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 Groups 1 2 Groups 2 3 Finite Groups; Subgroups 4 4 Cyclic Groups 8 5 Permutation Groups 12 6 Isomorphisms 17 7 Cosets and Lagrange s Theorem 21 8 External Direct Products 24 9 Normal Subgroups and Factor Groups 27 10 More on Group Homomorphisms 30 11 Introduction to Rings 34 12 Integral Domains 36 13 Ideals and Factor Rings 39 14 Ring Homomorphisms 42 ii

Chapter 1 Introduction to Groups Homework: # 1, 3, 5, 7. 1

Chapter 2 Groups Homework: # 1, 3, 5(a, d), 7, 9, 15, 19, 23, 25. Definition 2.1. Let G be a set. A binary operation on G is a function : G G G. The set G together with the binary operation will be written (G, ). Definition 2.2. Let G be a set together with a binary operation. We say (G, ) is a group if i. a b G for all a, b G. (Closure Property) (This is redundant) ii. (a b) c = a (b c) for all a, b, c G. (Associative Property) iii. There is an element e G such that a e = a = e a for every a G. (Identity) iv. For every a G there is an element b G such that a b = e = b a. (Inverse) Elementary Properties of Groups Theorem 2.1. The identity element of a group is unique. Proof. Suppose that a group G has two identity elements e and e. Then a e = e a = a and a e = e a = a for all a G. Now, e = e e = e. Theorem 2.2. Let G be a group and a, b, c G. Then i. a b = a c implies b = c. (left cancellation) ii. b a = c a implies b = c. (right cancellation) Proof. Let a be the inverse of a. Multiply on the left in (i) and on the right in (ii) by a. Theorem 2.3. The inverse of each element in a group is unique. 2

CHAPTER 2. GROUPS 3 Proof. Let G be a group and a G. Suppose that b and c are inverses of a, that is, a b = b a = e and a c = c a = e. Now, Hence the inverse is unique. Analogy of operators b = b e = b (a c) = (b a) c = e c = c. Multiplication Addition a b ab a + b Identitye 1 or e 0 Inverse a 1 a a a a(n a s) a n a + a + + a that is na ab 1 a b Theorem 2.4. For group elements a and b, (ab) 1 = b 1 a 1. Proof. We have (ab)(ab) 1 = e and (ab)(b 1 a 1 ) = a(bb 1 )a 1 = aea 1 = aa 1 = e. Thus (ab) 1 and b 1 a 1 are both inverses of ab. Since the inverse is unique, we have to have (ab) 1 = b 1 a 1. Definition 2.3. A group (G, ) is called an abelian group if a b = b a for all a, b G. Example 2.1. Prove that a group G is abelian if and only if (ab) 1 = a 1 b 1. Proof. Suppose that G is abelian. Then, (ab) 1 = b 1 a 1 = a 1 b 1. Conversely, assume that (ab) 1 = a 1 b 1. This implies (ab)(a 1 b 1 ) = e. Multiply from right by ba. We get, (ab)(a 1 b 1 )(ba) = e(ba) = ab = ab Example 2.2. Let G be a group and (b 1 ) n = (b n ) 1, where n is a positive integer. Proof. b n (b n ) 1 = e. And, b n (b 1 ) n = (bb b)(b 1 b 1 b 1 ) = e. By the uniqueness of the inverse (b n ) 1 = (b 1 ) n.

Chapter 3 Finite Groups; Subgroups Homework: # 2, 4, 6, 12, 18, 20, 22, 26, 42, 44. Definition 3.1. The number of elements of a group G is called the order of G and is denoted by G. Definition 3.2. The order of a group element g, denoted by g, is the smallest positive integer n such that g n = 3 (or ng = 0 if the operation is addition). If there is no such integer, then we say that the element has infinite order, that is, g =. Example 3.1. Take G = (Z, +). Then G =. Also, the order of any element k Z has infinite order. Example 3.2. Consider U(10) = {1, 3, 7, 9} under multiplication modulo 10. So, U(10) = 4. The order of the elements: 1 = 1, 3 = 4, 7 = 4, 9 = 2. Example 3.3. Consider Z 10 = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} under addition modulo 10. Then Z 10 = 10. The order of the elements: 0 = 1 1 = 10 2 = 5 3 = 10 4 = 5 5 = 2 6 = 5 7 = 10 8 = 5 9 = 10 Definition 3.3. Let H be a subset of a group G. If H itself is a group under the operation of G, then we say that H is a subgroup of G and write H G (or H < G if H is a proper subgroup.) The set {e} is a subgroup called trivial subgroup. Example 3.4. SL 2 (R) < GL 2 (R). Example 3.5. Z 10 under addition modulo 10 is not a subgroup of Z under addition. 4

CHAPTER 3. FINITE GROUPS; SUBGROUPS 5 Subgroup Tests Theorem 3.1. Let H be a non-empty subset of a group G. Then H is a subgroup of G if and only if ab 1 H for all a, b H. Proof. Let H G and a, b H. Since H is a group in itself, b 1 H and hence ab 1 H. Conversely, assume that ab 1 H for all a, b H. We want to show that H is a group. i. Associativity: H is associative because G is. ii. Identity: We have ab 1 H for every a, b H. Take b = a. We get e = aa 1 H. iii. Inverse: Let x H. Since e H, x 1 = ex 1 H. iv. Closure: Let x, y H. Since y = (y 1 ) 1 and y 1 H (by iii.), xy = x(y 1 ) 1 H. Example 3.6. Let G be an abelian group with identity e. Then H = {x G : x 2 = e} is a subgroup of G. Proof. Since e 2 = e, e H. So H is nonempty. Let a, b H. Then by Theorem 3.1 it suffices to show that ab 1 H. For this we have to show that (ab 1 ) 2 = e. Since G is abelian, (ab 1 ) 2 = (ab 1 )(ab 1 ) = a 2 (b 1 ) 2 = e(b 1 ) 2 = (b 2 ) 1 = e 1 = e. Corollary 3.2. Let H be a nonempty subset of a group G. Then H is a subgroup of G if and only if a, b H and a 1 H whenever a, b H. Proof. If H is a subgroup and a, b H then ab H and a 1 H. To prove the other way around, it suffices to show that ab 1 H whenever a, b H. Assume that a, b H. Then by hypothesis, b 1 H and ab 1 H. Example 3.7. Let G be an abelian group and H = {x G : x is finite }. Then H is a subgroup of G. Proof. e = 1 implies e H. Let a, b H and let a = m and b = n. Now, since G is abelian, (ab 1 ) mn = a mn (b 1 ) mn = (a m ) n ((b n ) m ) 1 = e n (e m ) 1 = e. Thus ab 1 is finite. Hence ab 1 H. Example 3.8. Let G be an abelian group and H and K be subgroups of G. Then HK = {hk : h H, k K} is a subgroup of G.

