A. (Groups of order 8.) (a) Which of the five groups G (as specified in the question) have the following property: G has a normal subgroup N such that


 Joy Warner
 4 years ago
 Views:
Transcription
1 MATH 402A  Solutions for the suggested problems. A. (Groups of order 8. (a Which of the five groups G (as specified in the question have the following property: G has a normal subgroup N such that N = Z/2Z and G/N = Z/2Z Z/2Z? Answer: Among the abelian groups, if G = Z/2Z Z/2Z Z/2Z or if G = Z/4Z Z/2Z, then G has the stated property, as is easily verified. However, if G is cyclic, then one of the homework problems shows that every quotient group of G is also cyclic. Therefore G does not have the stated property. If G is nonabelian and has order 8, then G has the stated property. Thus, both Q 8 and D 8 have the stated property. We will justify this statement. One helpful observation is that if G is nonabelian, then G/Z(G cannot be cyclic. This is because of the following result which was proved in class. Lemma. If G is any group and G/Z(G is cyclic, then G is abelian. It follows that the order of G/Z(G must be at least 4 if G is nonabelian. Now if G = Q 8 or G = D 8, one can verify that Z(G = 2. Hence Z(G = Z/2Z. Also, G/Z(G has order 4 and is not cyclic. Hence G/Z(G = Z/2Z Z/2Z. Combining these observations, if we let N = Z(G, then the above stated property holds. (b Which of the five specified groups G have the following property: G has a normal subgroup N such that N = Z/2Z and G/N = Z/4Z? Answer: Only two of the five groups have this property. We must have N Z(G. (See question E from problem set 4. If N = Z(G, then G/Z(G will be cyclic. Hence, the lemma stated above shows that G must be abelian (in which case we actually have Z(G = G. Thus, G must be isomorphic to one of the following two groups (up to isomorphism Z/8Z or Z/4Z Z/2Z. since Z/2Z Z/2Z Z/2Z clearly doesn t have the stated property. The above two groups do have the stated property, as is easily verified. (c Which of the five groups G have the following property: G has a normal subgroup N such that N = Z/2Z Z/2Z and G/N = Z/2Z?
2 Answer: It suffices to have a subgroup N of G isomorphic to Z/2Z Z/2Z. Such a subgroup would have index 2 and hence would automatically be a normal subgroup of G. The corresponding quotient group would have order 2 and would then be isomorphic to Z/2Z. Groups of order 8 having such a subgroup are isomorphic to Z/2Z Z/2Z Z/2Z, Z/4Z Z/2Z, or D 8. Now Q 8 has only one element of order 2 and therefore cannot contain any subgroups isomorphic to Z/2Z Z/2Z. Also, Z/8Z is cyclic and hence does not contain a subgroup isomorphic to the group Z/2Z Z/2Z because that group is not cyclic. (d Determine all the conjugacy classes in each of the five groups of order 8. Answer: For the three abelian groups, the conjugacy classes are just the singletons consisting of the individual elements. For the quaternionic group Q 8, the conjugacy classes are: {1}, { 1}, {i, i}, {j, j}, {k, k}. For D 8, the conjugacy classes are: {i}, {r 2 }, {r, r 3 }, {s, sr 2 }, {sr, sr 3 } where r = ( (which corresponds to a 90 o rotation of a square with vertices labeled 1, 2, 3, 4 and s = (12(34 (which corresponds to one of the reflections of that square. B. Suppose that G is a finite group and that G contains an abelian subgroup A such that [G : A] = 2. (a Suppose that a A. Prove that the conjugacy class of a in G has cardinality equal to 1 or 2. Proof. The cardinality of the conjugacy class for a is equal to the index [G : H], where H = C G (a, the centralizer of a in G. Obviously, since A is abelian and a A, we have A H G. Thus, by Lagrange s theorem, A divides H and so H = t A for some positive integer t. But, H divides G = 2 A and so we have t = 1 or 2. Therefore, the cardinality of the conjugacy
3 class of a in G is G / H = 2 A /t A = 2/t, which is either 1 or 2. Notice that we have H = A or H = G. In the first case, the conjugacy class of a has cardinality 2. In the second case, that cardinality is 1. (b Suppose that b G, but b A. Show that the cardinality of the conjugacy class of b in G divides A. Give an example of a group G and a subgroup A where the conjugacy class of any such b has cardinality equal to A. Solution: Note that A is a normal subgroup of G since [G : A] = 2, G/A has order 2, and ba is a nontrivial element of the group G/A. Hence ba has order 2 in G/A. Suppose that m is the order of b. Then b m = e, the identity element of G. Hence (ba m = b m A = A and so the order of ba in the quotient group G/A divides m. That is, m is divisible by 2. Now, let H = C G (b, the centralizer of b in G. Note that H contains the cyclic subgroup b, which has order m, and hence m divides H. Thus, H is even. The cardinality of the conjugacy class of b is equal to G / H. Since H is even, G / H divides 1 2 G. However A has index 2 in G and hence A = 1 2 G. Therefore, the cardinality of the conjugacy class of b in G divides A, as stated. As an example, let G = D 10. Then G = 10 and G is nonabelian. Let A be the subgroup of G consisting of rotations, which is a cyclic subgroup of order 5. If b G, but b A, then b is a reflection and has order 2. The number of conjugates of b in G divides G / b = 5. However, since G is a product of two primes and G is nonabelian, we know that Z(G is trivial. (This uses homework problem 4 on page 95. Hence b Z(G. Hence the cardinality of the conjugacy class of G is 5, which is indeed equal to A. C. Prove that if G is