Math 210A: Algebra, Homework 5


 Maurice Harrison
 2 years ago
 Views:
Transcription
1 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 further that σ and τ are conjugate so that σ = ρτρ 1. By earlier considerations, if τ = (a 1... a m ), where a i {1,..., n}, then σ = ρτρ 1 = (ρ(a 1 )... ρ(a m )). Hence σ and τ have the same number of symbols in their cycle, so type σ = type τ. Conversely, if type σ = type τ, then write σ = (a 1... a m ) and τ = (b 1... b m ). Define ρ S n so that ρ(a i ) = b i for each i {1,..., m}. Such a ρ is guaranteed to exist since we do not have ρ(a i ) = ρ(a j ) for any i j as b i b j for any i j. BY the above, ρσρ 1 = (ρ(a 1 )... ρ(a m )) = τ, so these cycles are conjugate. Now suppose that σ and τ are two permutations. Then we may write σ = σ 1 σ l, τ = τ 1 τ m where σ i, τ j are disjoint cycles. Then if σ = ρτρ 1, we have σi = ρ ( τj ) ρ 1 = ρτ j ρ 1. By the above reasoning, we must have type σ i = type τ j for some reordering of the cycles, so that σ and τ have the same type. Conversely, if type σ = type τ, then we have l = m and type σ i = type τ i for an appropriate rearrangement. Therefore For each there exists ρ i such that σ i = ρ i τ i ρ 1 i. Since the τ i are disjoint, the ρ i may be chosen disjoint so that ρ i ρ j = ρ j ρ i and ρ j τ i ρ 1 j = τ i for i j. Let ρ = ρ i. Then ρτρ 1 = ρτ j ρ 1 = ρ j τ j ρ 1 j = σ j = σ. Therefore the cycles are conjugate, so we are done. Problem 2. Prove that S 4 acts by conjugation on the set of nontrivial elements of the normal subgroup N = {e, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)}. Using this action prove that S 4 /N = S 3. 1
2 It is clear that N is a subgroup of S 4 and, since it is comprised of full conjugacy classes, it is normal. Therefore the action of S 4 on N is welldefined. Further, since conjugation preserves cycle type (see problem 1), the conjugate of any 22cycle is also a 22cycle, so S 4 acts in particular on the nontrivial elements of N, which are the only 22cycles. Therefore there is a homomorphism ϕ from S 4 to S(X) = = S 3, where X = N \ {e}. First, we calculate the kernel of ϕ. We claim that it is N. Since N is an abelian subgroup, we have στσ = τ for every στ N. Therefore we have N ker ϕ. Further, ϕ is not trivial since S 4 itself is not abelian, so ker ϕ S 4. By previous considerations A 4 and N are the only proper normal subgroups of S 4. Suppose that ker ϕ = A 4. Then conjugation by (1 2 3) would be trivial for every element of N. However, (1 2 3)(1 2)(3 4)(3 2 1) = (1 4)(2 3). Therefore ker ϕ A 4, so we must have ker ϕ = N. Then since S 4 /N = 24/4 = 6 = S 3, we must have im ϕ = S 3. By the first isomorphism theorem, S 4 /N = S 3, and we are done. Problem 3. Let σ = (1 2 n) S n. Show that the conjugacy class of σ has (n 1)! elements. Show that the centraliser of σ is the cyclic subgroup generated by σ. As in problem 1, all ncycles are conjugate in S n. An ncycle is uniquely determined by ordering the symbols 1,..., n up to rearrangement. There are n! orderings of the symbols and n rearrangements for each ncycle, so in total there are (n 1)! ncycles. Therefore the conjugacy class of σ has (n 1)! elements. We now use the orbitstabiliser theorem. Let G = S n. Let Gσ be the orbit of σ and G σ its stabiliser. Then Gσ = G : G σ = (n 1)! = n!/ G σ = G σ = n. Since the stabiliser under conjugation is the centraliser, we know the centraliser of σ has n elements. Since σ commutes with itself, we have σ G σ. But since σ = n, we must have σ = G σ, so we are done. Problem 4. Let H be a proper subgroup of a finite group G. Suppose that G does not divide G : H!. Then G contains a nontrivial normal subgroup N such that N is a subgroup of H. In particular, G is not simple. Let G act on the cosets G/H by left translation. This induces a homomorphism ϕ : G S(G/H). If ϕ were injective, then im ϕ = G would be a subgroup of S(G/H). By Cayley s theorem, we would have G G : H!, which is false by assumption. Hence ϕ is not an injection, so ker ϕ is a nontrivial normal subgroup of G. Further, if g ker ϕ, we have gh = H, hence g H. Therefore ker ϕ is a normal subgroup of G contained in H. This completes the proof. 2