CHAPTER 3. FINITE GROUPS; SUBGROUPS 6 Proof. e = ee implies e HK. So HK. Let a = h 1 k 1 and b = h 2 k 2 be any two elements of HK. Using the abelian property of G, ab = (h 1 k 1 )(h 2 k 2 ) = (h 1 h 2 )(k 1 k 2 ) HK. Similarly, a 1 = (h 1 k 1 ) 1 = k1 1 h 1 1 = h 1 1 k1 1 HK. Hence by Corollary 3.2, HK is a subgroup. Theorem 3.3. Let H be a nonempty subset of a group G. If i. H is finite and ii. H is closed under the operation of G, then H is a subgroup of G. Proof. Let a, b H. Since H is closed, ab H. It suffices to show that a 1 H. Since H is finite, the elements in the sequence a, a 2, a 3, H(H being closed) will repeat after finite number of elements. Take any two elements so that a m = a n. Suppose m > n. Then a m n = e. If m n = 1, then a = e. This implies that a 1 = a H. If m n 2, aa m n 1 = a m n = e. Hence a 1 = a m n 1 H. We are done. Theorem 3.4. Let G be a group and a G. Then a = {a n : n Z} is a subgroup of G. Notation: For n > 0, a n = (a 1 ) n. Proof. Since a a, a. Let a m, a n a. Then a m (a n ) 1 = a m n a. Hence a is a subgroup. a is called a cyclic subgroup. Definition 3.4. A group G is called cyclic group if G = a for some a G. The element a is called a generator of G and G is said to be generated by a. Remark 3.1. The cyclic groups are automatically abelian. Example 3.9. Consider U(10) = {1, 3, 7, 9}. Then 7 = {7, 9, 3, 1} = U(10). Similarly find 3 and 9. Example 3.10. Take Z 10 = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. Find 4. Solution: 4 = {4, 8, 2, 6, 0}. Example 3.11. Z = 1 = 1. Therefore Z under addition is a cyclic group. Definition 3.5. Let G be a group. The set is called the center of G. Z(G) = {a G : ax = xa for all x in G}

CHAPTER 3. FINITE GROUPS; SUBGROUPS 7 Theorem 3.5. Z(G) is a subgroup of G. Proof. e Z(G). Let a, b Z(G). Then (ab)x = a(bx) = a(xb) = (ax)b = (xa)b = x(ab) = ab Z(G). a Z(G) = ax = xa for all x G. = xa 1 = a 1 x = a 1 Z(G). We are done. Definition 3.6. Let G be a group and g G. The centralizer of g is the set C(g) = {x G : xg = gx}. Theorem 3.6. The centralizer C(g) of a group element g G is a subgroup of G. Proof. e C(g). If ab C(g) then (ab)g = a(bg) = a(gb) = (ag)b = (ga)b = g(ab) = ab C(g). ag = ga = ga 1 = a 1 g = a 1 C(g).

Chapter 4 Cyclic Groups Homework: #2, 4, 8(a), 10, 11, 12, 18, 22. Definition 4.1. A group G is called cyclic if G is generated by one element, that is, We write G = a. Example 4.1. Z = 1 = 1. G = {a n : n Z}. Example 4.2. Z 10 = 1 = 3 = 7 = 9. Quiz problem: Is U(10) cyclic? If it is find a generator. Theorem 4.1. Let G be a group and a G. i. If a has infinite order, then a i = a j if and only if i = j. ii. If a = n then a = {e, a, a 2, a 3,, a n 1 } and a i = a j if and only if n (i j). Proof. i. a i = a j = a i a j = e = a i j = e. Since a has infinite order i j = 0 = i = j. Other implication is obvious. ii. Let a k a. By the division algorithm, there exist integers q and r such that k = qn + r where 0 r < n. So a k = a qn+r = a qn a r = (a n ) q a r = e q a r = a r {e, a, a 2,, a n 1 } = a {e, a, a 2,, a n 1 }. The other way around is for free. Let a i = a j = a i j = e. i j = qn + r, 0 r < n. Now, e = a i j = a qn+r = a qn a r = (a n ) q a r = a r = r = 0 = i j = qn = n (i j). Coversely, suppose n (i j) = i j = nq for some q Z = a i j = a nq = e = a i = a j. 8

CHAPTER 4. CYCLIC GROUPS 9 Corollary 4.2. For any group element a, a = a. Corollary 4.3. Let G be a group and let a G with a = n. If a k = e, then n k. Proof. Write k = nq + r with 0 r < n. Then e = a k = a nq+r = (a n )qa r = a r = r = 0 = k = nq = n q. Theorem 4.4. Let a be a group element with a = n and k be a positive integer. Then a k = a gcd(n,k) and a k = n. gcd(n,k) Proof. Let gcd(n, k) = d. Then k = dr for some integer r. a k = a dr = (a d ) k = a k a d. There exist s, t Z such that d = ns + kt. So a d = a ns+kt = a ns a kt = (a n ) s (a k ) t = (a k ) t a k = a d a k. Thus we have a k = a d. Since (a d ) n d = a n = e = a d n d. For any positive integer i < n d = id < n = (a d ) i = a di e. Therefore a d = n d. Now ak = a k = a d = a d = n d. If a = 20, then a 6 = a 2, a 11 = a. Corollary 4.5. Let G be a finite cyclic group. The order of an element in G divides G. Proof. Let G = a and G = a = n. Let a k G. Then a k = n and n n. gcd(n,k) gcd(n,k) Corollary 4.6. Let a = n. Then 1. a i = a j if and only if gcd(n, i) = gcd(n, j). 2. a i = a j if and only if gcd(n, i) = gcd(n, j). Proof. 1. a i = a j = a i = a j = n gcd(n,i) = n gcd(n,j) = gcd(n, i) = gcd(n, j). Conversely, gcd(n, i) = gcd(g, j) = a gcd(n,i) = a gcd(n,j) = a i = a j. 2. a i = a j if and only if n gcd(n,i) = n gcd(n,j) if and only if gcd(n, i) = gcd(n, j). Corollary 4.7. Let a = n. Then a = a j if and only if gcd(n, j) = 1. Also, a = a j if and only if gcd(n, j) = 1. Corollary 4.8. Z n = k if and only if gcd(n, k) = 1.

CHAPTER 4. CYCLIC GROUPS 10 Classification of Subgroups of Cyclic Groups Theorem 4.9 (Fundamental Theorem of Cyclic Groups). i. Every subgroup of a cyclic group is cyclic. ii. If a = n, then order of any subgroup of a divides n. iii. If a = n, then for each positive divisor k of n there is exactly one subgroup of a of order k. Proof. i. Let H be a subgroup of a cyclic group a. If H = {e}. Then H is cyclic automatically. So let H {e}. Let m be the smallest positive integer such that a m H. We show that H = a m. Clearly, a m H. Next, let a t H. By division algorithm, t = qm + r, 0 r < m. Now a t = a qm+r = a qm a r = a r = a t a qm = a t (a m ) q H = r = 0 = a t = a mq = (a m ) q a m. Thus H a m. Hence H = a m. ii. Let H be any subgroup of a. Then H = a m. Therefore H = a m = n gcd(n,m) = H divides n. iii. Let k n. Then a n k is a subgroup or order k. To see this, a n k = n = k. n/k n = gcd(n, n k ) Let H be any subgroup of a and H = k. By (i), H = a m. Now k = H = a m n = gcd(n,m) = gcd(n, m) = n. Therefore H = k am = a gdc(n,m) = a n/k. Example 4.3. The list of subgroups of Z 20 is order 1 20 = {0} order 2 10 = {0, 10} order 4 5 = {0, 5, 10, 15} order 5 4 = {0, 4, 8, 12, 16} order 10 2 = {0, 2, 4, 6, 8, 10, 12, 14, 16, 18} order 20 1 = {0, 1, 2, 3,, 19}. Explanation: Since 4 20, Z 20 = 1 has a subgroup of order 4 which is generated by 1 20/4 = 1 5 = 5. Question 4.1. Does Z 24 have a subgroup of order 9?