a group of order 35, then G = Z/35Z. Proof. First of all, suppose that G is a group and G = 35. We will prove that G has at least one element of order 5 and at least one element of order 7.. If G is abelian, then we have already proved this. (It is the special case of Cauchy s theorem for finite abelian groups, which we proved one day in class. So now assume that G is a nonabelian group of order 35. According to one of the homework problems (page 95, problem 4, we have Z(G = {e}, where e is the identity element in G. Thus, if a G and a e, then a Z(G. Hence the conjugacy class of a in G has cardinality > 1. That cardinality must divide G / a = 35/ a. Therefore, if a = 5, then the conjugacy class of a has cardinality 7. If a = 7, then the conjugacy class of a has cardinality 5. Let x denote the number of distinct conjugacy classes in G for elements of order 5 and let y denote the number of distinct conjugacy classes in G for elements of order 7. Then we have 7x + 5y = G 1 = 34. This equation implies that the nonnegative integers x and y must actually be positive. Therefore, G indeed must contain elements of order 5 and elements of order 7. Thus, we have proved that if G
4 is a group (abelian or nonabelian and G = 35, then G contains elements of order 5 and of order 7. Let b be an element of G of order 7 and let N = b. Thus, N is a subgroup of G of order 7. According to problem C in problem set 5, it follows that N must be a normal subgroup of G. We will prove that N Z(G. To prove this, suppose that a N, a e. Then a = 7. Then the the conjugacy class of a in G must be contained in N. As explained in class, the cardinality of that conjugacy class divides G / a = 5. Hence the cardinality of that conjugacy class is 1 or 5. Since N has six elements, apart from e, it is clear that at least one of those elements a has just one conjugate. This means that N contains an element a of order 7 such that a Z(G. Hence Z(G {e}. Using problem4 on page 95 again, it follows that G is abelian. We have proved that if G is a group of order 35, then G is abelian. To complete the proof, we now know that G is abelian and that G contains an element b of order 7 and an element c of order 5. (We are using the special case of Cauchy s theorem again. We have bc = cb. Note that (bc 5 = b 5 c 5 = b 5 e, (bc 7 = b 7 c 7 = c 7 = c 2 e and so the order of bc cannot divide 5 or 7. It must divide 35. Hence bc has order 35. It follows that bc is a subgroup of G and has order 35, as does G itself. Therefore, G bc, and so G is indeed a cyclic group. D. Let A = Z/2Z Z/2Z. For each of the following groups G, determine if G has a subgroup isomorphic to A. Justify your answers fully. The group G = S 3 has no subgroup isomorphic to A. Justification: Since G = 6, any subgroup of G has order dividing 6. Hence G cannot have a subgroup of order 4. But A = 4 and any group isomorphic to A must have order 4. The group G = S 4 has a subgroup isomorphic to A, namely the Klein 4group K = { i, (12(34, (13(24, (14(23 } We know that K is a subgroup of S 4, that K = 4, and that none of the elements of K has order 4. Hence K is not cyclic. Hence K is isomorphic to A, using a theorem proved in class about groups of order 4. (We actually proved a theorem describing the isomorphism classes of groups of order p 2, where p is any prime. The quaternionic group G = Q 8 has no subgroup isomorphic to A. In fact, G has only one element of order 2, namely 1. Therefore, if H is a subgroup of G of order 4, then H must contain an element of order 4. Thus, H must be a cyclic group of order 4 and cannot be isomorphic to A.
5 The group G = D 8 has a subgroup isomorphic to A. Using the definition given in class, we can regard D 8 as a subgroup of S 4. That subgroup contains K. As pointed out above, K is isomorphic to A. Hence D 8 contains a subgroup isomorphic to A. The group G = Z/4Z Z/2Z contains a subgroup isomorphic to A. We will use additive notation for G. Since G is abelian, the map φ : G G defined by φ(g = 2g for all g G is a homomorphism. Let H = Ker(φ = { g G 2g = e } where e is the identity element of G. Of course, K is a subgroup of G. We will use the following notation. If a Z, we let [a] 2 = a+2z, which is an element of Z/2Z. We let [a] 4 = a+4z, which is an element of Z/4Z. Using this notation for congruence classes, this subgroup H can be described as follows: H = { ( [0] 4, [0] 2, ( [0]4, [1] 2, ( [2]4, [0] 2, ( [2]4, [1] 2 } Since H has order 4 and every element of H has order 1 or 2, we know that H is isomorphic to A. The group G = Z/48Z has no subgroup isomorphic to A. Justification: Since G is cyclic, every subgroup of G must also be cyclic. But A is not cyclic. Any group isomorphic to A will also fail to be cyclic. Hence no subgroup of G can be isomorphic to A. E. Suppose that G is a group and that H is a subgroup of G such that [G : H] = 2. Suppose that a, b G, but a H and b H. Prove that ab H. Solution: Since [G : H] = 2, it follows that H is a normal subgroup of G. Consider the quotient group G/H. It is a group of order 2. The identity element in that group is H. The other element in that group is x = ah since a H. However, this element of G/H is also bh since b H. That is, x = ah = bh. Since G/H has order 2, the element x is its own inverse. That is, xx = ahbh = H. Therefore, abh = H. Therefore, ab H, as stated. F. This is a question about the group S 5. It concerns the element ( σ = (a Write σ as a product of disjoint cycles. We have σ = (13(254.
