Advanced Abstract Algebra - Maharshi Dayanand University, Rohtak

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48 ADVANCED ABSTRACT ALGEBRA Conversely, if , then xNyN = xyN = (Θ −1 −1 = ( xyy x ) yxN = eyxN = yxN = yNxN ) (d) Given , to show H∆ G i.e. To show ∀ g ∈G , ∀h ∈H −1 −1 − − = ghg h h = ( ghg h ) h ∈H 1 1 (Θ ) ∴ Example 13 Final order of 1. (15 27) (284) in 2. (153) (284697) in S 9 Solution Both are product of disjoint cycles. Hence order of each would be l.c.m. of the lengths if its cycles. (i) 12 in S 8 (ii) 6 in S 9 Example 14 Write (12345) as a product of transpositions. It can be written in more than one way. (12345)= (54) (53) (52) (51) = (15) (14) (13) (12) = (54) (52) (51) (14) (32) (41) Q. 26. Let α = ( a 1 a 2 a 3 ........ a s ) be a cycle and let π be a permutation in . Then π (π π )..........π . Example 15 Compute aba −1, Where (i) a = (135) (12), b = (1579) (ii) a = (579), b = (123) Solution (i) a = (135) (12) = (1235) b a (1579) a −1 = a( 1) a( 5) a( 7) a( 9) = (2179) g is the cycle G (a a( S xy gh H απ n8 (ii) = ba ( 1) a( 2) a( 3) g, Where (579) = (123)
UNIT-IV 81 Definition. Let S be a subring of a ring R. If x ∈ S, a ∈ R ax ∈ S , then S is called left ideal of R. If x ∈ S , a ∈ R xa ∈ S , then S is called right ideal of R. Ideals and Quotient Rings If x ∈ S, a ∈ R xa ∈ S and ax ∈ S then S is called two sided ideal or simply ideal of R. * If R is a commutative ring then all the three notions are same since in that case ax = xa ∈ S . ** Every ring has two trivial ideals : (i) R itself and is called unit ideal. (ii) Zero ideal [0] consisting of zero element only. Any other ideal except these two trivial ideals is called proper ideal. Theorem. The intersection of any two left ideals of a ring is again a left ideal of the ring. Proof. Let S 1 and S 2 be two ideals of R. S 1 and S 2 being subring of R, S 1 ∩ S 2 is also a subring of R. Again let x ∈ S 1 ∩ S 2 . Let a ∈ R. x ∈ S 1 , x∈ S 2 . Then since S 1 and S 2 are left ideals, a ∈ R, x ∈ S 1 ax ∈ S 1 a ∈ R , x ∈ S 2 ax ∈ S 2 ax ∈ S 1 ∩ S 2 S 1 ∩ S 2 is a left ideal. Theorem :- Let K(T) be the kernel of a ring homomorphism T : R → S . Then K(T) is a two sided ideal of R. Proof. Let a, b ∈ K (T). Then T(a) = T(b) = 0 . Therefore, T(a+b) = T(a) + T(b) = 0 + 0 = 0 (by ring T(ab) = T(a).T(b) = 0.0 = 0 homomorphism) which implies that a+b, ab ∈ K(T). Hence K(T) is a subring of R. Now let a ∈ K(T) and r ∈ R . It suffices to prove that ar, ra ∈ K(T) T(ar) = T(a).T(r) = 0 . T(r) (Θ a ∈ K(T) T(a) = 0) = 0 This implies that ar ∈ K(T). Similarly, T(ra) = T(r) T(a) = T(r).0 = 0 Hence K(T) is an ideal of R. Theorem. A field has no proper ideal. ra ∈ K(T) .
Page 1 and 2: 1 Advanced Abstract Algebra M.A./M.