CHAPTER 4. CYCLIC GROUPS 11 Answer: No because Z 24 is a cyclic group, the order of it s subgroup must divide 24 but 9 24. Example 4.4. The subgroup of order 8 in Z 24 is generated by 24 8 = 3. So the subgroup of order 8 is 3 = {0, 3, 6, 9, 12, 15, 18, 21}. The other generators are 9, 15, and 21 because gcd(8, 3) = 1 and 9 = 3 3. Similarly others. Euler φ function φ(1) = 1 and for n > 1, φ(n) = number of positive integers less than n and relatively prime to n. Theorem 4.10. The number of generators in a cyclic group of order n is φ(n). Proof. Let a = {e, a, a 2,, a n 1 } be a cyclic group of order n. Then we know that a = a j if and only if gcd(n, j) = 1. It follows that # generators = φ(n). Theorem 4.11. Let G be a cyclic group of order n and d n. Then the number of elements of order d in G is φ(d). Proof. d n = there is exactly one subgroup of order d in G, say it is a = {e, a, a 2,, a d 1 } Then all other elements of order d are in a and they generate the same subgroup. Therefore # of elements of order d = # of generators of a subgroup of order d = φ(d). Corollary 4.12. In a finite group G, the number of elements of order d is a multiple of φ(d). Proof. If there is no element of order d then there is nothing to prove. If a G has order d then a has φ(d) elements of order d. If all the elements of order d are in a, we are done. If there is another element b G of order d and b / a, then b contains φ(d) elements of order d. Note that a and b have no elements of order d in common because if c is an element of order d that is in both a and b, then a = c = b = b a. Thus we have 2φ(d) elements of order d. Continuing in this fashion, we see that the number of elements of order d is multiple of φ(d).

Chapter 5 Permutation Groups Homework: # 1, 2, 6, 8, 16, 24, 28 Definition 5.1. A permutation of a set A is a bijective function from A to itself. The set of all permutations of A forms a group under composition. The group is called the permutation group of A. Example 5.1. Let A = {1, 2, 3}. Then α : A A defined by α(1) = 2, α(2) = 3, α(3) = 1 (also show this by a diagram) is a permutation of A. A common way to write this permutation is [ ] 1 2 3 n α = α(1) α(2) α(3) α(n) So the α in the example is written If α = β = then the composition βα is given by βα = [ 1 2 3 2 3 1 [ 1 2 3 1 3 2 [ 1 2 3 3 2 1 To see this βα(1) = β(α(1)) = β(2) = 3 and so on. ] ] ] Symmetric Group S 3 S 3 is a group of permutation group of A = {1, 2, 3}. So {[ ] [ ] [ ] [ 1 2 3 1 2 3 1 2 3 1 2 3 S 3 =,,, 1 2 3 2 3 1 3 1 2 1 3 2 = {e, α, α 2, β, αβ, α 2 β} 12 ] [ 1 2 3, 2 1 3 ] [ 1 2 3, 3 2 1 ]}

CHAPTER 5. PERMUTATION GROUPS 13 Symmetry Group S n The set of all permutations of {1, 2, 3,, n} is called the symmetric group of degree n and is denoted by S n. An element of S n looks like [ ] 1 2 n α = α(1) α(2) α(n) Order of S n is n!. S n is an-abelian. Example 5.2. D 4 is a subgroup of S 4. Cycle Notation The permutation α = can be written in cycle notation as ( 1 2 3 4 5 ) 6 2 4 3 1 6 5 α = (1, 2, 4)(3)(5, 6) = (124)(3)(56) (1, 2, 4), (3) and (5, 6) are called the cycles. Then number of elements in a cycle is called the length of the cycle. So (a 1, a 2,, a m ) is called a cycle of length m or an m-cycle. Example 5.3. Let α = ( 1 2 3 4 ) 5 3 2 1 5 4 and β = ( 1 2 3 4 ) 5 4 5 1 3 2 In cycle notations α = (13)(2)(45) and β = (143)(25). The cycle notation for the composition is: αβ = (13)(2)(45)(143)(25) = (1524)(3) and βα = (143)(25)(13)(2)(45) = (1)(2534) They are written as the multiplication of disjoint cycles. It is preferred not to write cycles that have only one entry. Then the permutations in above example are written as α = (13)(45) αβ = (1524) β = (143)(25) βα = (2534) The identity permutation ε consists of only one cycle with one entry.

CHAPTER 5. PERMUTATION GROUPS 14 Properties of Permutations Theorem 5.1. Every permutation of a finite set can be written as a product of disjoint cycles. Proof. Let A = {1, 2,, n} and α be a permutation of A. Let a A. Since A is finite, the sequence a, α(a), α 2 (a), α 3 (a), must repeat. Let m be the smallest positive integer such that α m (a) = a. Such m exist because if α i (a) = α j (a), i > j, then α i j (a) = a. Hence one of the cycles of α is (a, α(a), α 2 (a),, α m 1 (a)). If there is an element b A not appearing in this cycle, we similarly construct a cycle of α containing b, say (b, α(b), α 2 (b),, α k 1 (b)). These two cycles are disjoint because α i (a) = α j (b), i j( we can assume this) = b = α i j (a). Continue this process until the set A is exhausted. Theorem 5.2. Disjoint cycles of a permutation commute. Proof. Let α be a permutation. It s enough to consider that α has only two disjoint cycles, say α = (a 1, a 2,, a m )(b 1, b 2,, b n ) Let α 1 = (a 1, a 2,, a m ) and α 2 = (b 1, b 2,, b n ). We want to show that α 1 α 2 = α 2 α 1. Here α 1 α 2 (a i ) = α 1 (α 2 (a i )) = α 1 (a i ) = a i+1 where 1 i m and a m+1 = a 1, and α 2 α 1 (a i ) = α 2 (α 1 (a i )) = α 2 (a i+1 ) = a i+1 α 1 α 2 (b j ) = α 1 (α 2 (b j )) = α 1 (b j+1 ) = b j+1 α 2 α 1 (b j ) = α 2 (α 1 (b j )) = α 2 (b j ) = b j+1 where 1 j n and b n+1 = b 1. For any element x not in the list a 1,, a m, b 1,, b n we have α 1 α 2 (x) = α 2 α 1 (x) = x. Therefore, α 1 α 2 = α 2 α 1. Theorem 5.3. A cycle of length n has order n. Proof. Let α = (a 1, a 2,, a n ) be a cycle of length n. Then { α j a i+j if i + j n (a i ) = a i+j n if i + j > n. It follows that α n (a i ) = a i+n n = a i and for j < n, { α j a i+j if i + j n = a i+j a i (a i ) = a i+j n if i + j > n = a i+j n a i. Thus α has order n.

CHAPTER 5. PERMUTATION GROUPS 15 Theorem 5.4. The order of a permutation of a finite set is the least common multiple of the lengths of its disjoint cycles. Proof. We prove this theorem for a permutation which is a multiplication of two disjoint cycles. Other permutations are similarly handled. Let α be a permutation of a finite set which is a multiplication of two disjoint cycles β and γ of length m and n, respectively. Let lcm(m, n) = l. Then l = mx = ny for some positive integers x and y. Now, α l = (βγ) l = β l γ l = (β m ) x (γ n ) y = ε For k < l either m k or n k. Assume m k. Then k = qm + r, 0 r < m. It follows that α k = (βγ) k = β k γ k = β qm+r γ k = β r γ k ε. Therefore the order of α is l. Example 5.4. S 5 has 5! = 120 elements. They have the following cycle structure. Cycle Structure Order of the Elements Number of elements 5 5 24 4, 1 4 30 3, 2 6 20 3, 1, 1 3 20 2, 2, 1 2 15 2, 1, 1, 1 2 10 1, 1, 1, 1, 1 1 1 Theorem 5.5. Every permutation in S n, n > 1, is a product of 2-cycles. Proof. An identity permutation can be written as (12)(21). Since a permutations can be written as a product of disjoint cycles, it suffices to show that a cycle is a product of 2-cycles. Let (a 1 a 2 a k ) be a k-cycle. Then (a 1 a 2 a k ) = (a 1 a k )(a 1 a k 1 )(a 1 a k 2 ) (a 1 a 3 )(a 1 a 2 ) = (a 1 a 2 )(a 2 a 3 )(a 3 a 4 ) (a k 1 a k ). Example 5.5. (12345) = (15)(14)(13)(12) = (12)(23)(34)(45) = (54)(53)(52)(51) Lemma 5.6. If β 1 β 2 β r = ε, where β s are 2-cycles, then r is even. Proof. Take for granted. Theorem 5.7. If a permutatin α can be expressed as a product of even (odd) number of 2-cycles, then every decomposition of α into a product of 2-cycles must have even (odd) number of 2-cycles.