6 (b Determine the order of σ. The order of σ is equal to 6, the least common multiple of the lengths of the cycles in the disjoint cycle decomposition for σ. (c Is σ A 5? We have σ = (13(254 = (13(24(25. Hence σ is an odd permutation. Hence σ A 5. (d Find an element τ S 5 such that τστ 1 = σ 1. We have σ 1 = (13 1 (254 1 = (13(245. Let τ S 5 be defined as follows: τ(1 = 1, τ(3 = 3, τ(2 = 2, τ(5 = 4, τ(4 = 5. Then, by the Conjugacy Principle, we have as we wanted. τστ 1 = (τ(1 τ(3 ( τ(2 τ(5 τ(4 = (13(245 = σ 1, (e True or False: Every element of S 5 which has the same order as σ is conjugate to σ in S 5. Justify your answer. The statement is true. The element σ has order 6. However, let σ be any element of order 6 in S 5. We can express σ as a product of disjoint cycles of various lengths. The sum of the lengths must be 5 and their least common multiple must be 6, the order of σ. Thus, σ must be a product of two disjoint cycles, one of length 2 and one of length 3. Therefore, σ has the same cycle decomposition type as σ. As explained in class, this implies that σ is a conjugate of σ in S 5.
Math 210A: Algebra, Homework 5
Math 210A: Algebra, Homework 5 Ian Coley November 5, 2013 Problem 1. Prove that two elements σ and τ in S n are conjugate if and only if type σ = type τ. Suppose first that σ and τ are cycles. Suppose
More informationAnswers to Final Exam
Answers to Final Exam MA441: Algebraic Structures I 20 December 2003 1) Definitions (20 points) 1. Given a subgroup H G, define the quotient group G/H. (Describe the set and the group operation.) The quotient
More informationName: Solutions Final Exam
Instructions. Answer each of the questions on your own paper. Be sure to show your work so that partial credit can be adequately assessed. Put your name on each page of your paper. 1. [10 Points] All of
More informationGroup Theory
Group Theory 2014 2015 Solutions to the exam of 4 November 2014 13 November 2014 Question 1 (a) For every number n in the set {1, 2,..., 2013} there is exactly one transposition (n n + 1) in σ, so σ is
More informationHOMEWORK Graduate Abstract Algebra I May 2, 2004
Math 5331 Sec 121 Spring 2004, UT Arlington HOMEWORK Graduate Abstract Algebra I May 2, 2004 The required text is Algebra, by Thomas W. Hungerford, Graduate Texts in Mathematics, Vol 73, Springer. (it
More informationHomework #11 Solutions
Homework #11 Solutions p 166, #18 We start by counting the elements in D m and D n, respectively, of order 2. If x D m and x 2 then either x is a flip or x is a rotation of order 2. The subgroup of rotations
More informationAlgebra SEP Solutions
Algebra SEP Solutions 17 July 2017 1. (January 2017 problem 1) For example: (a) G = Z/4Z, N = Z/2Z. More generally, G = Z/p n Z, N = Z/pZ, p any prime number, n 2. Also G = Z, N = nz for any n 2, since
More informationPROBLEMS FROM GROUP THEORY
PROBLEMS FROM GROUP THEORY Page 1 of 12 In the problems below, G, H, K, and N generally denote groups. We use p to stand for a positive prime integer. Aut( G ) denotes the group of automorphisms of G.
More information7 Semidirect product. Notes 7 Autumn Definition and properties
MTHM024/MTH74U Group Theory Notes 7 Autumn 20 7 Semidirect product 7. Definition and properties Let A be a normal subgroup of the group G. A complement for A in G is a subgroup H of G satisfying HA = G;
More informationIntroduction to Groups
Introduction to Groups HongJian Lai August 2000 1. Basic Concepts and Facts (1.1) A semigroup is an ordered pair (G, ) where G is a nonempty set and is a binary operation on G satisfying: (G1) a (b c)
More informationDISCRETE MATH (A LITTLE) & BASIC GROUP THEORY  PART 3/3. Contents
DISCRETE MATH (A LITTLE) & BASIC GROUP THEORY  PART 3/3 T.K.SUBRAHMONIAN MOOTHATHU Contents 1. Cayley s Theorem 1 2. The permutation group S n 2 3. Center of a group, and centralizers 4 4. Group actions
More informationDefinitions. Notations. Injective, Surjective and Bijective. Divides. Cartesian Product. Relations. Equivalence Relations
Page 1 Definitions Tuesday, May 8, 2018 12:23 AM Notations " " means "equals, by definition" the set of all real numbers the set of integers Denote a function from a set to a set by Denote the image of
More informationElements of solution for Homework 5
Elements of solution for Homework 5 General remarks How to use the First Isomorphism Theorem A standard way to prove statements of the form G/H is isomorphic to Γ is to construct a homomorphism ϕ : G Γ
More informationSelected exercises from Abstract Algebra by Dummit and Foote (3rd edition).
