DOCSLIB.ORG

MATH 4361 (Modern Algebra) – Assignment 4 Solutions Due on December 4, 2015

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
MATH 4361 (Modern Algebra) – Assignment 4 Solutions Due on December 4, 2015 MATH 4361 (Modern Algebra) – Assignment 4 Solutions Due on December 4, 2015 Please write in a neat and organized fashion. In addition to being mathematically correct, your answers should also be easy for me to read and contain enough detail for me to be able to follow your thought process. Correct use of words and notation are essential. 1. The dihedral group D4 is the group of symmetries of a square. The elements of D4 are R0 do nothing ∘ R1 rotate clockwise 90 ∘ R2 rotate clockwise 180 ∘ R3 rotate clockwise 270 FA reflect across line A FB reflect across line B FC reflect across line C FD reflect across line D. a. Write the elements of D4 as permutations. b. Construct the Cayley table for D4. c. Find all of the subgroups of D4. d. Determine which of the subgroups of D4 are normal subgroups and which are not. Solution: We write the elements of D4 as permutations as follows: 1 R0 1 R1 1234 R2 1324 R3 1432 FA 24 FB 13 FC 1234 FD 1423. Here is the Cayley Table for D4. ∘ R0 R1 R2 R3 FA FB FC FD R0 R0 R1 R2 R3 FA FB FC FD R1 R1 R2 R3 R0 FC FD FB FA R2 R2 R3 R0 R1 FB FA FD FC R3 R3 R0 R1 R2 FD FC FA FB . FA FA FD FB FC R0 R2 R3 R1 FB FB FC FA FD R2 R0 R1 R3 FC FC FA FD FB R1 R3 R0 R2 FD FD FB FC FA R3 R1 R2 R0 There are 10 distinct subgroups of D4. They are R0 which is normal R0,R2 which is normal R0,FA which is not normal R0,FB which is not normal R0,FC which is not normal R0,FD which is not normal R0,R1,R2,R3 which is normal R0,R2,FA,FB which is normal R0,R2,FC,FD which is normal D4 which is normal. 2. The dihedral group D5 is the group of symmetries of a regular pentagon. 2 The elements of D5 are R0 do nothing ∘ R1 rotate clockwise 72 ∘ R2 rotate clockwise 144 ∘ R3 rotate clockwise 216 ∘ R4 rotate clockwise 288 FA reflect across line A FB reflect across line B FC reflect across line C FD reflect across line D FE reflect across line E. a. Write the elements of D5 as permutations. b. Construct the Cayley table for D5. c. Find all of the subgroups of D5. d. Determine which of the subgroups of D5 are normal subgroups and which are not. Solution: We write the elements of D5 as permutations as follows: 3 R0 1 R1 12345 R2 13524 R3 14253 R4 15432 FA 2534 FB 1235 FC 1345 FD 1423 FE 1524. Here is the Cayley Table for D5. ∘ R0 R1 R2 R3 R4 FA FB FC FD FE R0 R0 R1 R2 R3 R4 FA FB FC FD FE R1 R1 R2 R3 R4 R0 FB FC FD FE FA R2 R2 R3 R4 R0 R1 FC FD FE FA FB R3 R3 R4 R0 R1 R2 FD FE FA FB FC R4 R4 R0 R1 R2 R3 FE FA FB FC FD . FA FA FE FD FC FB R0 R4 R3 R2 R1 FB FB FA FE FD FC R1 R0 R4 R3 R2 FC FC FB FA FE FD R2 R1 R0 R4 R3 FD FD FC FB FA FE R3 R2 R1 R0 R4 FE FE FD FC FB FA R4 R3 R2 R1 R0 The subgroups of D5 are R0 which is normal R0,FA which is not normal R0,FB which is not normal R0,FC which is not normal R0,FD which is not normal R0,FE which is not normal R0,R1,R2,R3,R4 which is normal D5 which is normal. 3. If G is a group and b ∈ G, then a square root of b is an element a ∈ G such 4 that a2 b. For example 3 is a square root of 6 in 〈Z, because 2 3 3 3 6 and 123 is a square root of 132 in 〈S3,∘ because 1232 123123 132. a. Suppose that G and H are groups and that f : G → H is a homomorphism of G onto H. Prove that if every element of G has a square root then every element of H also has a square root. b. Let 〈Q, be the group of rational numbers with the operation of addition and let Qpos, be the group of positive rational numbers with the operation of multiplication. Use the fact that you proved in part a to explain why there does not exist any homomorphism of Q onto Qpos. Proof of Part a: Suppose that G and H are groups and that f : G → H is a homomorphism of G onto H. Also suppose that every element of G has a square root. We want to show that every element of H has a square root. To this end, let h ∈ H. Then, since f is onto H, there exists g ∈ G such that h fg.Sinceg has a square root, there exists a ∈ G such that g a2. Then, since f is a homomorphism, we see that y fg fa2 faa fafa. Since fa ∈ H and fa2 y, we see that y has a square root. This proves the