CHAPTER 5. PERMUTATION GROUPS 16 Proof. If α = β 1 β 2 β r = γ 1 γ 2 γ s be two decompositions of α into the product of 2-cycles. Then ε = γ 1 γ 2 γ s β r β r 1 β 2 β 1 On the right is the product of r + s 2-cycles. By the lemma r + s is even. Thus r and s are either both even or both odd. Definition 5.2. A permutation that can be expressed as a product of an even number of 2-cycles is called an even permutation. A permutation that can be expressed as a product of an odd number of 2-cycles is called an odd permutation. Theorem 5.8. The set of even permutations in S n forms a subgroup of S n. Proof. Let H be a subset of S n consisting of even permutations. Then ε H. If α, β H, then αβ 1 H. (Show how!) So H is a subgroup. Definition 5.3. The group of even permutations of n objects is denoted by A n and is called the alternating group of degree n. Theorem 5.9. For n > 1, A n has order n!/2. Proof. If α is an odd permutation, then (12)α is even permutation and if α β are two distinct odd permutations then (12)α (12)β are two distinct even permutations. It follows that, number of even permutations number of odd permutations By the similar argument with even and odd switched we get, Thus we have number of even permutations number of odd permutations number of even permutations = number of odd permutations = n!/2.

Chapter 6 Isomorphisms Homework: # 4, 10, 22, 48, 50 Definition 6.1. Let (G, ) and (G, ) be two groups. A group homomorphism from G to G is a mapping φ : G G that preserves the group operation. That is, φ(a b) = φ(a) φ(b) for all a, b G. If a homomorphism φ : G G is bijective, then it is called an isomorphism and we say that the groups G and G are isomorphic. Example 6.1. Let G = a = {a n : n Z} be an infinite cyclic group. Then G Z. The map φ : G Z given by φ(a k ) = k is an isomorphis. To see this, 1. φ(a k a l ) = φ(a k+l ) = k + l = φ(a k ) + φ(a l ). So φ is a homomorphism. 2. φ(a k ) = φ(a l ) = k = l = a k = a l. So φ is one-to-one. 3. For k Z there is a k G such that φ(a k ) = k. So φ is onto. Hence φ is an isomorphism. Example 6.2. Any finite cyclic group of order n is isomorphic to Z n. U(10) and U(5) are isomorphic to Z 4. Example 6.3. The mapping φ : (R, +) (R, +) given by φ(x) = x 2 is not an isomorphism because φ(x + y) = (x + y) 2 and φ(x) + φ(y) = x 2 + y 2 are not equal for all x, y R. Example 6.4. U(10) U(12), where U(10) = {1, 3, 7, 9} and U(12) = {1, 5, 7, 11}. Observe that x 2 = 1 for every x U(12). Now φ(9) = φ(3 3) = φ(3)φ(3) = 1 and φ(1) = φ(1 1) = φ(1)φ(1) = 1. Thus φ(9) = φ(1) and hence no homomorphism can be one-to-one. 17

CHAPTER 6. ISOMORPHISMS 18 Example 6.5. U(8) = {1, 3, 5, 7} and U(12) = {1, 5, 7, 11} are isomorphic. φ(1) = 1, φ(3) = 5, φ(5) = 7, φ(7) = 11 is an isomorphism. φ is automatically bijective. Show that it is a homomorphism. Theorem 6.1 (Caley s Theorem). Evert group is isomorphic to a group of permutations. Proof. Let G = {T g : g G and T g : G G is defined by T g (a) = ga for all a G}. 1. T g is a permutation of G. (a) T g (a) = T g (b) = ga = gb = a = b = T g is one-to-one. (b) For every a G there is g 1 a G such that T g (g 1 a) = g(g 1 a) = a = T g is onto. Hence T g is a permutation of G. 2. G is a group under function composition. (a) Closure: T g T h (a) = (gh)a = T gh (a) for all a G = T g T h = T gh G for all g, h G. (b) Associativity: T g (T h T k ) = T g T hk = T g(hk) = T (gh)k = T gh T k = (T g T h )T k. (c) Identity: T e T g = T eg = T g = T ge = T g T e = T e is the identity in G. (d) Inverse: T g T g 1 = T gg 1 = T e = (T g ) 1 = T g 1 G. 3. φ : G G defined by φ(g) = T g is an isomorphism. (a) φ is a homomorphism: φ(gh) = T gh = T g T h = φ(g)φ(h) for all g, h G. (b) φ is One-to-one: φ(g) = φ(h) = T g = T h = T g (e) = T h (e) = ge = he = g = h. (c) φ is onto: For every T g G automatically g G and φ(g) = T g. Properties of Isomorphisms Theorem 6.2. Let φ : G G be a group homomorphism. Then 1. If e is the identity of G, then φ(e) is the identity of G. 2. For every a G, φ(a 1 ) = φ(a) 1. 3. For every a G, φ(a n ) = (φ(a)) n. Proof.

CHAPTER 6. ISOMORPHISMS 19 1. Let the identity of G be e. Then φ(e) = φ(e) = φ(e)φ(e) = (φ(e)) 1 φ(e) = φ(e) = e = φ(e). 2. e = φ(e) = φ(aa 1 ) = φ(a)φ(a 1 ) = φ(a 1 ) = φ(a) 1. 3. For n > 0, φ(a n ) = φ(aa a) = φ(a)φ(a) φ(a) = (φ(a)) n (Use induction). For n = 0, it is obvious. For n < 0, φ(a n ) = φ((a 1 ) n ) = (φ(a 1 )) n = (φ(a) 1 ) n = φ(a) n. Theorem 6.3. Let φ : G G be an isomorphism. Then 1. For a, b G, ab = ba if and only if φ(a)φ(b) = φ(b)φ(a). 2. G = a if and only if G = φ(a). 3. a = φ(a) for all a G. (isomorphisms preserve orders) Proof. 1. ab = ba = φ(ab) = φ(ba) = φ(a)φ(b) = φ(b)φ(a). Conversely, φ(a)φ(b) = φ(b)φ(a) = φ(ab) = φ(ba) = ab = ba. The last implication follows because φ is one-to-one. 2. Let G = a. Then φ(a) G = φ(a) G. Let b G. Since φ is onto, there is a n a such that φ(a n ) = b = b = φ(a n ) = (φ(a)) n φ(a) = G φ(a). Thus G = φ(a). Conversely, let G = φ(a). Since a G, a G. Let x G. Since φ(x) φ(a), φ(x) = φ(a) m = φ(a m ) for some m. Since φ is one-to-one x = a m = x a = G a. Thus G = a. 3. Let a = n. Then φ(a) n = φ(a n ) = φ(e) = e. If for 0 < i < n, φ(a) i = e, then φ(a i ) = φ(e) = a i = e which is a contradiction. Hence φ(a) i e. Hence φ(a) = n. Theorem 6.4. Suppose that φ : G G is an isomorphism. Then 1. φ 1 : G G is an isomorphism. 2. If H is a subgroup of G, then φ(h) = {φ(h) : h H} is a subgroup of G. Proof.