Selected exercises from Abstract Algebra by Dummit Foote (3rd edition). Bryan Félix Abril 12, 2017 Section 4.1 Exercise 1. Let G act on the set A. Prove that if a, b A b = ga for some g G, then G b = gg
More informationNormal Subgroups and Quotient Groups
Normal Subgroups and Quotient Groups 3202014 A subgroup H < G is normal if ghg 1 H for all g G. Notation: H G. Every subgroup of an abelian group is normal. Every subgroup of index 2 is normal. If H
More informationFall /29/18 Time Limit: 75 Minutes
Math 411: Abstract Algebra Fall 2018 Midterm 10/29/18 Time Limit: 75 Minutes Name (Print): Solutions JHUID: This exam contains 8 pages (including this cover page) and 6 problems. Check to see if any pages
More informationSolutions to oddnumbered exercises Peter J. Cameron, Introduction to Algebra, Chapter 3
Solutions to oddnumbered exercises Peter J. Cameron, Introduction to Algebra, Chapter 3 3. (a) Yes; (b) No; (c) No; (d) No; (e) Yes; (f) Yes; (g) Yes; (h) No; (i) Yes. Comments: (a) is the additive group
More informationSimple groups and the classification of finite groups
Simple groups and the classification of finite groups 1 Finite groups of small order How can we describe all finite groups? Before we address this question, let s write down a list of all the finite groups
More informationMATH 28A MIDTERM 2 INSTRUCTOR: HAROLD SULTAN
NAME: MATH 28A MIDTERM 2 INSTRUCTOR: HAROLD SULTAN 1. INSTRUCTIONS (1) Timing: You have 80 minutes for this midterm. (2) Partial Credit will be awarded. Please show your work and provide full solutions,
More informationThe Outer Automorphism of S 6
Meena Jagadeesan 1 Karthik Karnik 2 Mentor: Akhil Mathew 1 Phillips Exeter Academy 2 Massachusetts Academy of Math and Science PRIMES Conference, May 2016 What is a Group? A group G is a set of elements
More informationMA441: Algebraic Structures I. Lecture 26
MA441: Algebraic Structures I Lecture 26 10 December 2003 1 (page 179) Example 13: A 4 has no subgroup of order 6. BWOC, suppose H < A 4 has order 6. Then H A 4, since it has index 2. Thus A 4 /H has order
More informationAssigment 1. 1 a b. 0 1 c A B = (A B) (B A). 3. In each case, determine whether G is a group with the given operation.
1. Show that the set G = multiplication. Assigment 1 1 a b 0 1 c a, b, c R 0 0 1 is a group under matrix 2. Let U be a set and G = {A A U}. Show that G ia an abelian group under the operation defined by
More informationGroup Theory. 1. Show that Φ maps a conjugacy class of G into a conjugacy class of G.
Group Theory Jan 2012 #6 Prove that if G is a nonabelian group, then G/Z(G) is not cyclic. Aug 2011 #9 (Jan 2010 #5) Prove that any group of order p 2 is an abelian group. Jan 2012 #7 G is nonabelian nite
More informationSection 10: Counting the Elements of a Finite Group
Section 10: Counting the Elements of a Finite Group Let G be a group and H a subgroup. Because the right cosets are the family of equivalence classes with respect to an equivalence relation on G, it follows
More informationANALYSIS OF SMALL GROUPS
ANALYSIS OF SMALL GROUPS 1. Big Enough Subgroups are Normal Proposition 1.1. Let G be a finite group, and let q be the smallest prime divisor of G. Let N G be a subgroup of index q. Then N is a normal
More informationTeddy Einstein Math 4320
Teddy Einstein Math 4320 HW4 Solutions Problem 1: 2.92 An automorphism of a group G is an isomorphism G G. i. Prove that Aut G is a group under composition. Proof. Let f, g Aut G. Then f g is a bijective
More information(5.11) (Second Isomorphism Theorem) If K G and N G, then K/(N K) = NK/N. PF: Verify N HK. Find a homomorphism f : K HK/N with ker(f) = (N K).
Lecture Note of Week 3 6. Normality, Quotients and Homomorphisms (5.7) A subgroup N satisfying any one properties of (5.6) is called a normal subgroup of G. Denote this fact by N G. The homomorphism π
More informationSection 15 Factorgroup computation and simple groups
Section 15 Factorgroup computation and simple groups Instructor: Yifan Yang Fall 2006 Outline Factorgroup computation Simple groups The problem Problem Given a factor group G/H, find an isomorphic group
More informationProblem 1. Let I and J be ideals in a ring commutative ring R with 1 R. Recall
I. TakeHome Portion: Math 350 Final Exam Due by 5:00pm on Tues. 5/12/15 No resources/devices other than our class textbook and class notes/handouts may be used. You must work alone. Choose any 5 problems
More informationYale University Department of Mathematics Math 350 Introduction to Abstract Algebra Fall Midterm Exam Review Solutions
Yale University Department of Mathematics Math 350 Introduction to Abstract Algebra Fall 2015 Midterm Exam Review Solutions Practice exam questions: 1. Let V 1 R 2 be the subset of all vectors whose slope
More informationBasic Definitions: Group, subgroup, order of a group, order of an element, Abelian, center, centralizer, identity, inverse, closed.
