# Solutions for Homework Assignment 5

Size: px
Start display at page:

Transcription

1 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 to the inequality a 2 +b 2 > 0. Let { ( ) } a b H a,b R, and a 2 +b 2 > 0. b a For a,b R, the condition a 2 + b 2 > 0 means that a 2 + b 2 0, which in turn means that the determinant of any matrix in H is nonzero. Hence H is a subset of GL 2 (R). We will see that it is a subgroup as a consequence of the proof below. Define a map ϕ : C H as follows: ϕ(a+bi) ( ) a b b a for all a+bi C. The above remarks make it clear that ϕ is one-to-one and onto. Thus, ϕ is a bijective map from C to H. We now show that ϕ is a homomorphism. Since ϕ is bijective, it follows that ϕ is indeed an isomorphism. Suppose that a+bi and c+di are elements of C. Then ϕ ( (a+bi)(c+di) ) ϕ ( (ac bd)+(ad+bc)i ). ( ) ac bd ad+bc ad bc ac bd. On the other hand, we have ϕ(a+bi)ϕ(c+di) ( )( ) a b c d b a d c ( ) ac bd ad+bc ad bc ac bd, where the second equality is simply matrix multiplication in GL 2 (R). It is the operation defining the group GL 2 (R). Thus, we see that if a+bi and c+di are elements of C, then ϕ ( (a+bi)(c+di) ) ϕ(a+bi)ϕ(c+di) and hence ϕ is indeed a homomorphism from C to H. We have proved that C and H are isomorphic. 1

2 Page 154, Problem 3. The groups U(8) and Z 4 are not isomorphic. The reason is that Z 4 is a cyclic group of order 4 and hence has an element of order 4. In contrast, every element of U(8) has order 1 or 2. The identity element is 1+8Z and has order 1. The other three elements of U(8) are 3+8Z, 5+8Z, and 7+8Z, and satisfy (3+8Z) 2 1+8Z, (5+8Z) 2 1+8Z, (7+8Z) 2 1+8Z and hence have order 2 (since their orders aren t equal to 1). Therefore, U(8) has no element of order 4. Therefore, since an isomorphism preserves the order of elements, it follows that U(8) is not isomorphic to Z 4. Page 154, Problem 9. Let G {r R r 1}. We define an operation on G by a b a + b + ab for all a,b G. Note that G is a group. This was proved in problem set 1. Define a map ϕ : G R by ϕ(a) 1+a Since a 1 means that 1 + a 0, it is clear that ϕ is a bijective map from G to R. Furthermore, we have ϕ(a b) ϕ(a+b+ab) 1+a+b+ab (1+a)(1+b) ϕ(a)ϕ(b). Hence ϕ is a homomorphism. Therefore, ϕ is indeed an isomorphism from G to R and so those two groups are indeed isomorphic. Page 156, Problem 24. No such group exists. To see this, suppose that G is an abelian group and that G 51. We proved in class (using Problem C below) that if G is a finite abelian group and p is a prime dividing G, then G has at least one element of order p. Applying this to the case where G 51, it follows that G has at least one element a of order 3 and at least one element b of order 17. Let c ab, an element of G. We will prove that c has order 51. It follows that c is a subgroup of G of order 51 and hence we must have G c. This implies that G is cyclic. Since a has order 3 and 3 51, it follows that a 51 e, where e denotes the identity element of G. Since b has order 17 and 17 51, it follows that b 51 e. To see that c has order 51, we use the fact that G is abelian. We have c 51 (ab) 51 a 51 b 51 ee e 2