CHAPTER 6. ISOMORPHISMS 20 1. Let a, b G. Then there exist a, b G such that φ(a) = a and φ(b) = b.φ 1 (a b ) = φ 1 (φ(a)φ(b)) = φ 1 (φ(ab)) = ab = φ 1 (a )φ 1 (b ). Since φ is bijective, φ 1 is bijective. Hence φ 1 is an isomorphism. 2. e = φ(e) φ(h). Let h, k φ(h). Then there exist h, k H such that φ(h) = h and φ(k) = k. Now h k 1 = φ(h)φ(k) 1 = φ(h)φ(k 1 ) = φ(hk 1 ) φ(h). Hence φ(h) is a subgroup of G. Definition 6.2. An automorphism of a group G is an isomorphism from G to itself. Example 6.6. φ : R 2 R 2 defined by (a, b) (a, b) is an automorphism of the group R 2 under componentwise addition. We denote the set of all automorphisms of a group G by Aut(G). Theorem 6.5. Let G be a group. Aut(G) is a group under function composition. Proof. 1. Closure: Let φ, ψ Aut(G) then φ ψ Aut(G). 2. Associative: For any φ, ψ, χ Aut(G), (φ ψ) χ = φ (ψ χ). 3. Identity: I d : G G, I d (g) = g for all g G is the identity in Aut(G). 4. Inverse: For every φ Aut(G), φ 1 Aut(G). Therefore Aut(G) is a group.

Chapter 7 Cosets and Lagrange s Theorem Homework: # 2, 6, 8, 16, 30, 40. Definition 7.1. Let G be a group and H be a subgroup of G. For a G, the set ah = {ah : h H} is called the left coset of H in G containing a. The set Ha = {ha : h H} is called the right coset of H in G containing a. The element a is called the coset representative of ah and Ha. Example 7.1. Let H = {0, 2, 4, 6, 8} be a subgroup of Z 10. Let s determine all the cosets of H in Z 10. 0 + H = 2 + H = 4 + H = 6 + H = 8 + H = {0, 2, 4, 6, 8} = H 1 + H = 3 + H = 5 + H = 7 + H = 9 + H = {1, 3, 5, 7, 9} So there are only two left cosets H and 1 + H. Example 7.2. Let G = S 3 and H = {(1), (12)}. Let s find all the cosets of H in G. Recall that S 3 = {(1), (12), (23), (13), (123), (132)}. (1)H = {(1), (12)} (12)H = {(12), (1)} (13)H = {(13), (123)} (23)H = {(23), (132)} (123)H = {(123), (13)} (132)H = {(132), (23)} We see that (1)H = (12)H, (13)H = (123)H, (23)H = (132)H. There are only 3 cosets. 21

CHAPTER 7. COSETS AND LAGRANGE S THEOREM 22 Properties of Cosets Theorem 7.1. Let H be a subgroup of a group G, and let a, b G. Then, 1. a ah. Proof. a = ae ah. 2. ah = H if and only if a H. Proof. Since a ah, ah = H = a H. Conversely, let a H. Then for any h H, ah H = ah H. Again for any h H, a 1 h H = h = (aa 1 )h = a(a 1 h) ah = H ah. Hence H = ah. 3. (ab)h = a(bh) and H(ab) = (Ha)b. Proof. For all h H, (ab)h = a(bh) = (ab)h = a(bh). 4. ah = bh if and only if a bh. Proof. Assume that ah = bh. Since a ah, a bh. Conversely, let a bh = a = bh for some h H = ah = (bh)h = b(hh) = bh. 5. either ah = bh or ah bh =. Proof. If ah bh then x ah bh = ah = xh = bh. 6. ah = bh if and only if a 1 b H. Proof. ah = bh if and only if H = a 1 (bh) = (a 1 b)h = a 1 b H. 7. ah = bh. Proof. ah bh is a one-to-one and onto map from ah to bh. To see this bh 1 = bh 2 = h 1 = h 2 = ah 1 = ah 2. By definition it is clearly onto. 8. ah = Ha if and only if H = aha 1. Proof. ah = Ha if and only if (ah)a 1 = (Ha)a 1 if and only if aha 1 = H. 9. ah is a subgroup of G if and only if a H.

CHAPTER 7. COSETS AND LAGRANGE S THEOREM 23 Proof. If a H, then ah = H. So ah is a subgroup. Conversely, suppose that ah is a subgroup. Then the identity e ah = e = ah for some h H = a = h 1 H. Theorem 7.2 (Lagrange s Theorem). Let G be a finite group and H is a subgroup of G, then H divides G. Moreover the number of distinct left (right) cosets of H in G is G / H. Proof. Let a 1 H, a 2 H,, a k H be the distinct left cosets of H in G. For any a G, ah is one of the cosets in the list, say ah = a i H. Then a a i H. Therefore Since these cosets are mutually disjoint, G = a 1 H a k H G = a 1 H + + a k H = k H Thus H divides G. From this it also follows that k = the number of distinct left (right) cosets = G / H. Definition 7.2. The index of a subgroup H in G is the number of distinct left cosets of H in G and is denoted by G : H ( also [G : H]) Corollary 7.3. If H is a subgroup of a finite group G, then G : H = G / H. Corollary 7.4. Let G be a finite group and a G. Then a divides G. Proof. Since a is a subgroup of G, a divides G. Since a = a, a divides G. Corollary 7.5. A group of prime order is cyclic. Proof. Let G be prime and a G, a e. Then a > 1 and a divides G. Since G is prime a = G. Thus we must have G = a. Corollary 7.6. Let G be a finite group and a G. Then a G = e. Proof. Since a divides G, G = a k for some k Z. Therefore a G = a a k = e k = e. Theorem 7.7 (Fermat s Little Theorem). For every integer a and every prime p, a p mod p = a mod p. Proof. Write a = pq + r, 0 r < p. Then a mod p = r and a p mod p = r p mod p. Let us show that r p mod p = r. Then we will be done. Since gcd(r, p) = 1, r U(p) = {1, 2,, p 1}. By the previous corollary r p 1 mod p = 1. Therefore r p mod p = r.

Chapter 8 External Direct Products Homework: 4, 8, 9, 10, 22 Definition 8.1. Let G 1, G 2,, G n be a finite collection of groups. The external direct product of G 1, G 2,, G n is the set G 1 G 2 G n = {(g 1, g 2,, g n ) : g i G i }, where (g 1, g 2,, g n )(g 1, g 2, g n) = (g 1 g 1, g 2 g 2,, g n g n). It is clear that G 1, G 2,, G n = G 1 G 2 G n. Theorem 8.1. The external direct product of groups G 1, G 2,, G n is a group. Proof. By definition it is closed. Associativity follows from ((g 1, g 2,, g n )(h 1, h 2,, h n ))(k 1, k 2,, k n ) =(g 1 h 1, g 2 h 2,, g n h n )(k 1, k 2,, k n ) =((g 1 h 1 )k 1, (g 2 h 2 )k 2,, (g n h n )k n ) =(g 1 (h 1 k 1 ), g 2 (h 2 k 2 ),, g n (h n k n )) =(g 1, g 2,, g n )(h 1 k 1, h 2 k 2,, h n k n ) =(g 1, g 2,, g n )((h 1, h 2,, h n )(k 1, k 2,, k n )). (e 1, e 2,, e n ) is the identity. Inverse of (g 1, g 2,, g n ) is (g 1 1, g 1 2,, g 1 n ). According to this definition R 2 = R R and R 3 = R R R. Example 8.1. U(5) = {1, 2, 3, 4} and U(8) = {1, 3, 5, 7}. U(5) U(8) = {(1, 1), (1, 3), (1, 5), (1, 7), (2, 1), (2, 3), (2, 5), (2, 7), (3, 1), (3, 3), (3, 5), (3, 7), (4, 1), (4, 3), (4, 5), (4, 7)} For instance, (3, 3)(3, 3) = (4, 1) and inverse of (3, 5) is (2, 5). Example 8.2. A group of order 4 is isomorphic to Z 4 or Z 2 Z 2. 24