Math 546 Review Exam 2 NOTE: An (*) at the end of a line indicates that you will not be asked for the proof of that specific item on the exam But you should still understand the idea and be able to apply
More informationMODEL ANSWERS TO THE FIFTH HOMEWORK
MODEL ANSWERS TO THE FIFTH HOMEWORK 1. Chapter 3, Section 5: 1 (a) Yes. Given a and b Z, φ(ab) = [ab] = [a][b] = φ(a)φ(b). This map is clearly surjective but not injective. Indeed the kernel is easily
More information1. Group Theory Permutations.
1.1. Permutations. 1. Group Theory Problem 1.1. Let G be a subgroup of S n of index 2. Show that G = A n. Problem 1.2. Find two elements of S 7 that have the same order but are not conjugate. Let π S 7
More informationDIHEDRAL GROUPS II KEITH CONRAD
DIHEDRAL GROUPS II KEITH CONRAD We will characterize dihedral groups in terms of generators and relations, and describe the subgroups of D n, including the normal subgroups. We will also introduce an infinite
More informationALGEBRA HOMEWORK SET 2. Due by class time on Wednesday 14 September. Homework must be typeset and submitted by as a PDF file.
ALGEBRA HOMEWORK SET 2 JAMES CUMMINGS (JCUMMING@ANDREW.CMU.EDU) Due by class time on Wednesday 14 September. Homework must be typeset and submitted by email as a PDF file. (1) Let H and N be groups and
More informationModern Algebra I. Circle the correct answer; no explanation is required. Each problem in this section counts 5 points.
1 2 3 style total Math 415 Please print your name: Answer Key 1 True/false Circle the correct answer; no explanation is required. Each problem in this section counts 5 points. 1. Every group of order 6
More informationMATH 101: ALGEBRA I WORKSHEET, DAY #3. Fill in the blanks as we finish our first pass on prerequisites of group theory.
MATH 101: ALGEBRA I WORKSHEET, DAY #3 Fill in the blanks as we finish our first pass on prerequisites of group theory 1 Subgroups, cosets Let G be a group Recall that a subgroup H G is a subset that is
More informationCHAPTER 9. Normal Subgroups and Factor Groups. Normal Subgroups
Normal Subgroups CHAPTER 9 Normal Subgroups and Factor Groups If H apple G, we have seen situations where ah 6= Ha 8 a 2 G. Definition (Normal Subgroup). A subgroup H of a group G is a normal subgroup
More informationCONSEQUENCES OF THE SYLOW THEOREMS
CONSEQUENCES OF THE SYLOW THEOREMS KEITH CONRAD For a group theorist, Sylow s Theorem is such a basic tool, and so fundamental, that it is used almost without thinking, like breathing. Geoff Robinson 1.
More information( ) 3 = ab 3 a!1. ( ) 3 = aba!1 a ( ) = 4 " 5 3 " 4 = ( )! 2 3 ( ) =! 5 4. Math 546 Problem Set 15
Math 546 Problem Set 15 1. Let G be a finite group. (a). Suppose that H is a subgroup of G and o(h) = 4. Suppose that K is a subgroup of G and o(k) = 5. What is H! K (and why)? Solution: H! K = {e} since
More information3. G. Groups, as men, will be known by their actions.  Guillermo Moreno
3.1. The denition. 3. G Groups, as men, will be known by their actions.  Guillermo Moreno D 3.1. An action of a group G on a set X is a function from : G X! X such that the following hold for all g, h
More informationSolutions for Homework Assignment 5
Solutions for Homework Assignment 5 Page 154, Problem 2. Every element of C can be written uniquely in the form a + bi, where a,b R, not both equal to 0. The fact that a and b are not both 0 is equivalent
More informationMaximal noncommuting subsets of groups
Maximal noncommuting subsets of groups Umut Işık March 29, 2005 Abstract Given a finite group G, we consider the problem of finding the maximal size nc(g) of subsets of G that have the property that no
More informationJohns Hopkins University, Department of Mathematics Abstract Algebra  Spring 2009 Midterm
Johns Hopkins University, Department of Mathematics 110.401 Abstract Algebra  Spring 2009 Midterm Instructions: This exam has 8 pages. No calculators, books or notes allowed. You must answer the first
More informationENTRY GROUP THEORY. [ENTRY GROUP THEORY] Authors: started Mark Lezama: October 2003 Literature: Algebra by Michael Artin, Mathworld.
ENTRY GROUP THEORY [ENTRY GROUP THEORY] Authors: started Mark Lezama: October 2003 Literature: Algebra by Michael Artin, Mathworld Group theory [Group theory] is studies algebraic objects called groups.
More informationMath 120. Groups and Rings Midterm Exam (November 8, 2017) 2 Hours
Math 120. Groups and Rings Midterm Exam (November 8, 2017) 2 Hours Name: Please read the questions carefully. You will not be given partial credit on the basis of having misunderstood a question, and please
More informationCourse 311: Abstract Algebra Academic year
Course 311: Abstract Algebra Academic year 200708 D. R. Wilkins Copyright c David R. Wilkins 1997 2007 Contents 1 Topics in Group Theory 1 1.1 Groups............................... 1 1.2 Examples of Groups.......................
More informationCosets. gh = {gh h H}. Hg = {hg h H}.