3 and therefore c divides 51. Thus c {1, 3, 17, 51}. On the other hand, c 3 a 3 b 3 eb 3 b 3 e, c 17 a 17 b 17 a 17 e a 17 e. We have used the facts that b 17 does not divide 3, and a 3 does not divide 17, respectively. It follows that c does not divide 3 and c does not divide 17. Therefore, the only possibility is that c 51. As pointed out above, the fact that c 51 implies that G is a cyclic group. Page 156, Problem 25. A non-cyclic abelian group G of order 52 does indeed exist. Let A be the Klein 4-group. Thus, A 4 and every element of A has order 1 or 2. Thus, a 2 e for all a A, where e is the identity element in A. Let B be a cyclic group of order 13. Let f be the identity element in B. Then b 13 f for all b B. Let G A B. Then G A B The identity element in the group G is (e, f). Consider an element g G. Then g (a, b), where a A and b B. We will verify that g 26 (e, f). To see this, note that since a 2 e, we have a 26 e and since b 13 f, we also have b 26 f. Therefore, g 26 (a, b) 26 (a 26, b 26 ) (e, f), as stated. It follows that g divides 26. Therefore, no element of G has order 52. Hence G is not a cyclic group. Page 157, Problem 48. By definition, G H { (g, h) g G, h H }, H G { (h, g) h H, g G }. The group operations on these sets were defined in class one day. We can define a map ϕ from G H to H G by ϕ ( (g, h) ) (h, g). Thus, any element (h, g) in H G is the image under the map ϕ of the element (g, h) in G H, and of no other element in G H. That is, the map ϕ is a bijective map. It remains to verify that ϕ is a homomorphism. To see this, suppose that (g 1, h 1 ) and (g 2, h 2 ) are elements of G H. Then, by definition, (g 1, h 1 )(g 2, h 2 ) (g 1 g 2, h 1 h 2 ). Hence ϕ ( (g 1, h 1 )(g 2, h 2 ) ) ϕ ( (g 1 g 2, h 1 h 2 ) ) (h 1 h 2, g 1 g 2 ) (h 1, g 1 )(h 2, g 2 ). On the other hand, we have ϕ ( (g 1, h 1 ) ) ϕ ( (g 2, h 2 ) ) (h 1, g 1 )(h 2, g 2 ). 3

4 Hence we have ϕ ( (g 1, h 1 )(g 2, h 2 ) ) ϕ ( (g 1, h 1 ) ) ϕ ( (g 2, h 2 ) ). This means that ϕ is indeed a homomorphism from G H to H G. Since ϕ is also bijective, ϕ is an isomorphism. Therefore, G H is isomorphic to H G, as stated. Page 166, Problem 4. This question concerns various subgroups of GL 2 (R), namely { ( ) } { ( ) } a b 1 x T a,b,c R,ac 0, U x R. 0 c This notation is used because the elements of T are triangular matrices (more precisely, upper triangular) and the elements of U are unipotent matrices (referring to the fact that the eigenvalues are equal to 1). The identity matrix I 2 is in both U and T. To see that U and T are subgroups of GL 2 (R), we have (by matrix algebra) ( )( ) ( ) ( ) ( ) ( ) a b d e ad ae+bf 1 x 1 y 1 x+y,. 0 c 0 f 0 cf Note also that ac 0 and df 0 implies that (ad)(cf) 0. Thus, both T and U are closed under matrix multiplication, which is the group operation in GL 2 (R). Furthermore, concerning inverses, we have ( ) 1 a b 0 c ( c 1 b(ac) 1 0 a 1 ), ( ) 1 1 x ( ) 1 x where we have used a standard formula for computing the inverse of a 2 2 matrix with nonzero determinant. It follows that if τ T, then τ 1 T and that if η U, then η 1 U., The above discussion shows that T and U are subgroups of GL 2 (R). Also, U T and hence U is a subgroup of T. The fact that U is abelian follows from observing that ( ) ( ) ( ) ( ) ( ) ( ) 1 x 1 y 1 x+y 1 y +x 1 y 1 x for all x,y R. For the rest of this question, it will be useful to note that if ac 0, then a 0 and we have ( ) ( )( ) a b a ba 1. 0 c 0 c 4