CHAPTER 8. EXTERNAL DIRECT PRODUCTS 25 Properties of External Direct Products Theorem 8.2. Let G 1, G 2,, G n be finite groups and (g 1, g 2,, g n ) G 1 G 2 G n. Then (g 1, g 2,, g n ) = lcm( g 1, g 2,, g n ). Proof. Let (g 1, g 2,, g n ) = m and lcm( g 1, g 2,, g n ) = l. We want to show that m = n. (g 1, g 2,, g n ) = m = (g1 m, g2 m,, gn m ) = (e 1, e 2,, e n ) = g i divides m for every i = 1, 2,, n. This implies m is a multiple of g 1, g 2,, g n = l m. On the other hand, lcm( g 1, g 2,, g n ) = l = l is multiple of g 1, g 2,, g n. Therefore (g 1, g 2,, g n ) l = (g1, l g2, l, gn) l = (e 1, e 2,, e n ) = m l. Thus m = l. Example 8.3. Determine the number of elements of order 3 in Z 6 Z 12. Solution: If (a, b) Z 6 Z 12 has (a, b) = 3, then lcm( a, b ) = 3. Case i. a = 3, b = 1. There are two choices for a, namely, 2 and 4 and one choice for b namely 0. This gives 2 elements of order 3. Case ii. a = 3, b = 3. There are two choices for a and two choices for b, namely, 4 or 8. This gives 4 elements of order 3. Case iii. a = 1, b = 3. There is one choice for a and two choices for b. This gives 2 elements of order 3. Therefore, there are 8 elements of order 3 in Z 6 Z 12. Example 8.4. Determine the number of cyclic subgroups of order 10 in Z 30 Z 60. Example 8.5. Determine the number of cyclic subgroups of order 10 in Z 25 Z 30. Solution: First, let us determine the number of elements of order 10 in Z 25 Z 30. If (a, b) Z 25 Z 30, then lcm( a, b ) = 10. Case 1. a = 1 and b = 10. There is only one choice for a, namely 0. By Theorem 4.11, there are φ(10) = 4 choices for b, namely 3, 9, 21, and 27. This gives 4 elements of order 10. Case 2. a = 5 and b = 2 or 10. There are φ(5) = 4 choices for a and φ(2)+φ(10) = 1 + 4 = 5 choices for b. This gives 20 elements of order 10. Thus, there are 24 elements of order 10 in Z 25 Z 30. Since each cyclic subgroup of order 10 contains φ(10) = 4 elements of order 10, there must be 24/4 = 6 cyclic subgroups of order 10. Example 8.6. Does Z 25 Z 30 have a subgroups isomorphic to Z 5 Z 3?

CHAPTER 8. EXTERNAL DIRECT PRODUCTS 26 Solution: Yes, it is 6 10. Theorem 8.3. Let G and H be finite cyclic groups. Then G H is cyclic if and only if gcd( G, H ) = 1. Proof. Let G = m, H = n. Assume that G H is cyclic, say G H = (g, h). Let gcd(m, n) = d. Observe that (g, h) mn/d = ((g m ) n/d, (h n ) m/d ) = (e, e) = mn = (g, h) mn/d = d = 1. Conversely, assume that gcd(m, n) = 1. Let G = g and H = h. Then (g, h) = lcm(m, n) = G H. mn gcd(m,n) = mn = G H. This implies that (g, h) generates Corollary 8.4. Let G 1, G 2,, G n be finite cyclic groups. Then G 1 G 2 G n is cyclic if and only if gcd( G i, G j ) = 1 when i j. Proof. Use induction. Corollary 8.5. Let m = n 1 n 2 n k. Then Z m is isomorphic to Z n1 Z n 2 Z nk if and only if gcd(n i, n j ) = 1 when i j. Proof. Follows directly from Corollary 8.4. We have Z 2 Z 3 Z 6. Similarly, Z 2 Z 2 Z 3 Z 5 Z 2 Z 6 Z 5 Z 2 Z 30. But Z 2 Z 6 Z 12.

Chapter 9 Normal Subgroups and Factor Groups Homework: #2, 4, 12, 14, 18 Normal Subgroups Definition 9.1. A subgroup H of a group G is called a normal subgroup of G, written H G, if ah = Ha for all a G. Question 9.1. What is the meaning of ah = Ha? Answer: For every h H there is h H such that ah = h a. Theorem 9.1. A subgroup H of G is normal in G if and only if xhx 1 H for all x G. Proof. Assume that H G. Let x G. We want to show that xhx 1 H. Let xhx 1 xhx 1. Since xh = Hx, xh = h x for some h H = xhx 1 = h H. Conversely, assume that xhx 1 H for all x G. Since x 1 G, it follows that x 1 Hx H. xhx 1 H = xh Hx and x 1 Hx H = Hx xh. Therefore xh = Hx for all x G. Example 9.1. Let H = {0, 3, 6, 9} and G = Z 12. Then for any a G, a+h = H +a. Hence H G. In general, every subgroup of an Abelian group is normal. Example 9.2. The center Z(G) of a group G is normal. It is obvious that az(g) = Z(G)a for all a G. Example 9.3. The alternating group A n of even permutations is a normal subgroup of S n. If α S n is an even permutation, then αa n = A n α (why?). Same is true if α is odd. Example 9.4. Let H G and K be any subgroup of G. Then HK = {hk : h H, k K} is a subgroup of G. 27

CHAPTER 9. NORMAL SUBGROUPS AND FACTOR GROUPS 28 Solution: e = ee HK = HK. Let a = h 1 k 1 and b = h 2 k 2 HK where h 1, h 2 H and k 1, k 2 K. Now, ab 1 = (h 1 k 1 )(k1 2 h 1 2 ) = h 1 (k 1 k2 1 )h 1 2 = h 1 h (k 1 k2 1 ) for some h H. Thus, ab 1 = (h 1 h )(k 1 k2 1 ) HK. Hence HK is a subgroup of G. Example 9.5. Recall that GL(2, R) is a group of 2 2 invertible matrices, that is, the matrices with non-zero determinant and SL(2, R), the set of 2 2 matrices with determinant 1 is a subgroup of GL(2, R). Notice that the operation here is matrix multiplication. Solution: For every x GL(2, R) we want to show that xsl(2, R)x 1 SL(2, R). Let xsx 1 xsl(2, R)x 1. det(xsx 1 ) = det x det s det x 1 = det x det x 1 = 1 = xsx 1 SL(2, R). By Theorem 9.1, SL(2, R) GL(2, R). Factor Groups If H is a normal subgroup of a group G, by definition, the left cosets of H in G and right cosets of H in G are same. So the set of left cosets is same as the set of right cosets. That is, {ah : a G} = {Ha : a G} The set of left cosets of H in G which we denote by G/H = {ah : a G} Theorem 9.2. Let H be a normal subgroup of a group G. Then the G/H is a group under the operation (ah)(bh) = (ab)h. Proof. i. We have to show that the operation is well defined. Let ah = a H and bh = b H. We show that (ah)(bh) = (a H)(b H), that is (ab)h = (a b )H. ah = a H and bh = b H = a = a h 1 and b = b h 2. Now, (ab)h = (a h 1 b h 2 )H = (a h 1 b )(h 2 H) = (a h 1 b )H = (a h 1 )(b H) = (a h 1 )(Hb ) = (a h 1 )(Hb ) = a (h 1 H)b = a Hb = a b H. This proves that the operation is well defined. ii. Associativity is automatic: (ahbh)ch = (ab)hch = (ab)ch = a(bc)h = ah(bc)h = ah(bhch). iii. eh = H is the identity.