Cosets 1042006 If H is a subgroup of a group G, a left coset of H in G is a subset of the form gh = {gh h H}. A right coset of H in G is a subset of the form Hg = {hg h H}. The collection of left cosets
More informationABSTRACT ALGEBRA: REVIEW PROBLEMS ON GROUPS AND GALOIS THEORY
ABSTRACT ALGEBRA: REVIEW PROBLEMS ON GROUPS AND GALOIS THEORY John A. Beachy Northern Illinois University 2000 ii J.A.Beachy This is a supplement to Abstract Algebra, Second Edition by John A. Beachy and
More informationφ(xy) = (xy) n = x n y n = φ(x)φ(y)
Groups 1. (Algebra Comp S03) Let A, B and C be normal subgroups of a group G with A B. If A C = B C and AC = BC then prove that A = B. Let b B. Since b = b1 BC = AC, there are a A and c C such that b =
More informationMATH 113 FINAL EXAM December 14, 2012
p.1 MATH 113 FINAL EXAM December 14, 2012 This exam has 9 problems on 18 pages, including this cover sheet. The only thing you may have out during the exam is one or more writing utensils. You have 180
More informationSupplement. Dr. Bob s Modern Algebra Glossary Based on Fraleigh s A First Course on Abstract Algebra, 7th Edition, Sections 0 through IV.
Glossary 1 Supplement. Dr. Bob s Modern Algebra Glossary Based on Fraleigh s A First Course on Abstract Algebra, 7th Edition, Sections 0 through IV.23 Abelian Group. A group G, (or just G for short) is
More informationConverse to Lagrange s Theorem Groups
Converse to Lagrange s Theorem Groups Blain A Patterson Youngstown State University May 10, 2013 History In 1771 an Italian mathematician named Joseph Lagrange proved a theorem that put constraints on
More informationSUPPLEMENT ON THE SYMMETRIC GROUP
SUPPLEMENT ON THE SYMMETRIC GROUP RUSS WOODROOFE I presented a couple of aspects of the theory of the symmetric group S n differently than what is in Herstein. These notes will sketch this material. You
More informationIIT Mumbai 2015 MA 419, Basic Algebra Tutorial Sheet1
IIT Mumbai 2015 MA 419, Basic Algebra Tutorial Sheet1 Let Σ be the set of all symmetries of the plane Π. 1. Give examples of s, t Σ such that st ts. 2. If s, t Σ agree on three noncollinear points, then
More informationAbstract Algebra, Second Edition, by John A. Beachy and William D. Blair. Corrections and clarifications
1 Abstract Algebra, Second Edition, by John A. Beachy and William D. Blair Corrections and clarifications Note: Some corrections were made after the first printing of the text. page 9, line 8 For of the
More informationABSTRACT ALGEBRA 1, LECTURES NOTES 5: SUBGROUPS, CONJUGACY, NORMALITY, QUOTIENT GROUPS, AND EXTENSIONS.
ABSTRACT ALGEBRA 1, LECTURES NOTES 5: SUBGROUPS, CONJUGACY, NORMALITY, QUOTIENT GROUPS, AND EXTENSIONS. ANDREW SALCH 1. Subgroups, conjugacy, normality. I think you already know what a subgroup is: Definition
More informationSUMMARY ALGEBRA I LOUISPHILIPPE THIBAULT
SUMMARY ALGEBRA I LOUISPHILIPPE THIBAULT Contents 1. Group Theory 1 1.1. Basic Notions 1 1.2. Isomorphism Theorems 2 1.3. Jordan Holder Theorem 2 1.4. Symmetric Group 3 1.5. Group action on Sets 3 1.6.
More information1 Chapter 6  Exercise 1.8.cf
1 CHAPTER 6  EXERCISE 1.8.CF 1 1 Chapter 6  Exercise 1.8.cf Determine 1 The Class Equation of the dihedral group D 5. Note first that D 5 = 10 = 5 2. Hence every conjugacy class will have order 1, 2
More informationREMARKS 7.6: Let G be a finite group of order n. Then Lagrange's theorem shows that the order of every subgroup of G divides n; equivalently, if k is
FIRSTYEAR GROUP THEORY 7 LAGRANGE'S THEOREM EXAMPLE 7.1: Set G = D 3, where the elements of G are denoted as usual by e, a, a 2, b, ab, a 2 b. Let H be the cyclic subgroup of G generated by b; because
More informationMATH 420 FINAL EXAM J. Beachy, 5/7/97
MATH 420 FINAL EXAM J. Beachy, 5/7/97 1. (a) For positive integers a and b, define gcd(a, b). (b) Compute gcd(1776, 1492). (c) Show that if a, b, c are positive integers, then gcd(a, bc) = 1 if and only
More informationMath 345 Sp 07 Day 7. b. Prove that the image of a homomorphism is a subring.
Math 345 Sp 07 Day 7 1. Last time we proved: a. Prove that the kernel of a homomorphism is a subring. b. Prove that the image of a homomorphism is a subring. c. Let R and S be rings. Suppose R and S are
More informationSchool of Mathematics and Statistics. MT5824 Topics in Groups. Problem Sheet I: Revision and ReActivation
MRQ 2009 School of Mathematics and Statistics MT5824 Topics in Groups Problem Sheet I: Revision and ReActivation 1. Let H and K be subgroups of a group G. Define HK = {hk h H, k K }. (a) Show that HK
More informationThe Class Equation X = Gx. x X/G
The Class Equation 992012 If X is a Gset, X is partitioned by the Gorbits. So if X is finite, X = x X/G ( x X/G means you should take one representative x from each orbit, and sum over the set of representatives.
