 / Base !ayer e / ., ..... ,, 'B Figure 6 T hen, identify t he !ayer that has two of the tb ree corner pieces that we want to relocate (in this case, the down !ayer that has the C and B corner pieces). This will be called the base !ayer (identifiecl with blue in Figure 6), and it cannot be unsolvecl. Thinking of the face of the cube that we are trying to solve as our F (front) face, t he sequence :tvl= RUR-1 takes the corner from A to t he corner from B while leaving the rest of the base !ayer unchanged. 13 �Celerinet January - June 2020 In order to restore the rest of the cube that has been scrambled, we must undo what we ha.ve done, but doing 1.f1 now would be pointless. Instead, let 's take the corner C to t he place where t he corner B sta1ted (now occupied by A) by doing the sequence N=D. Notice that, if we now do Nr1 , the parts t hat were scrambled will now be unscran1bled, and the corner C will go to the place where the corner A started. i\iloreover, the base !ayer will not be affected. The only th ing left is to undo the move N, a.nd we would have successfully permuted three corners without a.ffecting the rest of the cube. As we can notice, this permutation is of the form: [RUR-1 ,DJ. T his process is way easier done t han explained, but it can be a.dapted to create new sequences for different purposes as we need it, for example, to relocate the remaining edges or to rotate them. As an example of using a commutator to rotate pieces, suppose that our final !ayer is now facing up, and consider t he sequence: 11=[D,L) (13) If we consider now t he upper fa.ce as our base !ayer, t he commutator: P= [1.if2,U] (14) has the effect of rotating corners. As a consequence of Theorem 5, the scrambled parts of the cube will be solved again after doing P3. Conclusions Several topics in moclern mathe1natics are often quite complicated to visualize, or very abstract to be applied to real life situations, wruch is a common situation in abstract algebra. However, mathematieians ha.ve found very int uitive applications of group theory, such as the study of Dihedral groups, and perrnutation groups. The results that have been shown in this article stand out for showing the app!icability of abstract concepts for making sense of diverse properties and provicling oolutions to real problems. This results could also turn out to be useful to strengthen previous knowledge of abstract algebra and could serve to an undergraduate student as an auxiliary intuitive approach to t he notions of perrnutations, cyclic groups, and commutators, since the study of mathematical concepts via puzzles is a tangible and accessible expression of abstract mathematics. Reaclers are highly encouraged to read this article with a Rubik's cube on their hands, so they can replicate the results and experience the transcendence of a bstract algebra with their own hands. A5 shown, modeling a pwzzle by relating pieces with numbers, and interpreting rnoves as permutations, turned out to be crucial to obtain conclusions about the puzzle' s behavior. Sünilar ideas have been used to ana lyie other puzzles, such as the fifteen puzzle [8), and, in general, it can be used to study the properties of a generalized NxNxN Rubi k 's cube, since this modeling is, clearly, not exclusively valid for the 3x3x3 version of the puzzle. Consequently, a solution for a generalized Rubik 's cube can be done using a procedure analogous to the one proposecl ü1 this article. It is worth to remark that a solution via commutators can be very ineffi<:ient, and in order to reduce t he solving time and cornplexity, better methods have been developed, such as the Fridrich method, explained by ,lessica Fridrich in [9), thus creating a highly competitive environment among those who have accepted the challenge to solve the puzzle. 14 �ACADEMIC / MATHEMATICS About the author Jordi Andrés l\11artínez Álvarez is a l\11athematics' student in the Facultad de Ciencias Físico Matemáticas at the Universidad Autó noma de Nuevo León. He participated as a co-delegate of Nuevo León for the 32 l\liexican Olympiad of Mathematics in 2018, and has collaborated in a couple of projects organized by the Mexican Society of l\lJathematics. Address: Pedro de Al ba, Niños Héroes, Ciudad Universitaria, 66451. San Nicolás de los Garza, Nuevo León. E mail: jordi.martinezlv@uanl.edu.mx References 11] Ja.yson D. (2008). Adventures in Group Theory: Ilubik's Cube, Merlin's l\fachlne, and Other 1fatheumtical Toys (2nd ed.). Dalt.imore, Maryland: Johus Hoµkins Uuivcrsit.y Press. 12] BergvaU, O., Hynning, E., Hedberg, M., l\·lickeliJ1, J. , & l\fasaw(i, P. (2010). On Ilubik 's Cube. Iletrieved from https:/ / peoµle.kth.sc/-boij/ka ndexjobb\/Tl1/1\fa t.erial/ rubikscube.pdf 131 Dummit, D. S., & Foote, R. M. (2003). Abstract Algebrn (3rd ed.). Hoboken, Ncw J ersey: John \1/iley & Sons. 14] Bossert, P . (1981). You Can Do t.he Cube. Da llas, Pennsylvania: Puffin Books . 15] Sing111astcr, D. (1981). Note-; [6] Herstein, l. N. (1996). Abstrnct AJgebra (3rd ed.). Upptlr Saddle lliver, New J ersey: Preutice Hall. 17] Snapp, R. (2012). Ilubik's Cube Magic. Ilet1ieved from ht.tp://www.emba .uvm.edu/-rsnapp/teaching/cs32/ lectures/rubik.pdf 18] Beeler, Il. A. (2015). The Fifteen Puzzle A Motivating Example for the Alternating Group (Supplemental Material for Intro to Modern Algebrn). Retrieved from ht.tp://fa<:ulty.ctsu.edu/ bcelerr/fifteen-supp.pdf 19] Frichich, J. (n.d.) . My system for http://ws2.biughamton.edu/ fridrich/systcm.ht.ml 011 Rubik 's Magic Cube. Hillside, New Jersey: Euslow Publishers. solving Ilubik's cube. Iletrieved J une 24, 2020, from IIO] Milewski, A., & Frohardt, D. (2020). Seeing Algebraic Structure: The Ilubik's Cubc, l\fathematics Teacher: Learniug ami Teaching PK-12 MTLT, 113(5), 397-403. 111] Mullhollaml, .J. (u.ti.). The Fuudament.al Tbtórem of Cubology. R.et.rievcd, from ht.tp://www~fu .<:af-:jtlnulholjn¡¡1tb302/ptl½l-k'& ro-cubology.html 112] Revilla, F. (2014). Conmutador y su bgrupo derivado. Ret.rieved from http://fernandoreviUa.csjblog/2014/03/02/<:onmut,l<ior-ysubgrupo-derivado/ [13] \1/ong, R. (2016). The Group St,rncture of the Ilubik's https:/ / web.ma..utexas.eclu/users/1ich11rcl.wong/expositor:)'/ Ilubiks_ Cube_ Group. pdf Cube. Ret.rieved from 15 � Dublin Core The Dublin Core metadata element set is common to all Omeka records, including items, files, and collections. For more information see, http://dublincore.org/documents/dces/. Title A name given to the resource Celerinet Description An account of the resource La revista Celerinet, inició en el 2012, sólo en formato digital, es semestral y se mantiene activa; ofrece información de las últimas investigaciones realizadas por docentes, estudiantes y egresados de la Facultad de Ciencias Físico Matemáticas, también se encarga de difundir las actividades institucionales más relevantes. La publicación incluye artículos de  investigación relacionados con las siguientes áreas: matemáticas, matemáticas aplicadas, física, ciencias computacionales, actuaría, multimedia y animación digital, y seguridad en tecnologías de información. Text A resource consisting primarily of words for reading. 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