3 Problem 5. Prove that all groups of order 2p n and 4p n are not simple. If p is an odd prime, then there exists a Sylow psubgroup P of order p n. Then if G = 2p n, then G : P = 2, so P is normal, hence the group is not simple. Now let G = 4p n and suppose P is not normal in G. Then there exists another Sylow psubgroup Q such that P Q. Let H = P Q. Then we have 4p n P Q = P Q H = p2n H. Therefore H p n /4 > p n 2. Further, H is a pgroup and H < p n so we must have H = p n 1. Therefore since P : H = Q : H = p is the smallest prime dividing p n, we must have H P and H Q. Hence P Q N G (H), the normaliser of H. But as above we have calculated P Q = p2n p n 1 = pn+1 > 2p n = P Q = 4p n. Therefore N G (H) = G, so H G. Therefore G is not simple. If p = 2, then groups of order 2p n and 4p n are 2groups. By previous considerations, pgroups have nontrivial centres. Therefore groups of order 2p n and 4p n are not simple. This completes the proof. Problem 6. (a) Let N G be a subgroup. Prove that if H is contained in the centre of G and G/N is cyclic, then G is abelian. (b) Prove that any group of order p 2 is abelian. We follow the solution for Problem 5 on Homework 2. (a) We use the definition N G if and only if xnx 1 N for all x G. Let g N. Then xgx 1 = xx 1 g = g, since g Z(G), and hence xgx 1 N for all x G, g N. Hence N is normal. We claim that if G/N is cyclic that G is abelian. Write G/N = xn. Now let a, b G such that a (xz(g)) m = x m Z(G) and b x n Z(G). Then a = x m g 1 and b = x n g 2 for g 1, g 2 N. Then ab = (x m g 1 )(x n g 2 ) = x m (g 1 x n )g 2 = x m (x n g 1 )g 2 = (x m x n )(g 1 g 2 ) = (x n x m )(g 2 g 1 ) = x n (x m g 2 )(g 1 ) = x n (g 2 x m )g 1 = (x n g 2 )(x m g 1 ) = ba. We use the fact that g 1, g 2 Z(G) and that x commutes with itself freely. This shows that G is abelian. 3
4 (b) Let G act on itself by conjugation. Then by the class equation p 2 = G = Z(G) + G : N G (g). nontrivial conjugacy classes By Lagrange s theorem, G : N G (g) = G / N G (g) and since these conjugacy classes are proper and nontrivial, p G : N G (g). Since p p 2, we must have p Z(G), so Z(G) is nontrivial. If Z(G) = p 2, we are done. Suppose Z(G) = p. By 5(a), Z(G) is normal so we may examine the factor group G/Z(G). Since G/Z(G) = G / Z(G) = p, by earlier considerations G/Z(G) must be the cyclic group of order p, which implies that G is abelian, contradicting Z(G) = p. Therefore Z(G) = G and we are done. Problem 7. Let G be a nonabelian group of order p 3. Prove that the centre of G coincides with [G, G]. By earlier considerations, pgroups have nontrivial centres. Therefore Z(G) has order p or p 2 (since it is nonabelian). Suppose that Z(G) has order p 2. Then G/Z(G) = Z/pZ is cyclic, so by an earlier problem, G is abelian, which is a contradiction. Therefore we must have Z(G) = p. As such, G/Z(G) = p 2, and by the preceding problem, G/Z(G) is abelian. Since [G, G] has the property that it is the smallest subgroup N such that G/N is abelian, we must have [G, G] Z(G). But since Z(G) = p and [G, G] p (since G is nonabelian), we must have [G, G] = Z(G). This completes the proof. Problem 8. Let G be a semidirect product of a cyclic normal subgroup N of order n and an abelian group K. Show that if K is relatively prime to ϕ(n), then G is abelian. The semidirect product G = N K is given by a homomorphism ϕ : K Aut N such that nk n k = n ϕ k (n )kk where ϕ(k) = ϕ k Aut N. By earlier considerations, we know that Aut N = Aut Z/nZ = (Z/nZ). Further, (Z/nZ) = ϕ(n). By Cayley s theorem, im ϕ divides ϕ(n), and further im ϕ divides K. Therefore im ϕ divides gcd( K, ϕ(n)) = 1, so ϕ must be the trivial homomorphism. Therefore we have nk n k = nn kk for every n N, k K. Therefore N and K commute with each other, and since N and K are abelian, the whole group is abelian. This completes the proof. Problem 9. Determine the centre of D 2n. 4