result. Explanation of Part b: Observe that every element of Q has a square root. (Remember that the operation of Q is addition.) To be specific, if b ∈ Q, then b p/q where p and q are integers. This means that a p/2q ∈ Q and we observe that p p p a2 a a b 2q 2q q which means that a is the square root of b in Q. Now observe that not every element of Qpos has a square root. (Remember that the operation of Qpos is multiplication.) In particular 2 ∈ Qpos but there is no number, x,inQpos that satisfies x2 2. Since every member of Q has a square root but not every member of Qpos has a square root, then we see by Part a that there does not exist any homomorphism of Q onto Qpos. 4. Let G D4 and let H R0,R2 . a. In problem 1, you should have found that H is a normal subgroup of G (so you don’t have to do this again). b. List the elements of the quotient group G/H. c. Construct the Cayley table for G/H. Solution: The cosets of H are 5 H R0,R2 R2H R1H R1,R3 R3H FAH FA,FB FBH FCH FC,FD FDH. Thus G H,R1H,FAH,FCH H with Cayley Table HR1HFAHFCH HHR1HFAHFCH R1HR1HH FCHFAH . FAHFAHFCHH R1H FCHFCHFAHR1HH Observe that G/H is the Klein 4–group. 5. a. Suppose that G and K are finite groups and suppose that f : G → K is a homomorphism. Prove that for each element a ∈ G, the order of the element fa in K is a divisor of the order of a. b. Suppose that G is a finite group and that H is a normal subgroup of G. Prove that for each element a ∈ G, the order of the element aH in G/H is a divisor of the order of the element a. (Hint: Use what you proved in part a.) Proof of Part a: Suppose that G and K are finite groups and that f : G → K is a homomorphism. Let a ∈ G and suppose that orda n. Also suppose that ordfa m. We want to prove that m is a divisor of n. First observe that since f is a homomorphism then n n fa fa feG eK. n Since fa eK, then by Theorem 5 in Chapter 10, n must be a multiple of ordfa. Thus n is a multiple of m (which is the same as saying that m is a divisor of n). Proof of Part b: Suppose that G is a finite group and that H is a normal subgroup of G and let a ∈ G. We want to show that ordaH is a divisor of orda.Todothiswedefinef : G → G/H by fx xH. We have shown in class that this mapping is well–defined and that it is a homomorphism. Since fa aH, then by Part a above, we can conclude that ordaH is a divisor of orda. 6.
Load more

Recommended publications
  • Complex Algebras of Semigroups Pamela Jo Reich Iowa State University
    Iowa State University Capstones, Theses and Retrospective Theses and Dissertations Dissertations 1996 Complex algebras of semigroups Pamela Jo Reich Iowa State University Follow this and additional works at: https://lib.dr.iastate.edu/rtd Part of the Mathematics Commons Recommended Citation Reich, Pamela Jo, "Complex algebras of semigroups " (1996). Retrospective Theses and Dissertations. 11765. https://lib.dr.iastate.edu/rtd/11765 This Dissertation is brought to you for free and open access by the Iowa State University Capstones, Theses and Dissertations at Iowa State University Digital Repository. It has been accepted for inclusion in Retrospective Theses and Dissertations by an authorized administrator of Iowa State University Digital Repository. For more information, please contact [email protected]. INFORMATION TO USERS This manuscript has been reproduced from the microfihn master. UMI fibns the text du-ectly from the original or copy submitted. Thus, some thesis and dissertation copies are in typewriter face, while others may be from any type of computer printer. The quality of this reproductioii is dependent upon the quality of the copy submitted. Broken or indistinct print, colored or poor quality illustrations and photographs, print bleedthrough, substandard margins, and unproper alignment can adversely affect reproduction. In the unlikely event that the author did not send UMI a complete manuscript and there are missing pages, these will be noted. Also, if unauthorized copyright material had to be removed, a note will indicate the deletion. Oversize materials (e.g., m^s, drawings, charts) are reproduced by sectioning the original, beginning at the upper left-hand comer and continuing from left to right in equal sections with small overiaps.