CHAPTER 9. NORMAL SUBGROUPS AND FACTOR GROUPS 29 iv. a 1 H is the inverse of ah. Example 9.6. Let 5Z = {0, ±5, ±10, }. Clearly 5Z is a subgroup of Z. Moreover it is a normal subgroup. The left cosets of 5Z in Z are 0 + 5Z = {, 10, 5, 0, 5, 10, } 1 + 5Z = {, 9, 4, 1, 6, 11, } 2 + 5Z = {, 8, 3, 2, 7, 12, } 3 + 5Z = {, 7, 2, 3, 8, 13, } 4 + 5Z = {, 6, 1, 4, 9, 14, } These are the only cosets because for any k Z, by division algorithm, k = 5q+r, 0 r < 5. Hence k + 5Z = r + 5q + 5Z = r + 5Z. Thus k + Z is one the above left cosets. Its Cayley table is, 0 + 5Z 1 + 5Z 2 + 5Z 3 + 5Z 4 + 5Z 0 + 5Z 0 + 5Z 1 + 5Z 2 + 5Z 3 + 5Z 4 + 5Z 1 + 5Z 1 + 5Z 2 + 5Z 3 + 5Z 4 + 5Z 0 + 5Z 2 + 5Z 2 + 5Z 3 + 5Z 4 + 5Z 0 + 5Z 1 + 5Z 3 + 5Z 3 + 5Z 4 + 5Z 0 + 5Z 1 + 5Z 2 + 5Z 4 + 5Z 4 + 5Z 0 + 5Z 1 + 5Z 2 + 5Z 3 + 5Z Clearly, Z/5Z is isomorphic to Z 5. Example 9.7. In general, for any n > 0, Z/nZ Z n.

Chapter 10 More on Group Homomorphisms Definition 10.1. A homomorphism φ from a group G to a group G is a mapping from G into G such that φ(ab) = φ(a)φ(b), for all a, b G. Definition 10.2. The kernel of a group homomorphism φ : G G is the set where e is the identity of G. ker φ = {g G : φ(g) = e } Example 10.1. If φ : G G is an isomorphism, then ker φ = {e}. Example 10.2. Let φ : GL(2, R) R be defined by φ(a) = det A, where GL(2, R) is a group of 2 2 invertible matrices under matrix multiplication and R is a group of nonzero real numbers under multiplication. Then implies φ is homomorphism. And, φ(ab) = det(ab) = det A det B = φ(a)φ(b) ker φ = {A GL(2, R) : φ(a) = det A = 1} = SL(2, R). Example 10.3. The set R[x] of all polynomials with real coefficients is a group under addition. The derivative D x : R[x] R[x] defined by D x (f) = df dx is a homomorphism. the kernel of D x is { ker D x = f R[x] : D x (f) = df } dx = 0 = All constant polynomials = R. Example 10.4. The kernel of φ : Z Z 10, n n mod 10 is 10. 30

CHAPTER 10. MORE ON GROUP HOMOMORPHISMS 31 Properties of Homomorphisms Theorem 10.1. Let φ : G G be a group homomorphism. Then i. ker φ is a subgroup of G. ii. If H is normal subgroup of G then φ(h) is a normal subgroup of φ(g). Proof. i. Since φ(e) = e, e ker φ = ker φ. Let a, b ker φ. Then φ(ab 1 = φ(a)φ(b 1 ) = e φ(b) 1 = φ(b) 1 = e 1 = e implies ab 1 ker φ. Hence ker φ is subgroup of G. ii. Recall that φ(h) is a subgroup of φ(g) and φ(g) is a subgroup of G. φ(g) φ(g) be an arbitrary element. For every φ(h) φ(h), Let φ(g)φ(h)φ(g) 1 = φ(ghg 1 ) φ(h) because ghg 1 ghg 1 H. Therefore φ(h) is normal in φ(g). Left Cosets of the Kernel Theorem 10.2. Let φ : G G be a group homomorphism. Then i. φ(a) = φ(b) if and only if a ker φ = b ker φ. ii. If φ(g) = g, then φ 1 (g ) = {x G : φ(x) = g } = g ker φ. Proof. i. Recall that the coset ah = bh if and only if a 1 b H. Here φ(a) = φ(b) φ(a) 1 φ(b) = e φ(a 1 b) = e a 1 b ker φ. Hence φ(a) = φ(b) if and only if a 1 b ker φ if and only if a ker φ = b ker φ. ii. Let x φ 1 (g ) = φ(x) = g = φ(g) = x ker φ = g ker φ = x g ker φ = φ 1 (g ) g ker φ. Again, take gk g ker φ where k ker φ. Now φ(gk) = φ(g)φ(k) = φ(g) = g = gk φ 1 (g ) = g ker φ φ 1 (g ). From the theorem above it is clear that if φ is a homomorphism from a group G onto G then all the left cosets of ker φ in G are exactly φ 1 (g ), g G.

CHAPTER 10. MORE ON GROUP HOMOMORPHISMS 32 Theorem 10.3. If ker φ = n then φ : G φ(g) is a n-to-1 mapping, that is, for every g φ(g), φ 1 (g ) = n. Proof. By Theorem 10.2 for every g φ(g), φ 1 (g ) is a left coset of ker φ in G. Therefore φ 1 (g ) = ker φ = n. Theorem 10.4. The group homomorphism φ : G G is one-to-one if and only if ker φ = {e}. Proof. This theorem is just a particular case of Theorem 10.3 with n = 1. However we can give an independent proof. Assume that φ is one-to-one. Let g ker φ. Then φ(g) = e = φ(e) = g = e = ker φ = {e}. Conversely, assume that ker φ = {e}. Let φ(a) = φ(b). Then φ(a)φ(b) 1 = e = φ(ab 1 ) = e = ab 1 ker φ = ab 1 = e = a = b. Thus φ is one-to-one. The First Isomorphism Theorem Theorem 10.5. Let φ : G G be a group homomorphism. Then the factor group G/ ker φ is isomorphic to φ(g). The isomorphism is given by g ker φ φ(g).

CHAPTER 10. MORE ON GROUP HOMOMORPHISMS 33 Homework 1. Let φ : G G be a group homomorphism and a G. a. Prove that φ(a n ) = (φ(a)) n. b. If a is finite, prove that φ(a) divides a. c. If H is cyclic, prove that φ(h) is cyclic. d. Prove that ker φ is a normal subgroup of G. 2. Let φ : Z 15 Z 5 be a homomorphism defined by φ(k) = k mod 5. a. Find ker φ. b. Find all the left cosets of ker φ. c. Find φ 1 (0), φ 1 (1), φ 1 (2), φ 1 (3) and φ 1 (4). d. Compare your answers in b. and c.

Chapter 11 Introduction to Rings Homework: # 5, 8, 12, 18. Definition 11.1. A ring R is a set with two binary operations addition and multiplication such that for all a, b, c R the following properties are satisfied: 1. a + b = b + a. 2. (a + b) + c = a + (b + c) 3. There is a 0 R such that a + 0 = a. 4. There is a R such that a + ( a) = 0. 5. a(bc) = (ab)c. 6. a(b + c) = ab + ac and (b + c)a = ba + ca. So (R, +) is an abelian group. A ring R is called commutative if ab = ba for all a, b R. A unity or identity is a nonzero element e such that ea = ae = a for all a R. A nonzero element a R is called unit if it has a multiplicative inverse. Let R be a commutative ring and a, b R, a 0. We say that a divides b and write a b if there is an element c R such that b = ac. Otherwise we say a does not divide b and write a b. Example 11.1. The set of integers Z under the usual addition and multiplication is a ring. It is commutative. 1 is the identity (unity). The units of Z are 1 and 1. Example 11.2. The set Z n is a ring under addition and multiplication modulo n. It is commutative, 1 is unity. The units are the elements in U(n). Example 11.3. The set 2Z of even integers under ordinary addition and multiplication is a ring. It is a commutative ring without unity. Example 11.4. The set M[ 2 (Z) of ] 2 2 matrices of with integer entries is a noncommutative ring with unity. 1 0 0 1 34