More information15 Permutation representations and Gsets
15 Permutation representations and Gsets Recall. If C is a category and c C then Aut(c) =the group of automorphisms of c 15.1 Definition. A representation of a group G in a category C is a homomorphism
More informationSection III.15. FactorGroup Computations and Simple Groups
III.15 FactorGroup Computations 1 Section III.15. FactorGroup Computations and Simple Groups Note. In this section, we try to extract information about a group G by considering properties of the factor
More informationStab(t) = {h G h t = t} = {h G h (g s) = g s} = {h G (g 1 hg) s = s} = g{k G k s = s} g 1 = g Stab(s)g 1.
1. Group Theory II In this section we consider groups operating on sets. This is not particularly new. For example, the permutation group S n acts on the subset N n = {1, 2,...,n} of N. Also the group
More informationAlgebra Exercises in group theory
Algebra 3 2010 Exercises in group theory February 2010 Exercise 1*: Discuss the Exercises in the sections 1.11.3 in Chapter I of the notes. Exercise 2: Show that an infinite group G has to contain a nontrivial
More informationMATH 436 Notes: Cyclic groups and Invariant Subgroups.
MATH 436 Notes: Cyclic groups and Invariant Subgroups. Jonathan Pakianathan September 30, 2003 1 Cyclic Groups Now that we have enough basic tools, let us go back and study the structure of cyclic groups.
More informationMath 451, 01, Exam #2 Answer Key
Math 451, 01, Exam #2 Answer Key 1. (25 points): If the statement is always true, circle True and prove it. If the statement is never true, circle False and prove that it can never be true. If the statement
More information120A LECTURE OUTLINES
120A LECTURE OUTLINES RUI WANG CONTENTS 1. Lecture 1. Introduction 1 2 1.1. An algebraic object to study 2 1.2. Group 2 1.3. Isomorphic binary operations 2 2. Lecture 2. Introduction 2 3 2.1. The multiplication
More information1 Finite abelian groups
Last revised: May 16, 2014 A.Miller M542 www.math.wisc.edu/ miller/ Each Problem is due one week from the date it is assigned. Do not hand them in early. Please put them on the desk in front of the room
More informationCosets, factor groups, direct products, homomorphisms, isomorphisms
Cosets, factor groups, direct products, homomorphisms, isomorphisms Sergei Silvestrov Spring term 2011, Lecture 11 Contents of the lecture Cosets and the theorem of Lagrange. Direct products and finitely
More informationLecture 5.6: The Sylow theorems
Lecture 5.6: The Sylow theorems Matthew Macauley Department of Mathematical Sciences Clemson University http://www.math.clemson.edu/~macaule/ Math 4120, Modern Algebra M. Macauley (Clemson) Lecture 5.6:
More informationKevin James. pgroups, Nilpotent groups and Solvable groups
pgroups, Nilpotent groups and Solvable groups Definition A maximal subgroup of a group G is a proper subgroup M G such that there are no subgroups H with M < H < G. Definition A maximal subgroup of a
More informationPhysics 251 Solution Set 1 Spring 2017
Physics 5 Solution Set Spring 07. Consider the set R consisting of pairs of real numbers. For (x,y) R, define scalar multiplication by: c(x,y) (cx,cy) for any real number c, and define vector addition
More informationS11MTH 3175 Group Theory (Prof.Todorov) Final (Practice Some Solutions) 2 BASIC PROPERTIES
S11MTH 3175 Group Theory (Prof.Todorov) Final (Practice Some Solutions) 2 BASIC PROPERTIES 1 Some Definitions For your convenience, we recall some of the definitions: A group G is called simple if it has
More informationMAT534 Fall 2013 Practice Midterm I The actual midterm will consist of five problems.