    [Show full text]
  • Group Isomorphisms MME 529 Worksheet for May 23, 2017 William J
    Group Isomorphisms MME 529 Worksheet for May 23, 2017 William J. Martin, WPI Goal: Illustrate the power of abstraction by seeing how groups arising in different contexts are really the same. There are many different kinds of groups, arising in a dizzying variety of contexts. Even on this worksheet, there are too many groups for any one of us to absorb. But, with different teams exploring different examples, we should { as a class { discover some justification for the study of groups in the abstract. The Integers Modulo n: With John Goulet, you explored the additive structure of Zn. Write down the addition table for Z5 and Z6. These groups are called cyclic groups: they are generated by a single element, the element 1, in this case. That means that every element can be found by adding 1 to itself an appropriate number of times. The Group of Units Modulo n: Now when we look at Zn using multiplication as our operation, we no longer have a group. (Why not?) The group U(n) = fa 2 Zn j gcd(a; n) = 1g ∗ is sometimes written Zn and is called the group of units modulo n. An element in a number system (or ring) is a \unit" if it has a multiplicative inverse. Write down the multiplication tables for U(6), U(7), U(8) and U(12). The Group of Rotations of a Regular n-Gon: Imagine a regular polygon with n sides centered at the origin O. Let e denote the identity transformation, which leaves the poly- gon entirely fixed and let a denote a rotation about O in the counterclockwise direction by exactly 360=n degrees (2π=n radians).
    [Show full text]
  • Arxiv:2001.06557V1 [Math.GR] 17 Jan 2020 Magic Cayley-Sudoku Tables
    Magic Cayley-Sudoku Tables∗ Rosanna Mersereau Michael B. Ward Columbus, OH Western Oregon University 1 Introduction Inspired by the popularity of sudoku puzzles along with the well-known fact that the body of the Cayley table1 of any finite group already has 2/3 of the properties of a sudoku table in that each element appears exactly once in each row and in each column, Carmichael, Schloeman, and Ward [1] in- troduced Cayley-sudoku tables. A Cayley-sudoku table of a finite group G is a Cayley table for G the body of which is partitioned into uniformly sized rectangular blocks, in such a way that each group element appears exactly once in each block. For example, Table 1 is a Cayley-sudoku ta- ble for Z9 := {0, 1, 2, 3, 4, 5, 6, 7, 8} under addition mod 9 and Table 3 is a Cayley-sudoku table for Z3 × Z3 where the operation is componentwise ad- dition mod 3 (and ordered pairs (a, b) are abbreviated ab). In each case, we see that we have a Cayley table of the group partitioned into 3 × 3 blocks that contain each group element exactly once. Lorch and Weld [3] defined a modular magic sudoku table as an ordinary sudoku table (with 0 in place of the usual 9) in which the row, column, arXiv:2001.06557v1 [math.GR] 17 Jan 2020 diagonal, and antidiagonal sums in each 3 × 3 block in the table are zero mod 9. “Magic” refers, of course, to magic Latin squares which have a rich history dating to ancient times.
    [Show full text]
  • Cosets and Cayley-Sudoku Tables
    Cosets and Cayley-Sudoku Tables Jennifer Carmichael Chemeketa Community College Salem, OR 97309 [email protected] Keith Schloeman Oregon State University Corvallis, OR 97331 [email protected] Michael B. Ward Western Oregon University Monmouth, OR 97361 [email protected] The wildly popular Sudoku puzzles [2] are 9 £ 9 arrays divided into nine 3 £ 3 sub-arrays or blocks. Digits 1 through 9 appear in some of the entries. Other entries are blank. The goal is to ¯ll the blank entries with the digits 1 through 9 in such a way that each digit appears exactly once in each row and in each column, and in each block. Table 1 gives an example of a completed Sudoku puzzle. One proves in introductory group theory that every element of any group appears exactly once in each row and once in each column of the group's op- eration or Cayley table. (In other words, any Cayley table is a Latin square.) Thus, every Cayley table has two-thirds of the properties of a Sudoku table; only the subdivision of the table into blocks that contain each element ex- actly once is in doubt. A question naturally leaps to mind: When and how can a Cayley table be arranged in such a way as to satisfy the additional requirements of being a Sudoku table? To be more speci¯c, group elements labeling the rows and the columns of a Cayley table may be arranged in any order. Moreover, in de¯ance of convention, row labels and column labels need not be in the same order.