5 First, the case of n = 1 is trivial since D 2 = Z/2Z. in the case that n = 2, we have D 4 = Z/2Z Z/2Z, so the group itself is abelian. Now let D 2n = r, s : r n = s 2 = 1, rs = sr 1. Let σ Z = Z(D 2n ). Suppose that σ = r m. Then r m sr k = r m k s? = r (m+k) s = sr m+k = sr k r m. This is only true if (k + m) m k (mod n) for all k, i.e. m m (mod n), which would mean m = n/2 (which only exists for even n). It is also clear that r m r k = r k r m for any k. Thus r m commutes with every element if m = n/2. Now suppose σ = s. Then s r r s in the case that n > 2. Hence s / Z. Finally, suppose that σ = sr m. Then sr m sr k = r m k? = r k m = sr k sr m. This is only true if m k k m (mod n) for all k, i.e. 2k = 2m (mod n) for all k, which is impossible if n > 2. Therefore no element of this form exists. Therefore we summarise the result: D 2n n = 1, 2 Z(D 2n ) = {e} n > 2, even {e, r n/2 } n > 2, odd Problem 10. Show that the exact sequence is not split. 0 Z Q Q/Z 0 Let i : Z Q and π : Q Q/Z be the injective and surjective maps in the exact sequence. We claim that Q = Q/Z Z as groups, so the sequence is not split. If the sequence is split, the isomorphism ϕ : Q Q/Z Z must be given by p/q (p /q, n) where 0 p /q < 1 and n + p /q = p/q so that i(n) = (0, n) and π(p/q) = p /q. Every element q Q has infinite order, so Q is torsionfree. However, every element of Q/Z is torsion, since an arbitrary element p/q has order q. In particular, let 1/n Q. Then (0, 1) = ϕ(1) = ϕ(n (1/n)) = n ϕ(1/n) = n (1/n, 0) = (0, 0), but this is absurd. Therefore such a ϕ cannot exist, so the sequence is not split. 5
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
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 =
More informationMath 210A: Algebra, Homework 6
Math 210A: Algebra, Homework 6 Ian Coley November 13, 2013 Problem 1 For every two nonzero integers n and m construct an exact sequence For which n and m is the sequence split? 0 Z/nZ Z/mnZ Z/mZ 0 Let
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 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 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 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 informationGroups and Symmetries
Groups and Symmetries Definition: Symmetry A symmetry of a shape is a rigid motion that takes vertices to vertices, edges to edges. Note: A rigid motion preserves angles and distances. Definition: Group
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 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 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 informationTwo subgroups and semidirect products
Two subgroups and semidirect products 1 First remarks Throughout, we shall keep the following notation: G is a group, written multiplicatively, and H and K are two subgroups of G. We define the subset
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 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 informationGroup Theory (Math 113), Summer 2014
Group Theory (Math 113), Summer 2014 George Melvin University of California, Berkeley (July 8, 2014 corrected version) Abstract These are notes for the first half of the upper division course Abstract
More informationExercises on chapter 1
Exercises on chapter 1 1. Let G be a group and H and K be subgroups. Let HK = {hk h H, k K}. (i) Prove that HK is a subgroup of G if and only if HK = KH. (ii) If either H or K is a normal subgroup of G
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 informationHomework #5 Solutions
Homework #5 Solutions p 83, #16. In order to find a chain a 1 a 2 a n of subgroups of Z 240 with n as large as possible, we start at the top with a n = 1 so that a n = Z 240. In general, given a i we will
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 informationMath 210B: Algebra, Homework 4
Math 210B: Algebra, Homework 4 Ian Coley February 5, 2014 Problem 1. Let S be a multiplicative subset in a commutative ring R. Show that the localisation functor RMod S 1 RMod, M S 1 M, is exact. First,
More informationits image and kernel. A subgroup of a group G is a nonempty subset K of G such that k 1 k 1
10 Chapter 1 Groups 1.1 Isomorphism theorems Throughout the chapter, we ll be studying the category of groups. Let G, H be groups. Recall that a homomorphism f : G H means a function such that f(g 1 g
More informationALGEBRA QUALIFYING EXAM PROBLEMS
ALGEBRA QUALIFYING EXAM PROBLEMS Kent State University Department of Mathematical Sciences Compiled and Maintained by Donald L. White Version: August 29, 2017 CONTENTS LINEAR ALGEBRA AND MODULES General
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 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 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 informationCSIR  Algebra Problems
