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Theorem 6.21: Let H be a subgroup of G. H is a normal subgroup of G iff g-1hgH for gG and hH. Proof: (1) H is a normal subgroup of G (2) g-1hgH for gG and hH For gG , Hg?=gH Let H be a normal subgroup of G, and let G/H={Hg|gG} For Hg1 and Hg2G/H, Let Hg1Hg2=H(g1*g2) Lemma 3: Let H be a normal subgroup of G. Then [G/H; ] is a algebraic system. Proof: is a binary operation on G/H. For Hg1=Hg3 and Hg2=Hg4G/H, Hg1Hg2=H(g1*g2), Hg3Hg4=H(g3*g4), Hg1Hg2?=Hg3Hg4? H(g1*g2)=?H(g3*g4) g3*g4?H(g1*g2), i.e. (g3g4)(g1*g2)-1?H. Theorem 6.22: Let [H;] be a normal subgroup of the group [G;]. Then [G/H;] is a group. Proof: associative Identity element: Let e be identity element of G. He=HG/H is identity element of G/H Inverse element: For HaG/H, Ha-1G/H is inverse element of Ha, where a-1G is inverse element of a. Definition 19: Let [H;*] be a normal subgroup of the group [G;*]. [G/H;] is called quotient group, where the operation is defined on G/H by Hg1Hg2= H(g1*g2). If G is a finite group, then G/H is also a finite group, and |G/H|=|G|/|H| 6.5 The fundamental theorem of homomorphism for groups 6.5.1.Homomorphism kernel and homomorphism image Lemma 4: Let [G;*] and [G';] be groups, and be a homomorphism function from G to G'. Then (e) is identity element of [G';]. Proof: Let x(G)G'. Then aG such that x=(a). Definition 20: Let be a homomorphism function from group G with identity element e to group G' with identity element e’. {xG| (x)= e'} is called the kernel of homomorphism function . We denoted by Ker( K(),or K). Example: [R-{0};*] and [{-1,1};*] are groups. 1 ( x) 1 x0 x0 Ker {x | x 0, x R} Theorem 6.23:Let be a homomorphism function from group G to group G'. Then following results hold. (1)[Ker;*] is a normal subgroup of [G;*]. (2) is one-to-one iff K={eG} (3)[(G); ] is a subgroup of [G';]. proof:(1)i) Ker is a subgroup of G For a,bKer, a*b?Ker, i.e.(a*b)=?eG‘ Inverse element: For aKer, a-1?Ker ii)For gG,aKer, g-1*a*g?Ker 6.5.2 The fundamental theorem of homomorphism for groups Theorem 6.24 Let H be a normal subgroup of group G, and let [G/H;] be quotient group. Then f: GG/H defined by f(g)=Hg is an onto homomorphism, called the natural homomorphism. Proof: homomorphism Onto Theorem 6.25 : Let be a homomorphism function from group [G;*] to group [G';]. Then [G/Ker();][(G);] isomorphism function f:G/ Ker()(G). Let K= Ker(). For KaG/K,f(Ka)=(a) f is an isomorphism function。 Proof: For KaG/K,let f(Ka)=(a) (1)f is a function from G/K to (G) For Ka=Kb,(a)=?(b) (2)f is a homomorphism function For Ka,KbG/K, f(KaKb)=?f(Ka)f(Kb) (3) f is a bijection One-to-one Onto Corollary 6.2: If is a homomorphism function from group [G;*] to group [G';], and it is onto, then [G/K;][G';] Example: Let W={ei|R}. Then [R/Z;][W;*]. Let (x)=e2ix is a homomorphism function from [R;+] to [W;*], is onto Ker={x|(x)=1}=Z 6.6 Rings and fields 6.6.1 Rings Definition 21: A ring is an Abelian group [R, +] with an additional associative binary operation(denoted · such that for all a, b, cR, (1) a · (b + c) = a · b + a · c, (2) (b + c) · a = b · a + c · a. We write 0R for the identity element of the group [R, +]. For a R, we write -a for the additive inverse of a. Remark: Observe that the addition operation is always commutative while the multiplication need not be. Observe that there need not be inverses for multiplication. Example: The sets Z,Q, with the usual operations of multiplication and addition form rings, [Z;+,],[Q;+,] are rings Let M={(aij)nn|aij is real number}, Then [M;+,]is a ring Example: S,[P(S);,∩], Commutative ring Definition 23: A ring R is a commutative ring if ab = ba for all a, bR . A ring R is an unitary ring if there is 1R such that 1a = a1 = a for all aR. Such an element is called a multiplicative identity. Example: If R is a ring, then R[x] denotes the set of polynomials with coefficients in R. We shall not give a formal definition of this set, but it can be thought of as: R[x] = {a0 + a1x + a2x2 + …+ anxn|nZ+, aiR}. Multiplication and addition are defined in the usual manner; if n m f ( x) ai x and g ( x) bi x i i i 0 f ( x) g ( x) then i 0 max{ n , m} (a b ) x i 0 i i i nm f ( x) g ( x) k ( a b ) x i j k 0 i j k One then has to check that these operations define a ring. The ring is called polynomial ring. Theorem 6.26: Let R be a commutative ring. Then for all a,bR, n (a b) n C (n, i )a i b n i i 0 where nZ+. Quiz:1.Let G be a cyclic group generated by the element g, where |G|=18. Then g6 is not a generator of G 2.Let Q be the set of all rational numbers. Define & on Q by a&b=a+b+ab (1)Prove[Q;&]is a monoid. (2)Is [Q;&] a group?Why? Exercise:P367 6 1.Prove Theorem 6.23(2)(3) 2.Let W={ei|R}. Then [C*/W;][R+;*]. 3.Let X be any non-empty set. Show that [P(X); ∪, ∩] is not a ring.