    [Show full text]
  • Midterm # 1 Solutions
    Midterm # 1 Solutions The Nintendo game “Baseball Stars” February 12, 2008 Hi baseball fans! We’re coming to you from video game land to give you the solutions to the first test. We’re lazy aging video game superstars and don’t feel the need to type out something that has already been typed out, or can be found verbatim from the book. We hope you enjoy them! Rock on! 1. All of these first ones are definition and can be found in the book. 2. (a) (i) U(12) is the set of all elements mod 12 that are relatively prime to 12. This would be the set containing the integer representatives {1, 5, 7, 11}. The multiplication table is as follows: 1 5 7 11 1 1 5 7 11 5 5 1 11 7 7 7 11 1 5 11 11 7 5 1 (ii) This group is not cyclic: you can see this from the multiplication table that < 1 >= {1}, < 5 >= {1, 5}, < 7 >= {1, 7}, < 11 >= {1, 11}. None of these elements generate U(12), so U(12) is not cyclic. 1 (b) (i) I’m sure at some point we produced a Cayley table for D3. We computed in class (ii) the the center of D3 is Z(D3) = {e}. Should your class notes be incomplete on this matter or we never did a Cayley table of D3, then speak with Corey. Nobody chose this problem to do, so we Baseball Stars feel okay about leaving it at that. (c) (i) The group Gl(n, R) = {A ∈ Mn×n(R)|det(A) 6= 0}.
    [Show full text]
  • Dihedral Group of Order 8 (The Number of Elements in the Group) Or the Group of Symmetries of a Square
    CHAPTER 1 Introduction to Groups Symmetries of a Square A plane symmetry of a square (or any plane figure F ) is a function from the square to itself that preserves distances, i.e., the distance between the images of points P and Q equals the distance between P and Q. The eight symmetries of a square: 22 1. INTRODUCTION TO GROUPS 23 These are all the symmetries of a square. Think of a square cut from a piece of glass with dots of di↵erent colors painted on top in the four corners. Then a symmetry takes a corner to one of four corners with dot up or down — 8 possibilities. A succession of symmetries is a symmetry: Since this is a composition of functions, we write HR90 = D, allowing us to view composition of functions as a type of multiplication. Since composition of functions is associative, we have an operation that is both closed and associative. R0 is the identity: for any symmetry S, R0S = SR0 = S. Each of our 1 1 1 symmetries S has an inverse S− such that SS− = S− S = R0, our identity. R0R0 = R0, R90R270 = R0, R180R180 = R0, R270R90 = R0 HH = R0, V V = R0, DD = R0, D0D0 = R0 We have a group. This group is denoted D4, and is called the dihedral group of order 8 (the number of elements in the group) or the group of symmetries of a square. 24 1. INTRODUCTION TO GROUPS We view the Cayley table or operation table for D4: For HR90 = D (circled), we find H along the left and R90 on top.
    [Show full text]
  • Elements of Quasigroup Theory and Some Its Applications in Code Theory and Cryptology
    Elements of quasigroup theory and some its applications in code theory and cryptology Victor Shcherbacov 1 1 Introduction 1.1 The role of definitions. This course is an extended form of lectures which author have given for graduate students of Charles University (Prague, Czech Republic) in autumn of year 2003. In this section we follow books [56, 59]. In mathematics one should strive to avoid ambiguity. A very important ingredient of mathematical creativity is the ability to formulate useful defi- nitions, ones that will lead to interesting results. Every definition is understood to be an if and only if type of statement, even though it is customary to suppress the only if. Thus one may define: ”A triangle is isosceles if it has two sides of equal length”, really meaning that a triangle is isosceles if and only if it has two sides of equal length. The basic importance of definitions to mathematics is also a structural weakness for the reason that not every concept used can be defined. 1.2 Sets. A set is well-defined collection of objects. We summarize briefly some of the things we shall simply assume about sets. 1. A set S is made up of elements, and if a is one of these elements, we shall denote this fact by a ∈ S. 2. There is exactly one set with no elements. It is the empty set and is denoted by ∅. 3. We may describe a set either by giving a characterizing property of the elements, such as ”the set of all members of the United State Senate”, or by listing the elements, for example {1, 3, 4}.