5 Thus, every element τ T can be expressed as τ δη, where η U and δ is a diagonal matrix. Let { ( ) } a 0 D a, c R,ac 0, 0 c which is another subgroup of GL 2 (R). (The verification that D is a subgroup is similar to the verification for T.) Thus, if τ T, then we can express τ as τ δη, where δ D, η U. To verify that U is a normal subgroup of T, suppose that τ T and u U. We must verify that τuτ 1 U. We can write τ as τ δη as in the previous paragraph. Then τuτ 1 (δη)u(δη) 1 ) (δη)u(η 1 δ 1 δ(ηuη 1 )δ 1 δvδ 1 where v ηuη 1. Note that v U because u U, η U and η 1 U. Thus, it is sufficient to verify that δvδ 1 U for all v U and all δ D. This fact becomes clear from the following matrix identity (valid for ac 0 and any x R): ( )( )( ) ( ) a x a 1 ac 0 c 0 c 1 1 x. And so we have shown that U is indeed a normal subgroup of T. Now we will prove that T/U is abelian. Suppose that τ 1 and τ 2 are in T. As above, we can write τ 1 δ 1 η 1 and τ 2 δ 2 η 2, where δ 1, δ 2 D and η 1, η 2 U. This will be very helpful because we then have τ 1 U δ 1 η 1 U δ 1 (η 1 U) δ 1 U, τ 2 U δ 2 η 2 U δ 2 (η 2 U) δ 2 U. We have used the fact that ηu U for all η U. Note that D is an abelian group. This is clear from the facts that ( )( ) ( ) ( ) ( )( ) a 0 d 0 ad 0 da 0 d 0 a 0 0 c 0 f 0 cf 0 fc 0 f 0 c Thus, δ 1 δ 2 δ 2 δ 1 for all δ 1, δ 2 D. It follows that. (τ 1 U)(τ 2 U) (δ 1 U)(δ 2 U) δ 1 δ 2 U δ 2 δ 1 U (δ 2 U)(δ 1 U) (τ 2 U)(τ 1 U) and this shows that T/U is indeed an abelian group. Finally, we will show that T is not a normal subgroup of GL 2 (R). Consider the following matrices: ( ) ( ) τ, γ

6 Then we have τ T and γ GL 2 (R), but ( )( )( ) γτγ ( )( ) ( ) 1 2 and hence γτγ 1 T. Hence, γtγ 1 T. Thus, T is not a normal subgroup of GL 2 (R). Page 167, Problem 5. Suppose that H and K are normal subgroups of a group G. We have proved previous that H K is a subgroup of G. We now prove that H K is a normal subgroup. Suppose that g G and a H K. Consider gag 1. First of all, we have a H. Since H is a normal subgroup of G, we also have gag 1 H. Furthermore, a K. Since K is a normal subgroup of G, we have gag 1 H. Therefore, gag 1 H and gag 1 K. It follows that gag 1 H K for all g G and all a H K. This implies that the subgroup H K is indeed a normal subgroup of G. Page 167, Problem 7. The statement is not true. Consider G Q 8. Let H j. Then H is a subgroup of G. Furthermore, H 4 and [G : H] G / H 8/4 2. Thus, H is a subgroup of G which has index 2. As proved in class, this implies that H is a normal subgroup of G. Now G/H is a group of order 2 and must be cyclic, and hence abelian. Also, H is a group of order 4. Recall that groups of order 4 must be abelian. Thus, both H and G/H are abelian. However, G itself is a nonabelian group. Page 167, Problem 9. This statement is not true. Let G be the Klein 4-group. Recall that every element of G has order 1 or 2. But G 4. Hence G is not a cyclic group. However, if a G and a is not the identity element of G, then let H a. Then H is a cyclic group of order 2. Also, H is a normal subgroup of G because G is abelian. Now G/H has order equal to G / H 4/2 2. Hence G/H is a group of order 2 and must be cyclic. In summary, G is not cyclic, but H is cyclic and G/H is also cyclic. The example given in problem 7 would also be an example disproving the statement in this problem. Page 176, Problems 2a, c. First we consider part (a). If a,b R, then ( ) ( )( ) 1 0 φ(ab) φ(a)φ(b) 0 ab 0 a 0 b 6