CSIR  Algebra Problems N. Annamalai DST  INSPIRE Fellow (SRF) Department of Mathematics Bharathidasan University Tiruchirappalli 620024 Email: algebra.annamalai@gmail.com Website: https://annamalaimaths.wordpress.com
More informationLecture 7 Cyclic groups and subgroups
Lecture 7 Cyclic groups and subgroups Review Types of groups we know Numbers: Z, Q, R, C, Q, R, C Matrices: (M n (F ), +), GL n (F ), where F = Q, R, or C. Modular groups: Z/nZ and (Z/nZ) Dihedral groups:
More informationMath 581 Problem Set 8 Solutions
Math 581 Problem Set 8 Solutions 1. Prove that a group G is abelian if and only if the function ϕ : G G given by ϕ(g) g 1 is a homomorphism of groups. In this case show that ϕ is an isomorphism. Proof:
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 informationMath 121 Homework 5: Notes on Selected Problems
Math 121 Homework 5: Notes on Selected Problems 12.1.2. Let M be a module over the integral domain R. (a) Assume that M has rank n and that x 1,..., x n is any maximal set of linearly independent elements
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 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 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 informationAlgebraI, Fall Solutions to Midterm #1
AlgebraI, Fall 2018. Solutions to Midterm #1 1. Let G be a group, H, K subgroups of G and a, b G. (a) (6 pts) Suppose that ah = bk. Prove that H = K. Solution: (a) Multiplying both sides by b 1 on the
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 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 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 informationMODEL ANSWERS TO HWK #4. ϕ(ab) = [ab] = [a][b]
MODEL ANSWERS TO HWK #4 1. (i) 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 seen to be nz. (ii) No. Suppose that
More informationAlgebra Qualifying Exam Solutions. Thomas Goller
Algebra Qualifying Exam Solutions Thomas Goller September 4, 2 Contents Spring 2 2 2 Fall 2 8 3 Spring 2 3 4 Fall 29 7 5 Spring 29 2 6 Fall 28 25 Chapter Spring 2. The claim as stated is false. The identity
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 informationMath 3140 Fall 2012 Assignment #3
Math 3140 Fall 2012 Assignment #3 Due Fri., Sept. 21. Remember to cite your sources, including the people you talk to. My solutions will repeatedly use the following proposition from class: Proposition
More informationTheorems and Definitions in Group Theory
Theorems and Definitions in Group Theory Shunan Zhao Contents 1 Basics of a group 3 1.1 Basic Properties of Groups.......................... 3 1.2 Properties of Inverses............................. 3
More information1.5 Applications Of The Sylow Theorems
14 CHAPTER1. GROUP THEORY 8. The Sylow theorems are about subgroups whose order is a power of a prime p. Here is a result about subgroups of index p. Let H be a subgroup of the finite group G, and assume
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 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 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 informationQuiz 2 Practice Problems
Quiz 2 Practice Problems Math 332, Spring 2010 Isomorphisms and Automorphisms 1. Let C be the group of complex numbers under the operation of addition, and define a function ϕ: C C by ϕ(a + bi) = a bi.
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 informationHOMEWORK 3 LOUISPHILIPPE THIBAULT
HOMEWORK 3 LOUISPHILIPPE THIBAULT Problem 1 Let G be a group of order 56. We have that 56 = 2 3 7. Then, using Sylow s theorem, we have that the only possibilities for the number of Sylowp subgroups
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 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 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 informationMath 581 Problem Set 9
Math 581 Prolem Set 9 1. Let m and n e relatively prime positive integers. (a) Prove that Z/mnZ = Z/mZ Z/nZ as RINGS. (Hint: First Isomorphism Theorem) Proof: Define ϕz Z/mZ Z/nZ y ϕ(x) = ([x] m, [x] n
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 informationDefinition List Modern Algebra, Fall 2011 Anders O.F. Hendrickson
Definition List Modern Algebra, Fall 2011 Anders O.F. Hendrickson On almost every Friday of the semester, we will have a brief quiz to make sure you have memorized the definitions encountered in our studies.