    [Show full text]
  • 4 Direct Products
    4 Direct products 4.1 Definition and Examples Direct products are a straightforward way to define new groups from old. The simplest example is the Cartesian product Z2 × Z2. Writing Z2 = f0, 1g, we see that Z2 × Z2 = f(0, 0), (0, 1), (1, 0), (1, 1)g has four elements. This set can be given a group structure in an obvious way. Simply define (p, q) + (r, s) := (p + r, q + s) where p + r and q + s are computed in Z2. This is clearly a binary operation on Z2 × Z2. It is straightforward to check that this operation defines a group. Indeed, we obtain the following Cayley table: + (0, 0) (0, 1) (1, 0) (1, 1) (0, 0) (0, 0) (0, 1) (1, 0) (1, 1) (0, 1) (0, 1) (0, 0) (1, 1) (1, 0) (1, 0) (1, 0) (1, 1) (0, 0) (0, 1) (1, 1) (1, 1) (1, 0) (0, 1) (0, 0) ∼ This should look familiar: it is exactly the Cayley table for the Klein 4-group: Z2 × Z2 = V. This construction works in general. Theorem 4.1. Let G1,..., Gn be multiplicative groups. The binary operation (a1,..., an) · (b1,..., bn) := (a1b1,..., anbn) induces a group structure on the set G1 × · · · × Gn. Definition 4.2. The group G1 × · · · × Gn is the direct product of the groups G1,..., Gn. If the groups G1,..., Gn are additive, then the operation will also be written additively. It is immediate that a direct product has cardinality equal to the product of the cardinalities of G1,..., Gn. Proof of Theorem.
    [Show full text]
  • Chapter 8 Cayley Theorem and Puzzles
    Chapter 8 Cayley Theorem and Puzzles \As for everything else, so for a mathematical theory: beauty can be perceived but not explained."(Arthur Cayley) We have seen that the symmetric group Sn of all the permutations of n objects has order n!, and that the dihedral group D3 of symmetries of the equilateral triangle is isomorphic to S3, while the cyclic group C2 is isomor- phic to S2. We now wonder whether there are more connections between finite groups and the group Sn. There is in fact a very powerful one, known as Cayley Theorem: Theorem 15. Every finite group is isomorphic to a group of permutations (that is to some subgroup of Sn). This might be surprising but recall that given any finite group G = fg1; g2; : : : ; gng, every row of its Cayley table g1 = e g2 g3 ··· gn g1 g2 . gr grg1 grg2 grg3 ··· grgn . gn is simply a permutation of the elements of G (grgs 2 fg1; g2; : : : ; gng). 171 172 CHAPTER 8. CAYLEY THEOREM AND PUZZLES Groups and Permutation Groups • We saw that D3=S3 and C2=S2. • Is there any link in general between a given group G and groups of permutations? • The answer is given by Cayley Theorem! Cayley Theorem Theorem Every finite group is isomorphic to a group of permutations. This means a subgroup of some symmetric group. One known link: for a group G, we can consider its multiplication (Cayley) table. Every row contains a permutation of the elements of the group. 173 Proof. Let (G; ·) be a group. We shall exhibit a group of permutations (Σ; ◦) that is isomorphic to G: We have seen that the Cayley table of (G; ·) has rows that are permutations of fg1; g2; : : : ; gng, the elements of G: Therefore let us define Σ = fσg : G ! G; σg(x) = gx; 8x 2 Gg for g 2 G: In words we consider the permutation maps given by the rows of the Cayley table.