7 and therefore φ is a homomorphism from R to GL 2 (R). ( ) 1 0 The identity element in GL 2 (R) is I 2. If a R and φ(a) I 2, then a 1, which is the identity element in R. Therefore, the kernel of φ is {1}, the trivial subgroup of R. Now consider part (c). The map φ is not a homomorphism. The easiest way to see this is to notice that φ(i 2 ) The identity element in GL 2 (R) is I 2. The identity element in R is 0. But φ(i 2 ) 0. If φ were a homomorphism, then this could not happen. Page 176, Problem 4. Suppose that n, m Z. Then φ(n+m) 7(n+m) 7n+7m φ(n)+φ(m) and hence φ is indeed a homomorphism from Z to Z. To find the kernel, note that 0 is the identity element in Z. We have φ(n) 0 7n 0 n 0 and therefore the kernel of φ is the subgroup {0} of Z. The image of φ is obviously the subgroup 7Z of Z. Page 177, Problem 9. WearegiventhatGisabelian. Itturnsoutthatφ(G)isasubgroup of H. To prove that φ(g) is abelian, suppose that a, b φ(g). This means that a φ(u) and b φ(v), where u, v G. We have uv vu because G is abelian. Therefore, ab φ(u)φ(v) φ(uv) φ(vu) φ(v)φ(u) ba. Thus, ab ba for all a, b φ(g). This proves that φ(g) is indeed abelian. Page 177, Problem 14. Recall the following fact. Suppose that G and G are groups and that φ : G G is an isomorphism. Suppose that g G. Then φ(g) has the same order as g. We proved this in class one day. It will be useful in this problem. Note for example that if G has an element g of order 2, then G will also have an element of order 2, namely φ(g). Let G Q / Z. Then G has an element of order 2, namely the element g Z. 7

8 This is a left coset of Z in Q. Furthermore, 1 Z and hence g is not the identity element 2 of Q / Z (which is the left coset 0+Z Z). However, 2g g +g ( ) Z + ( ) Z 1 + Z Z and so g has order 2. However, the identity element in Q is 0. If r Q and 2r 0, then r 0. Hence no element of Q can have order 2. In summary, Q / Z has an element of order 2, but Q has no elements of order 2. Hence those two groups cannot be isomorphic. Problem A. We must show that if σ S n, then σ and σ 1 are conjugate in S n. We know that two elements of S n are conjugate if and only if they have the same cycle decomposition type. And so one must show that σ and σ 1 have the same cycle decomposition type. Suppose first that σ is a k-cycle. Thus σ (i 1 i 2... i k ), where i 1,i 2,...,i k are distinct elements in the set {1,...,n}. But σ 1 is also a k-cycle. In fact, σ 1 (i k i k 1... i 1 ), which is indeed a k-cycle. Now if σ is a product of t disjoint cycles of lengths k 1,...,k t, then σ 1 will be a product of the inverses of those cycles, and so σ 1 will also be a product of t disjoint cycles of lengths k 1,...,k t. Thus, σ and σ 1 indeed have the same cycle decomposition type. Problem B. This problem is identical to problem 5 on page 167. Problem C. Suppose that G is a finite group, that N is a normal subgroup of G, and that G/N has an element of order m, where m is a positive integer. The elements of G/N are of the form an, where a G. Suppose that a is chosen so that an is an element of G/N which has order m. The rest of this proof will concern the element a. Since a G and G is finite, it follows that the subgroup a of G is a finite group. Thus a has finite order. Let n be the order of a. In particular, a n e, where e is the identity element of G. In the group G/N, we have (an)(bn) abn for all a,b G. In particular, we have (an) 2 a 2 N. A straightforward mathematical induction proof shows that (an) k a k N for all positive integers k. 8

9 Since a n e, it follows that (an) n a n N en N. Now we chose a at the beginning of this proof so that an is an element in the group G/N of order m. Therefore, the fact that (an) n e implies that m divides n. The subgroup a of G which is generated by a has order n. It is a cyclic group of order n. We proved in class that if d is a positive integer which divides n, then a cyclic group of order n must contain an element of order d. In particular, since m divides n, it follows that a contains an element of order m. Therefore, it follows that G contains an element of order m. This is what we wanted to prove. 9