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 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 430 Final Exam, Fall 2008
IIT Dept. Applied Mathematics, December 9, 2008 1 PRINT Last name: Signature: First name: Student ID: Math 430 Final Exam, Fall 2008 Grades should be posted Friday 12/12. Have a good break, and don t forget
More informationSolutions of exercise sheet 4
DMATH Algebra I HS 14 Prof. Emmanuel Kowalski Solutions of exercise sheet 4 The content of the marked exercises (*) should be known for the exam. 1. Prove the following two properties of groups: 1. Every
More informationFrank Moore Algebra 901 Notes Professor: Tom Marley Direct Products of Groups:
Frank Moore Algebra 901 Notes Professor: Tom Marley Direct Products of Groups: Definition: The external direct product is defined to be the following: Let H 1,..., H n be groups. H 1 H 2 H n := {(h 1,...,
More informationAlgebra Ph.D. Entrance Exam Fall 2009 September 3, 2009
Algebra Ph.D. Entrance Exam Fall 2009 September 3, 2009 Directions: Solve 10 of the following problems. Mark which of the problems are to be graded. Without clear indication which problems are to be graded
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 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 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 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 informationCONJUGATION IN A GROUP
CONJUGATION IN A GROUP KEITH CONRAD 1. Introduction A reflection across one line in the plane is, geometrically, just like a reflection across any other line. That is, while reflections across two different
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 informationMath 250A, Fall 2004 Problems due October 5, 2004 The problems this week were from Lang s Algebra, Chapter I.
Math 250A, Fall 2004 Problems due October 5, 2004 The problems this week were from Lang s Algebra, Chapter I. 24. We basically know already that groups of order p 2 are abelian. Indeed, pgroups have nontrivial
More informationx 2 = xn xn = x 2 N = N = 0
Potpourri. Spring 2010 Problem 2 Let G be a finite group with commutator subgroup G. Let N be the subgroup of G generated by the set {x 2 : x G}. Then N is a normal subgroup of G and N contains G. Proof.
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 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 informationAbstract Algebra II Groups ( )
Abstract Algebra II Groups ( ) Melchior Grützmann / melchiorgfreehostingcom/algebra October 15, 2012 Outline Group homomorphisms Free groups, free products, and presentations Free products ( ) Definition
More informationA SIMPLE PROOF OF BURNSIDE S CRITERION FOR ALL GROUPS OF ORDER n TO BE CYCLIC
A SIMPLE PROOF OF BURNSIDE S CRITERION FOR ALL GROUPS OF ORDER n TO BE CYCLIC SIDDHI PATHAK Abstract. This note gives a simple proof of a famous theorem of Burnside, namely, all groups of order n are cyclic
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 informationCosets and Normal Subgroups
Cosets and Normal Subgroups (Last Updated: November 3, 2017) These notes are derived primarily from Abstract Algebra, Theory and Applications by Thomas Judson (16ed). Most of this material is drawn from
More informationOn the characterization of the numbers n such that any group of order n has a given property P
On the characterization of the numbers n such that any group of order n has a given property P arxiv:1501.03170v1 [math.gr] 6 Jan 2015 Advisor: Professor Thomas Haines Honors Thesis in Mathematics University
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 informationExercises MAT2200 spring 2013 Ark 4 Homomorphisms and factor groups
Exercises MAT2200 spring 2013 Ark 4 Homomorphisms and factor groups This Ark concerns the weeks No. (Mar ) and No. (Mar ). Plans until Eastern vacations: In the book the group theory included in the curriculum
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 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 informationPh.D. Qualifying Examination in Algebra Department of Mathematics University of Louisville January 2018
Ph.D. Qualifying Examination in Algebra Department of Mathematics University of Louisville January 2018 Do 6 problems with at least 2 in each section. Group theory problems: (1) Suppose G is a group. The
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 informationMATH 323, Algebra 1. Archive of documentation for. Bilkent University, Fall 2014, Laurence Barker. version: 17 January 2015
Archive of documentation for MATH 323, Algebra 1 Bilkent University, Fall 2014, Laurence Barker version: 17 January 2015 Source file: arch323fall14.tex page 2: Course specification. page 4: Homeworks and
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 informationGroups Subgroups Normal subgroups Quotient groups Homomorphisms Cyclic groups Permutation groups Cayley s theorem Class equations Sylow theorems
Group Theory Groups Subgroups Normal subgroups Quotient groups Homomorphisms Cyclic groups Permutation groups Cayley s theorem Class equations Sylow theorems Groups Definition : A nonempty set ( G,*)
More informationHigher Algebra Lecture Notes
Higher Algebra Lecture Notes October 2010 Gerald Höhn Department of Mathematics Kansas State University 138 Cardwell Hall Manhattan, KS 665062602 USA gerald@math.ksu.edu This are the notes for my lecture
More informationThe group (Z/nZ) February 17, In these notes we figure out the structure of the unit group (Z/nZ) where n > 1 is an integer.