Page 3 and 4: Contents 3 Unit I 5 Unit II 49 Unit
Page 5 and 6: UNIT-I 5 Unit-I Definition Group A
Page 7 and 8: UNIT-I 7 2. In example 1, a 1 1 0 1
Page 9 and 10: UNIT-I 9 Solutions: 1. CGbg≠ a φ
Page 11 and 12: UNIT-I 11 ∴ Ha ⊆ H. To show let
Page 13 and 14: UNIT-I 13 Corollary 4: (Feremat's L
Page 15 and 16: UNIT-I 15 Definition: Group Homomor
Page 17 and 18: UNIT-I 17 Proof: Consider the diagr
Page 19 and 20: UNIT-I 19 ( G ) N G ( H ) ≅ H N D
Page 21 and 22: UNIT-1 21 Theorem 8. Let H be a sub
Page 23 and 24: UNIT-1 23 From (i) and (ii), we hav
Page 25 and 26: UNIT-I 25 1 Note that αβ = 2 2 1
Page 27 and 28: UNIT-1 27 In this case the length o
Page 29 and 30: UNIT-1 29 | n Theorem 18. The set o
Page 31 and 32: UNIT-I b = φ ( φ φ ) ( ) g h t x
Page 33 and 34: UNIT-I 33 This is impossible. So no
Page 35 and 36: UNIT-I 35 integers x and y. Note th
Page 37 and 38: UNIT-I 37 O(7) = 4, as 7 1 = 7, 7 2
Page 39 and 40: UNIT-I 39 Put a x mt , = b = x Then
Page 41 and 42: UNIT-I n iii. N = ∈Gs, show that
Page 43 and 44: UNIT-I 43 abab------ab = a n b n
Page 45 and 46: UNIT-I Q.20. Q.21. Q.22. Q.23. Q.24
Page 47: UNIT-I 47 Remark : If G is abelian,
Page 51 and 52: UNIT-IV 83 (r+s) − (r 1 +s 1 ) =
Page 53 and 54: UNIT-IV 85 and = ψ(r+K) + ψ(s+K)
Page 55 and 56: UNIT-IV 87 M ⊂ M′ ⊂ R M′ =
Page 57 and 58: UNIT-IV 89 We define addition and m
Page 59 and 60: UNIT-IV 91 be the set of the ordere
Page 61 and 62: UNIT-IV 93 (iii) a + b − a b
Page 63 and 64: UNIT-IV 95 and Z/A ~ Im(f) = f (z)
Page 65 and 66: UNIT-IV 97 If a n = 0 for all n ≥
Page 67 and 68: UNIT-IV 99 x 3 = (0, 0, 0, 1,…) .
Page 69 and 70: UNIT-IV 101 Then either (i) or n i
Page 71 and 72: UNIT-IV 103 deg f(x) = 0 and deg g(
Page 73 and 74: UNIT-IV 105 If x, y ∈ A , then th
Page 75 and 76: UNIT-IV 107 The possibility φ(r) <
Page 77 and 78: UNIT-IV 109 Lemma 4. If f(x) is an
Page 79 and 80: UNIT-IV 111 Proof. Let, if possible
Page 81 and 82: UNIT-II 49 Unit-II Definition: Comp
Page 83 and 84: UNIT-II 51 If r = 1, then G is simp
Page 85 and 86: UNIT-II 53 group of the Commutator
Page 87 and 88: UNIT-II di + 1 idi + 1 i di + 1 idi
Page 89 and 90: UNIT-II 57 we write the refinement
Page 91 and 92: UNIT-II 59 of length e in which eac
Page 93 and 94: UNIT-II 61 of nilpotency of G) , so
Page 95 and 96: UNIT-III 63 n.v = R i.e. abelian gr
Page 97 and 98: UNIT-III 65 Theorem 1 Let M, N be R
Page 99 and 100:
UNIT-III 67 F ( BA ad 1'= = , 1a d(
Page 101 and 102:
UNIT-III 69 Vector space over F. Le
Page 103 and 104:
UNIT-III 71 1. J = only one linearl
Page 105 and 106:
UNIT-III Hence J = −L NM 4 0 0 0
Page 107 and 108:
UNIT-III 75 Example 5 The direct su
Page 109 and 110:
UNIT-IV 77 A. B ≠ B. A. If IR, th
Page 111 and 112:
UNIT-IV 79 Example: (D, +, .) is a
Page 113 and 114:
UNIT-IV 81 Examples 1. The polynomi
Page 115 and 116:
UNIT-V 129 (Characteristic of a rin
Page 117 and 118:
UNIT-V 131 As F = p dΘ F ≅ Z p i
Page 119 and 120:
UNIT-V 133 In this case is called t
Page 121 and 122:
+ c c+ 0 0 1 c c+ 1 1 1 0 1+ c c c
Page 123 and 124:
UNIT-V 137 element of defines a per
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Advanced Abstract Algebra - Maharshi Dayanand University, Rohtak

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