### 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 =

### Algebra homework 6 Homomorphisms, isomorphisms

MATH-UA.343.005 T.A. Louis Guigo Algebra homework 6 Homomorphisms, isomorphisms Exercise 1. Show that the following maps are group homomorphisms and compute their kernels. (a f : (R, (GL 2 (R, given by

### MATH 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

### EXAM 3 MAT 423 Modern Algebra I Fall c d a + c (b + d) d c ad + bc ac bd

EXAM 3 MAT 23 Modern Algebra I Fall 201 Name: Section: I All answers must include either supporting work or an explanation of your reasoning. MPORTANT: These elements are considered main part of the answer

### Solution Outlines for Chapter 6

Solution Outlines for Chapter 6 # 1: Find an isomorphism from the group of integers under addition to the group of even integers under addition. Let φ : Z 2Z be defined by x x + x 2x. Then φ(x + y) 2(x

### Yale 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

### Solutions to Some Review Problems for Exam 3. by properties of determinants and exponents. Therefore, ϕ is a group homomorphism.

Solutions to Some Review Problems for Exam 3 Recall that R, the set of nonzero real numbers, is a group under multiplication, as is the set R + of all positive real numbers. 1. Prove that the set N of

### ENTRY 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.

### 3.4 Isomorphisms. 3.4 J.A.Beachy 1. from A Study Guide for Beginner s by J.A.Beachy, a supplement to Abstract Algebra by Beachy / Blair

3.4 J.A.Beachy 1 3.4 Isomorphisms from A Study Guide for Beginner s by J.A.Beachy, a supplement to Abstract Algebra by Beachy / Blair 29. Show that Z 17 is isomorphic to Z 16. Comment: The introduction

### Math 546, Exam 2 Information.

Math 546, Exam 2 Information. 10/21/09, LC 303B, 10:10-11:00. Exam 2 will be based on: Sections 3.2, 3.3, 3.4, 3.5; The corresponding assigned homework problems (see http://www.math.sc.edu/ boylan/sccourses/546fa09/546.html)

### Problem Set Mash 1. a2 b 2 0 c 2. and. a1 a

Problem Set Mash 1 Section 1.2 15. Find a set of generators and relations for Z/nZ. h 1 1 n 0i Z/nZ. Section 1.4 a b 10. Let G 0 c a, b, c 2 R,a6 0,c6 0. a1 b (a) Compute the product of 1 a2 b and 2 0

### Math 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:

### Assigment 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

### Abstract Algebra, HW6 Solutions. Chapter 5

Abstract Algebra, HW6 Solutions Chapter 5 6 We note that lcm(3,5)15 So, we need to come up with two disjoint cycles of lengths 3 and 5 The obvious choices are (13) and (45678) So if we consider the element

### Lecture 3. Theorem 1: D 6

Lecture 3 This week we have a longer section on homomorphisms and isomorphisms and start formally working with subgroups even though we have been using them in Chapter 1. First, let s finish what was claimed

### Lecture 4.1: Homomorphisms and isomorphisms

Lecture 4.: Homomorphisms and isomorphisms Matthew Macauley Department of Mathematical Sciences Clemson University http://www.math.clemson.edu/~macaule/ Math 4, Modern Algebra M. Macauley (Clemson) Lecture

### Ring Theory Problem Set 2 Solutions

Ring Theory Problem Set 2 Solutions 16.24. SOLUTION: We already proved in class that Z[i] is a commutative ring with unity. It is the smallest subring of C containing Z and i. If r = a + bi is in Z[i],

### Your 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.

### Name: 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

### Section 15 Factor-group computation and simple groups

Section 15 Factor-group computation and simple groups Instructor: Yifan Yang Fall 2006 Outline Factor-group computation Simple groups The problem Problem Given a factor group G/H, find an isomorphic group

### MATH 4107 (Prof. Heil) PRACTICE PROBLEMS WITH SOLUTIONS Spring 2018

MATH 4107 (Prof. Heil) PRACTICE PROBLEMS WITH SOLUTIONS Spring 2018 Here are a few practice problems on groups. You should first work through these WITHOUT LOOKING at the solutions! After you write your

