Advanced Abstract Algebra - Maharshi Dayanand University, Rohtak

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18 ADVANCED ABSTRACT ALGEBRA f is 1–1 since f ( kg ) f ( ) = 2 1 kg ( g ) f ( ), −1 f 1 = g 2 hence f ( g1 ) f ( g2 ) = e −1 and f ( g g ) = e ( Θ f is a homo). 1 2 − 1 So g g ∈ K ker f , which shows that kg 1 = kg 2 . Thus 1 2 = G k homorphism. So ≅ f ( G) ⊆G. Consider Again Example 4: ( R) GL( n,R) f : GLn = R* A f ( AB) = det ( AB) = det( A) det( B) is 1–1. By definition f is onto. Hence f is So is a homomorphism. the identity of R*. ⇔ A∈ SLn = ( R) SL( n, R), the subgroup of By above fundmental homomorphism theorem, we get ( n,R) GL ker f ≅ Im( f ) of all n×n matrices with determinant 1. Ga = ⇔ HA f i.e GL SL ( n,R) ( n,R) ≅ Im( f ) But f is onto, since for Hence a A = 0 1 1 1. . .1 ( , R) ( n, R) GL n SL 0 n× n ≅ R * ∈GL ( n,R) Theorem 5 (First Isomorphism Theorem) such that f ( A) = det A= a Let G be a group with normal subgroup N and H such that N ⊆ H Then and
UNIT-I 19 ( G ) N G ( H ) ≅ H N Define f : G N by Na Ha is well-defined, since Na = Nbfor we get . Since N ⊆ H, thus gives and so Ha = Hb. f is a homomorphism: the identity of ⇔ Ha = H ⇔ a ∈ H Hence ker . As ker f ∆ G , N so H ∆G N N The fundamental homomorphism theorem for groups implies that G N ker ≅ Im f = G f N ( ) ) ( ) ( ) ( ) ( H ) f ( Nb) ∈ ∀f G ⇔ker −1 ,b G, = H HN Na f Ha ∩,∀ , ∈Na bf Nker a = Nb ∆ f = , ⇔f fNab ( ) = HabH , = HaHb = f Na a b H ≅ ∈Nab HN ∩= NNaNb = f a f b ∀a, b∈ N ( H ∩ N ) G ⇔ a ∈ N and a ∈ H∴ N ≅ G H H ⇔ a ∈ H ∩ N N Theorem 6 (Second Isomorphism Theorem) Let G be a group, and let N ∆G, let H be any subgroup of g. Then HN is a subgroup of G, and Proof: Define f:H by a is a homomorphism since a ∈ ker f ⇔ f ( a) = N, the identity element of HN and a H So
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: UNIT-I 17 Proof: Consider the diagr
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 and 48: UNIT-I 47 Remark : If G is abelian,
Page 49 and 50: UNIT-IV 81 Definition. Let S be a s
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
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UNIT-IV 103 deg f(x) = 0 and deg g(
Page 73 and 74:
UNIT-IV 105 If x, y ∈ A , then th
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UNIT-IV 107 The possibility φ(r) <
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UNIT-IV 109 Lemma 4. If f(x) is an
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UNIT-IV 111 Proof. Let, if possible
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UNIT-II 49 Unit-II Definition: Comp
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UNIT-II 51 If r = 1, then G is simp
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UNIT-II 53 group of the Commutator
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UNIT-II di + 1 idi + 1 i di + 1 idi
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UNIT-II 57 we write the refinement
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UNIT-II 59 of length e in which eac
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UNIT-II 61 of nilpotency of G) , so
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UNIT-III 63 n.v = R i.e. abelian gr
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UNIT-III 65 Theorem 1 Let M, N be R
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UNIT-III 67 F ( BA ad 1'= = , 1a d(
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UNIT-III 69 Vector space over F. Le
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UNIT-III 71 1. J = only one linearl
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UNIT-III Hence J = −L NM 4 0 0 0
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UNIT-III 75 Example 5 The direct su
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UNIT-IV 77 A. B ≠ B. A. If IR, th
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UNIT-IV 79 Example: (D, +, .) is a
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UNIT-IV 81 Examples 1. The polynomi
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UNIT-V 129 (Characteristic of a rin
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UNIT-V 131 As F = p dΘ F ≅ Z p i
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UNIT-V 133 In this case is called t
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+ c c+ 0 0 1 c c+ 1 1 1 0 1+ c c c
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UNIT-V 137 element of defines a per
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Advanced Abstract Algebra - Maharshi Dayanand University, Rohtak

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