MAT534 Fall 2013 Practice Midterm I The actual midterm will consist of five problems. Problem 1 Find all homomorphisms a) Z 6 Z 6 ; b) Z 6 Z 18 ; c) Z 18 Z 6 ; d) Z 12 Z 15 ; e) Z 6 Z 25 Proof. a)ψ(1)
More informationModern Algebra Homework 9b Chapter 9 Read Complete 9.21, 9.22, 9.23 Proofs
Modern Algebra Homework 9b Chapter 9 Read 9.19.3 Complete 9.21, 9.22, 9.23 Proofs Megan Bryant November 20, 2013 First Sylow Theorem If G is a group and p n is the highest power of p dividing G, then
More informationAlgebraic Structures Exam File Fall 2013 Exam #1
Algebraic Structures Exam File Fall 2013 Exam #1 1.) Find all four solutions to the equation x 4 + 16 = 0. Give your answers as complex numbers in standard form, a + bi. 2.) Do the following. a.) Write
More informationREU 2007 Discrete Math Lecture 2
REU 2007 Discrete Math Lecture 2 Instructor: László Babai Scribe: Shawn Drenning June 19, 2007. Proofread by instructor. Last updated June 20, 1 a.m. Exercise 2.0.1. Let G be an abelian group and A G be
More informationMA441: Algebraic Structures I. Lecture 18
MA441: Algebraic Structures I Lecture 18 5 November 2003 1 Review from Lecture 17: Theorem 6.5: Aut(Z/nZ) U(n) For every positive integer n, Aut(Z/nZ) is isomorphic to U(n). The proof used the map T :
More informationFinite groups determined by an inequality of the orders of their elements
Publ. Math. Debrecen 80/34 (2012), 457 463 DOI: 10.5486/PMD.2012.5168 Finite groups determined by an inequality of the orders of their elements By MARIUS TĂRNĂUCEANU (Iaşi) Abstract. In this note we introduce
More informationLecture Note of Week 2
Lecture Note of Week 2 2. Homomorphisms and Subgroups (2.1) Let G and H be groups. A map f : G H is a homomorphism if for all x, y G, f(xy) = f(x)f(y). f is an isomorphism if it is bijective. If f : G
More informationChapter 5 Groups of permutations (bijections) Basic notation and ideas We study the most general type of groups  groups of permutations
Chapter 5 Groups of permutations (bijections) Basic notation and ideas We study the most general type of groups  groups of permutations (bijections). Definition A bijection from a set A to itself is also
More informationINTRODUCTION TO THE GROUP THEORY
Lecture Notes on Structure of Algebra INTRODUCTION TO THE GROUP THEORY By : Drs. Antonius Cahya Prihandoko, M.App.Sc email: antoniuscp.fkip@unej.ac.id Mathematics Education Study Program Faculty of Teacher
More informationFoundations of Cryptography
Foundations of Cryptography Ville Junnila viljun@utu.fi Department of Mathematics and Statistics University of Turku 2015 Ville Junnila viljun@utu.fi Lecture 7 1 of 18 Cosets Definition 2.12 Let G be a
More informationSolutions to Assignment 4
1. Let G be a finite, abelian group written additively. Let x = g G g, and let G 2 be the subgroup of G defined by G 2 = {g G 2g = 0}. (a) Show that x = g G 2 g. (b) Show that x = 0 if G 2 = 2. If G 2
More informationDefinitions, Theorems and Exercises. Abstract Algebra Math 332. Ethan D. Bloch
Definitions, Theorems and Exercises Abstract Algebra Math 332 Ethan D. Bloch December 26, 2013 ii Contents 1 Binary Operations 3 1.1 Binary Operations............................... 4 1.2 Isomorphic Binary
More informationBefore you begin read these instructions carefully.
MATHEMATICAL TRIPOS Part IA Tuesday, 4 June, 2013 1:30 pm to 4:30 pm PAPER 3 Before you begin read these instructions carefully. The examination paper is divided into two sections. Each question in Section
More informationMath 4400, Spring 08, Sample problems Final Exam.
Math 4400, Spring 08, Sample problems Final Exam. 1. Groups (1) (a) Let a be an element of a group G. Define the notions of exponent of a and period of a. (b) Suppose a has a finite period. Prove that
More information0 Sets and Induction. Sets
0 Sets and Induction Sets A set is an unordered collection of objects, called elements or members of the set. A set is said to contain its elements. We write a A to denote that a is an element of the set
More informationA PROOF OF BURNSIDE S p a q b THEOREM
A PROOF OF BURNSIDE S p a q b THEOREM OBOB Abstract. We prove that if p and q are prime, then any group of order p a q b is solvable. Throughout this note, denote by A the set of algebraic numbers. We
More informationAlgebra. Travis Dirle. December 4, 2016
Abstract Algebra 2 Algebra Travis Dirle December 4, 2016 2 Contents 1 Groups 1 1.1 Semigroups, Monoids and Groups................ 1 1.2 Homomorphisms and Subgroups................. 2 1.3 Cyclic Groups...........................
More informationNormal Subgroups and Factor Groups
Normal Subgroups and Factor Groups Subject: Mathematics Course Developer: Harshdeep Singh Department/ College: Assistant Professor, Department of Mathematics, Sri Venkateswara College, University of Delhi
More informationSF2729 GROUPS AND RINGS LECTURE NOTES
SF2729 GROUPS AND RINGS LECTURE NOTES 20110301 MATS BOIJ 6. THE SIXTH LECTURE  GROUP ACTIONS In the sixth lecture we study what happens when groups acts on sets. 1 Recall that we have already when looking
More informationAlgebra Exam, Spring 2017
Algebra Exam, Spring 2017 There are 5 problems, some with several parts. Easier parts count for less than harder ones, but each part counts. Each part may be assumed in later parts and problems. Unjustified
More informationMATH 1530 ABSTRACT ALGEBRA Selected solutions to problems. a + b = a + b,
MATH 1530 ABSTRACT ALGEBRA Selected solutions to problems Problem Set 2 2. Define a relation on R given by a b if a b Z. (a) Prove that is an equivalence relation. (b) Let R/Z denote the set of equivalence
More informationYour Name MATH 435, EXAM #1
MATH 435, EXAM #1 Your Name You have 50 minutes to do this exam. No calculators! No notes! For proofs/justifications, please use complete sentences and make sure to explain any steps which are questionable.
More informationGROUP ACTIONS RYAN C. SPIELER
GROUP ACTIONS RYAN C. SPIELER Abstract. In this paper, we examine group actions. Groups, the simplest objects in Algebra, are sets with a single operation. We will begin by defining them more carefully
More information