### 5 Group theory. 5.1 Binary operations

5 Group theory This section is an introduction to abstract algebra. This is a very useful and important subject for those of you who will continue to study pure mathematics. 5.1 Binary operations 5.1.1

### Solutions for Assignment 4 Math 402

Solutions for Assignment 4 Math 402 Page 74, problem 6. Assume that φ : G G is a group homomorphism. Let H = φ(g). We will prove that H is a subgroup of G. Let e and e denote the identity elements of G

### MATH 3005 ABSTRACT ALGEBRA I FINAL SOLUTION

MATH 3005 ABSTRACT ALGEBRA I FINAL SOLUTION SPRING 2014 - MOON Write your answer neatly and show steps. Any electronic devices including calculators, cell phones are not allowed. (1) Write the definition.

### (Think: three copies of C) i j = k = j i, j k = i = k j, k i = j = i k.

Warm-up: The quaternion group, denoted Q 8, is the set {1, 1, i, i, j, j, k, k} with product given by 1 a = a 1 = a a Q 8, ( 1) ( 1) = 1, i 2 = j 2 = k 2 = 1, ( 1) a = a ( 1) = a a Q 8, (Think: three copies

### Section 18 Rings and fields

Section 18 Rings and fields Instructor: Yifan Yang Spring 2007 Motivation Many sets in mathematics have two binary operations (and thus two algebraic structures) For example, the sets Z, Q, R, M n (R)

### MODEL 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

### Two subgroups and semi-direct products

Two subgroups and semi-direct 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

### Name: Solutions - AI FINAL EXAM

1 2 3 4 5 6 7 8 9 10 11 12 13 total Name: Solutions - AI FINAL EXAM The first 7 problems will each count 10 points. The best 3 of # 8-13 will count 10 points each. Total is 100 points. A 4th problem from

### HOMEWORK 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

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

### B Sc MATHEMATICS ABSTRACT ALGEBRA

UNIVERSITY OF CALICUT SCHOOL OF DISTANCE EDUCATION B Sc MATHEMATICS (0 Admission Onwards) V Semester Core Course ABSTRACT ALGEBRA QUESTION BANK () Which of the following defines a binary operation on Z

### MA441: 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

### MATH 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

### 2MA105 Algebraic Structures I

2MA105 Algebraic Structures I Per-Anders Svensson http://homepage.lnu.se/staff/psvmsi/2ma105.html Lecture 7 Cosets once again Factor Groups Some Properties of Factor Groups Homomorphisms November 28, 2011

### 3.8 Cosets, Normal Subgroups, and Factor Groups

3.8 J.A.Beachy 1 3.8 Cosets, Normal Subgroups, and Factor Groups from A Study Guide for Beginner s by J.A.Beachy, a supplement to Abstract Algebra by Beachy / Blair 29. Define φ : C R by φ(z) = z, for

### Abstract 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

### Section 13 Homomorphisms

Section 13 Homomorphisms Instructor: Yifan Yang Fall 2006 Homomorphisms Definition A map φ of a group G into a group G is a homomorphism if for all a, b G. φ(ab) = φ(a)φ(b) Examples 1. Let φ : G G be defined

### Cosets, 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

### Theorems 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

### Math 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

### Supplementary Notes: Simple Groups and Composition Series

18.704 Supplementary Notes: Simple Groups and Composition Series Genevieve Hanlon and Rachel Lee February 23-25, 2005 Simple Groups Definition: A simple group is a group with no proper normal subgroup.

### Lecture 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:

### book 2005/1/23 20:41 page 132 #146

book 2005/1/23 20:41 page 132 #146 132 2. BASIC THEORY OF GROUPS Definition 2.6.16. Let a and b be elements of a group G. We say that b is conjugate to a if there is a g G such that b = gag 1. You are

### First Semester Abstract Algebra for Undergraduates

First Semester Abstract Algebra for Undergraduates Lecture notes by: Khim R Shrestha, Ph. D. Assistant Professor of Mathematics University of Great Falls Great Falls, Montana Contents 1 Introduction to

### 1 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

### Supplement. 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

### Part IV. Rings and Fields

IV.18 Rings and Fields 1 Part IV. Rings and Fields Section IV.18. Rings and Fields Note. Roughly put, modern algebra deals with three types of structures: groups, rings, and fields. In this section we

### Modern Algebra Homework 9b Chapter 9 Read Complete 9.21, 9.22, 9.23 Proofs

Modern Algebra Homework 9b Chapter 9 Read 9.1-9.3 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

### Homework #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

### Definition 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.

### Quiz 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.

### Automorphism Groups Definition. An automorphism of a group G is an isomorphism G G. The set of automorphisms of G is denoted Aut G.