    [Show full text]
  • Homework 7 Solutions 1 Chapter 9, Problem 10
    Homework 7 Solutions March 17, 2012 1 Chapter 9, Problem 10 (graded) Let G be a cyclic group. That is, G = hai for some a 2 G. Then given any g 2 G, g = an for some integer n. Let H be any normal subgroup of G (actually, since G is cyclic, it is also Abelian, so all subgroups of G are normal), and consider the factor group G=H = fgH : g 2 Gg. G=H is the group whose elements are left cosets of H. Let gH be any element of G=H. Since g = an for some integer n, we have gH = anH: Next, by denition of multiplication in a factor group, gH = anH = (aH)n: Therefore, if gH is any element of G=H, then gH = (aH)n for some integer n. This implies that G=H = haHi. That is, G=H is a cyclic group generated by the element aH. 2 Chapter 9, Problem 16 (graded) Before presenting the solution, let me talk about computing order in a factor group G=H. Suppose gH is an element of G=H (so g 2 G) and I want to compute its order as an element of G=H. In other words, I want to nd an integer n such that (gH)n = eH = H and if 1 ≤ m < n, (gH)m 6= H: By denition of multiplication in a factor group, we need to nd n so that gnH = H and if 1 ≤ m < n, gmH 6= H: 1 By the Lemma on page 139, gnH = H i gn 2 H, and gmH 6= H i gm 2= H.
    [Show full text]
  • Homework #3 Solutions
    Homework #3 Solutions p 23, #4. s = −3 and t = 2 work since 7s + 11t = −21 + 22 = 1. These choices are not unique since s = 8, t = −5 also work: 7s + 11t = 56 − 55 = 1. p 24, #30. Given any integer n, at least one of the three consecutive numbers n−1, n, n+1 must be divisible by 2 and at least one of them must be divisible by 3. Therefore the product (n − 1)n(n + 1) = n(n2 − 1) = n3 − n must be divisible by 2 · 3 = 6. It follows that n3 and n must have the same remainder when divided by 6, i.e. n3 mod 6 = n mod 6. p 54, #8 We must verify that the 4 group axioms hold for the set S = {5, 15, 25, 35} together with the operation of multiplication modulo 40. Since we know that this operation is associative on all of Zn, it will be associative on S as well. We need only verify closure, the existence of an identity, and the existence of inverses. We can do this by building a Cayley table for S: 5 15 25 35 5 25 35 5 15 15 35 25 15 5 25 5 15 25 35 35 15 5 35 25 The table shows that S is closed under multiplication mod 40, that 25 is the identity of S, and, since 25 appears in each row, every element has an inverse. In fact, each element is its own inverse! To compare S to U(8) = {1, 3, 5, 7}, we examine the Cayley table of the latter: 3 5 1 7 3 1 7 3 5 5 7 1 5 3 1 3 5 1 7 7 5 3 7 1 If we swap symbols in this table according to the following rules 1 ↔ 25, 3 ↔ 5, 5 ↔ 15, 7 ↔ 35 then the Cayley table for U(8) is transformed into the Cayley table for S.
    [Show full text]
  • Introduction to Groups 1
    Contents Contents i List of Figures ii List of Tables iii 1 Introduction to Groups 1 1.1 Disclaimer .............................................. 1 1.2 Why Study Group Theory? .................................... 1 1.2.1 Discrete and continuous symmetries .......................... 1 1.2.2 Symmetries in quantum mechanics ........................... 2 1.3 Formal definition of a discrete group ............................... 3 1.3.1 Group properties ..................................... 3 1.3.2 The Cayley table ...................................... 4 1.3.3 Loopsand groups ..................................... 4 1.3.4 The equilateral triangle : C3v and C3 .......................... 5 1.3.5 Symmetry breaking .................................... 8 1.3.6 The dihedralgroup Dn .................................. 9 1.3.7 The permutation group Sn ................................ 10 1.3.8 Our friend, SU(2) ..................................... 12 1.4 Aspects of Discrete Groups .................................... 14 1.4.1 Basic features of discrete groups ............................. 14 i ii CONTENTS 1.4.2 Other math stuff ...................................... 18 1.4.3 More about permutations ................................. 19 1.4.4 Conjugacy classes of the dihedral group ........................ 21 1.4.5 Quaternion group ..................................... 21 1.4.6 Group presentations .................................... 23 1.5 Lie Groups .............................................. 25 1.5.1 DefinitionofaLiegroup ................................
    [Show full text]