CHAPTER 11. INTRODUCTION TO RINGS 35 Properties of Rings Theorem 11.1. Let R be a ring and a, b, c R. Then 1. a0 = 0a = 0. 2. a( b) = ( a)b = (ab). 3. ( a)( b) = ab. 4. a(b c) = ab ac and (b c)a = ba ca. If R has a unity element 1, then Proof. 1. ( 1)a = a 2. ( 1)( 1) = 1. 1. a0 + 0 = a0 = a(0 + 0) = a0 + a0 = a0 = 0. 2. a( b) + ab = a( b + b) = a0 = 0 = a( b) = (ab). Similarly ( a)b = (ab). 3. ( a)( b) = (a( b)) is inverse of a( b). But a( b) = (ab) is inverse of ab. Thus ( a)( b) and ab are inverses of a( b). By the uniqueness ( a)( b) = ab. 4. a(b c) = a(b + ( c)) = ab + a( c) = ab ac. Theorem 11.2. If a ring R has unity, it is unique. If a ring element has a multiplicative inverse, it is unique. Subrings Definition 11.2. A subset S of a ring R is a subring of R if S is itself a ring with the operations of R. Theorem 11.3. A nonempty subset S of a ring R is a subring if a b S and ab S whenever a, b S. Proof. For every a, b S, a b S implies S is a subgroup of R under addition. S is commutative under addition because R is commutative under addition. Thus S is an abelian group under addition. For every a, b, c S, a(bc) = (ab)c and a(b+c) = ab+ac because a, bc R. Example 11.5. For each positive integer n, is a subring of a the ring Z. nz = {0, ±n, ±2n, ±3n, }

Chapter 12 Integral Domains Homework: # 4, 8, 18, 22, 41. Definition 12.1. A zero-divisor is a nonzero element a of a commutative ring R such that there is a nonzero element b R with ab = 0. Definition 12.2. An integral domain is a commutative ring R with unity and no zero-divisors, that is, if ab = 0 then either a = 0 or b = 0. Example 12.1. The ring of integers Z is an integral domain. Example 12.2. The ring of Guassian integers Z[i] = {a + bi : a, b Z} and the ring Z[x] of polynomials with integer coefficients are integral domains. Example 12.3. The ring Z p of integers modulo p is an integral domain if p is prime otherwise it is not an integral domain. For instance Z 10 is not an integral domain because 5 2 = 0. Example 12.4. The ring M 2 (Z) of 2 2 matrices over integers is not an integral domain. The cancellation property holds in integral domain. Theorem 12.1. Let R be an integral domain and a, b, c R. If a 0 and ab = ac, then b = c. Proof. ab = ac = ab ac = 0 = a(b c) = 0. Since R is an integral domain and a 0 we must have b c = 0 = b = c. Definition 12.3. A field is a commutative ring with unity in which every nonzero element is a unit. Theorem 12.2. Every field is an integral domain. 36

CHAPTER 12. INTEGRAL DOMAINS 37 Proof. Let F be a field. Let a, b F such that ab = 0. We want to show that either a = 0 or b = 0. If a 0 then there a is unit. There exist a 1 F such that aa 1 = a 1 a = 1. Thus a 1 (ab) = 0 = b = 0. Hence F is an integral domain. Theorem 12.3. A finite integral domain is a field. Proof. Let D be a finite integral domain. Let a D, a 0. We want to show that a is a unit. If a = 1 then it is unit. So assume that a 1. Since D is finite there must be repetition in a, a 2, a 3,. There exist integers i > j such that a i = a j = a i j = 1. Since a 1, i j > 1. Hence aa i j 1 = 1 = a i j 1 is the inverse of a and a is unit. Corollary 12.4. For p prime, Z p is a field. Proof. We just have to show that Z p, p prime, is an integral domain. Suppose that a, b Z p and ab = 0 = ab = pk for some integer k. Since p is prime either p a or p b = either a = 0 or b = 0. Characteristic of a Ring Definition 12.4. The characteristic of a ring R, denoted by char R, is the least positive integer n such that nx = 0 for all x R, that is, x+x+ +x(n times ) = 0. If no such integer exists we say that the characteristic of R is zero. The ring Z n of integers modulo n has characteristic n. Theorem 12.5. Let R be a ring with unity 1. If 1 has order n under addition, then the characteristic of R is n. If 1 has infinite order under addition, then char R = 0. Proof. If 1 has order n, n is the smallest positive integer such that For any x R, 1 + 1 + + 1(n times ) = 0. x + x + + x = 1x + 1x + + 1x = (1 + 1 + + 1)x = 0x = 0. Therefore char R = n. If the order of 1 is infinity, then there is no n such that n1 = 0. Hence char R = 0. Theorem 12.6. The characteristic of an integral domain is 0 or prime.

CHAPTER 12. INTEGRAL DOMAINS 38 Proof. Let 1 be the unity of the integral domain. If 1 has infinite order, then we are done. So suppose that the order of 1 is finite, say n. We want to show that n is prime. Let n = st where 1 s, t n. Then n 1 = 0 (st) 1 = 0 (s 1)(t 1) = 0 = s 1 = 0 or t 1 = 0. Since n is the smallest positive integer with this property either s = n or t = n. Thus n is prime.

Chapter 13 Ideals and Factor Rings Homework: # 8, 10, 12, 14 Definition 13.1. A subring I of a ring R is called (i) a left ideal of R if for every r R, ri = {ra : a I} I, that is, ra I for every a I. (ii) a right ideal of R if for every r R, Ir = {ar : a I} I, that is, ar I for every a I. If I is both a left ideal and a right ideal, we simply call it an ideal of R or two-sided ideal of R for obvious reason. If an ideal I is a proper subset of R then it is called proper ideal. If a ring R is commutative then all the ideals are two-sided ideals. In this course we will be talking about the two-sided ideals only. Examples 1. For any ring R, I = {0} is an ideal of R. It is called the trivial ideal. I = R is also an ideals of R. 2. 2Z = {2n : n Z} is an ideal of Z. Similarly, nz for any positive integer n is an ideal of Z. 3. Let R be a commutative ring with unity and a R. The set a = {ra : r R} is an ideal of R called the principal ideal generated by a. The assumption that R is commutative is necessary otherwise there is nothing that guarantees that r 1 as 1 + r 2 as 2 a where r 1, r 2, s 1, s 2 R. 39

CHAPTER 13. IDEALS AND FACTOR RINGS 40 4. Let R be a commutative ring with unity and let a 1, a 2,, a n R. Then I = a 1, a 2,, a n = {r 1 a 1 + + r n a n : r i R} is an ideal of R called the ideal generated by a 1, a 2,, a n. Verify that I is an ideal. Factor Rings We have seen that if H is a normal subgroup of a group G, we can form a factor group G/H of left cosets of H in G. In the similar fashion we can form the factor rings. Theorem 13.1. Let R be a ring and let I be a subring of R. {r + I : r R} is a ring under the operations The set of cosets if and only if I is an ideal of R. Addition: (r + I) + (s + I) = (r + s) + I Multiplication: (r + I)(s + I) = rs + I Example 13.1. Let R = Z and I = 6Z. Then Z/6Z = {0 + 6Z, 1 + 6Z, 2 + 6Z, 3 + 6Z, 4 + 6Z, 5 + 6Z} is a factor ring. Discuss the addition and multiplication on Z/6Z. {[ ] } a b Example 13.2. Let R = M 2 (Z) = : a, b, c, d Z and I be the subset of c d R consisting of matrices with even entries. It is easy to see that I is an ideal of R. What is in the factor ring R/I? {[ ] } s t R/I = : s, t, u, v {0, 1} u v. How many elements does R/I have? It has only 16 elements count carefully. Prime Ideals and Maximal Ideals Definition 13.2. Let R be a commutative ring and P be a proper ideal of R. Then P is called prime ideal if whenever ab P where a, b R, then a P or b P. A maximal ideal M of a commutative ring R is a proper ideal of R such that if there is an ideal I of R containing M, that is, M I then I = M or I = R. Example 13.3. Let p > 1 be an integer. Then pz is a prime ideal of Z, the ring of integers, if and only if p is prime.