Automorphism Groups 9-9-2012 Definition. An automorphism of a group G is an isomorphism G G. The set of automorphisms of G is denoted Aut G. Example. The identity map id : G G is an automorphism. Example.

### Cosets 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

### IIT Mumbai 2015 MA 419, Basic Algebra Tutorial Sheet-1

IIT Mumbai 2015 MA 419, Basic Algebra Tutorial Sheet-1 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 non-collinear points, then

### φ(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 =

### Zachary Scherr Math 370 HW 7 Solutions

1 Book Problems 1. 2.7.4b Solution: Let U 1 {u 1 u U} and let S U U 1. Then (U) is the set of all elements of G which are finite products of elements of S. We are told that for any u U and g G we have

### Recall: Properties of Homomorphisms

Recall: Properties of Homomorphisms Let φ : G Ḡ be a homomorphism, let g G, and let H G. Properties of elements Properties of subgroups 1. φ(e G ) = eḡ 1. φ(h) Ḡ. 2. φ(g n ) = (φ(g)) n for all n Z. 2.

### Introduction to Groups

Introduction to Groups Hong-Jian 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)

### SUPPLEMENT 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

### GRE Subject test preparation Spring 2016 Topic: Abstract Algebra, Linear Algebra, Number Theory.

GRE Subject test preparation Spring 2016 Topic: Abstract Algebra, Linear Algebra, Number Theory. Linear Algebra Standard matrix manipulation to compute the kernel, intersection of subspaces, column spaces,

### January 2016 Qualifying Examination

January 2016 Qualifying Examination If you have any difficulty with the wording of the following problems please contact the supervisor immediately. All persons responsible for these problems, in principle,

### ALGEBRA 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

### SF2729 GROUPS AND RINGS LECTURE NOTES

SF2729 GROUPS AND RINGS LECTURE NOTES 2011-03-01 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

### Algebraic 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

### ABSTRACT ALGEBRA MODULUS SPRING 2006 by Jutta Hausen, University of Houston

ABSTRACT ALGEBRA MODULUS SPRING 2006 by Jutta Hausen, University of Houston Undergraduate abstract algebra is usually focused on three topics: Group Theory, Ring Theory, and Field Theory. Of the myriad

### Mathematics 331 Solutions to Some Review Problems for Exam a = c = 3 2 1

Mathematics 331 Solutions to Some Review Problems for Exam 2 1. Write out all the even permutations in S 3. Solution. The six elements of S 3 are a =, b = 1 3 2 2 1 3 c =, d = 3 2 1 2 3 1 e =, f = 3 1

### Fall /29/18 Time Limit: 75 Minutes

Math 411: Abstract Algebra Fall 2018 Midterm 10/29/18 Time Limit: 75 Minutes Name (Print): Solutions JHU-ID: This exam contains 8 pages (including this cover page) and 6 problems. Check to see if any pages

### MTH 411 Lecture Notes Based on Hungerford, Abstract Algebra

MTH 411 Lecture Notes Based on Hungerford, Abstract Algebra Ulrich Meierfrankenfeld Department of Mathematics Michigan State University East Lansing MI 48824 meier@math.msu.edu August 28, 2014 2 Contents

### SUMMARY ALGEBRA I LOUIS-PHILIPPE THIBAULT

SUMMARY ALGEBRA I LOUIS-PHILIPPE 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.

### Simple 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

### CONSEQUENCES 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.

### Teddy 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

### ALGEBRA 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

### Lecture 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

### Homomorphisms. The kernel of the homomorphism ϕ:g G, denoted Ker(ϕ), is the set of elements in G that are mapped to the identity in G.