The group (Z/nZ) February 17, 2016 1 Introduction In these notes we figure out the structure of the unit group (Z/nZ) where n > 1 is an integer. If we factor n = p e 1 1 pe, where the p i s are distinct
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 informationALGEBRA I (LECTURE NOTES 2017/2018) LECTURE 9  CYCLIC GROUPS AND EULER S FUNCTION
ALGEBRA I (LECTURE NOTES 2017/2018) LECTURE 9  CYCLIC GROUPS AND EULER S FUNCTION PAVEL RŮŽIČKA 9.1. Congruence modulo n. Let us have a closer look at a particular example of a congruence relation on
More informationBASIC GROUP THEORY : G G G,
BASIC GROUP THEORY 18.904 1. Definitions Definition 1.1. A group (G, ) is a set G with a binary operation : G G G, and a unit e G, possessing the following properties. (1) Unital: for g G, we have g e
More informationSelected Solutions to Math 4107, Set 4
Selected Solutions to Math 4107, Set 4 November 9, 2005 Page 90. 1. There are 3 different classes: {e}, {(1 2), (1 3), (2 3)}, {(1 2 3), (1 3 2)}. We have that c e = 1, c (1 2) = 3, and c (1 2 3) = 2.
More informationHomework Problems, Math 200, Fall 2011 (Robert Boltje)
Homework Problems, Math 200, Fall 2011 (Robert Boltje) Due Friday, September 30: ( ) 0 a 1. Let S be the set of all matrices with entries a, b Z. Show 0 b that S is a semigroup under matrix multiplication
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 informationSelected exercises from Abstract Algebra by Dummit and Foote (3rd edition).
Selected exercises from Abstract Algebra by Dummit and Foote (3rd edition). Bryan Félix Abril 12, 2017 Section 2.1 Exercise (6). Let G be an abelian group. Prove that T = {g G g < } is a subgroup of G.
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 602, Fall 2002, HW 2, due 10/14/2002
Math 602, Fall 2002, HW 2, due 10/14/2002 Part A AI) A Fermat prime, p, is a prime number of the form 2 α + 1. E.g., 2, 3, 5, 17, 257,.... (a) Show if 2 α + 1 is prime then α = 2 β. (b) Say p is a Fermat
More informationAlgebra Exam Fall Alexander J. Wertheim Last Updated: October 26, Groups Problem Problem Problem 3...
Algebra Exam Fall 2006 Alexander J. Wertheim Last Updated: October 26, 2017 Contents 1 Groups 2 1.1 Problem 1..................................... 2 1.2 Problem 2..................................... 2
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 informationPermutation groups H. W. Lenstra, Fall streng/permutation/index.html
Permutation groups H. W. Lenstra, Fall 2007 http://www.math.leidenuniv.nl/ streng/permutation/index.html Solvable groups. Let G be a group. We define the sequence G (0) G (1) G (2)... of subgroups of G
More informationAlgebra I Notes. Clayton J. Lungstrum. July 18, Based on the textbook Algebra by Serge Lang
Algebra I Notes Based on the textbook Algebra by Serge Lang Clayton J. Lungstrum July 18, 2013 Contents Contents 1 1 Group Theory 2 1.1 Basic Definitions and Examples......................... 2 1.2 Subgroups.....................................
More information