10. Homomorphisms 1 Homomorphisms Isomorphisms are important in the study of groups because, being bijections, they ensure that the domain and codomain groups are of the same order, and being operation-preserving,

### Math Introduction to Modern Algebra

Math 343 - Introduction to Modern Algebra Notes Rings and Special Kinds of Rings Let R be a (nonempty) set. R is a ring if there are two binary operations + and such that (A) (R, +) is an abelian group.

### INTRODUCTION TO THE GROUP THEORY

Lecture Notes on Structure of Algebra INTRODUCTION TO THE GROUP THEORY By : Drs. Antonius Cahya Prihandoko, M.App.Sc e-mail: antoniuscp.fkip@unej.ac.id Mathematics Education Study Program Faculty of Teacher

### Math 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

### Definitions, 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

### ABSTRACT ALGEBRA 1, LECTURE NOTES 5: HOMOMORPHISMS, ISOMORPHISMS, SUBGROUPS, QUOTIENT ( FACTOR ) GROUPS. ANDREW SALCH

ABSTRACT ALGEBRA 1, LECTURE NOTES 5: HOMOMORPHISMS, ISOMORPHISMS, SUBGROUPS, QUOTIENT ( FACTOR ) GROUPS. ANDREW SALCH 1. Homomorphisms and isomorphisms between groups. Definition 1.1. Let G, H be groups.

### 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

### Exercises 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

### Ideals, congruence modulo ideal, factor rings

Ideals, congruence modulo ideal, factor rings Sergei Silvestrov Spring term 2011, Lecture 6 Contents of the lecture Homomorphisms of rings Ideals Factor rings Typeset by FoilTEX Congruence in F[x] and

### Chapter 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

### Definitions. 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

### Frank 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,...,

### Problem 1. Let I and J be ideals in a ring commutative ring R with 1 R. Recall

I. Take-Home 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

### Ph.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

### Lecture 6 Special Subgroups

Lecture 6 Special Subgroups Review: Recall, for any homomorphism ϕ : G H, the kernel of ϕ is and the image of ϕ is ker(ϕ) = {g G ϕ(g) = e H } G img(ϕ) = {h H ϕ(g) = h for some g G} H. (They are subgroups

### Some practice problems for midterm 2

Some practice problems for midterm 2 Kiumars Kaveh November 14, 2011 Problem: Let Z = {a G ax = xa, x G} be the center of a group G. Prove that Z is a normal subgroup of G. Solution: First we prove Z is

### Algebra Exercises in group theory

Algebra 3 2010 Exercises in group theory February 2010 Exercise 1*: Discuss the Exercises in the sections 1.1-1.3 in Chapter I of the notes. Exercise 2: Show that an infinite group G has to contain a non-trivial

### Math 103 HW 9 Solutions to Selected Problems

Math 103 HW 9 Solutions to Selected Problems 4. Show that U(8) is not isomorphic to U(10). Solution: Unfortunately, the two groups have the same order: the elements are U(n) are just the coprime elements

### ( ) 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

### A Little Beyond: Linear Algebra

A Little Beyond: Linear Algebra Akshay Tiwary March 6, 2016 Any suggestions, questions and remarks are welcome! 1 A little extra Linear Algebra 1. Show that any set of non-zero polynomials in [x], no two

### Examples: The (left or right) cosets of the subgroup H = 11 in U(30) = {1, 7, 11, 13, 17, 19, 23, 29} are

Cosets Let H be a subset of the group G. (Usually, H is chosen to be a subgroup of G.) If a G, then we denote by ah the subset {ah h H}, the left coset of H containing a. Similarly, Ha = {ha h H} is the

### Solutions to odd-numbered exercises Peter J. Cameron, Introduction to Algebra, Chapter 3

Solutions to odd-numbered 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

### Math 594, HW2 - Solutions

Math 594, HW2 - Solutions Gilad Pagi, Feng Zhu February 8, 2015 1 a). It suffices to check that NA is closed under the group operation, and contains identities and inverses: NA is closed under the group