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Definition 16: Let H be a subgroup of a group G, and let a G. We define the left coset of H in G containing g,written gH, by gH ={g*h| h H}. Similarity we define the right coset of H in G containing g,written Hg, by Hg ={h*g| h H}.
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[E;+] Example:S 3 ={e, 1, 2, 3, 4, 5 } H 1 ={e, 1 }; H 2 ={e, 2 }; H 3 ={e, 3 }; H 4 ={e, 4, 5 } 。 H1H1
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Lemma 2 : Let H be a subgroup of the group G. Then |gH|=|H| and |Hg|=|H| for g G. Proof: :H Hg, (h)=h g
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6.4.3 Lagrange's Theorem Theorem 6.19: Let H be a subgroup of the group G. Then {gH|g G} and {Hg|g G} have the same cardinal number Proof : Let S= {Hg|g G} and T= {gH|g G} : S→T, (Ha)=a -1 H 。 (1) is an everywhere function. for Ha=Hb, a -1 H?=b -1 H [a] [b] iff [a] ∩[b]= (2) is one-to-one 。 For Ha,Hb,suppose that Ha Hb , and (Ha)= (Hb) (3)Onto
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Definition 17 : Let H is a subgroup of the group G. The number of all right cosets(left cofets) of H is called index of H in G. [E;+] is a subgroup of [Z;+]. E’s index? ? Theorem 6.20: Let G be a finite group and let H be a subgroup of G. Then |G| is a multiple of |H|. Example: Let G be a finite group and let the order of a in G be n. Then n| |G|.
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Example: Let G be a finite group and |G|=p. If p is prime, then G is a cyclic group.
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6.4.4 Normal subgroups Definition 18 : A subgroup H of a group is a normal subgroup if gH=Hg for g G. Example: Any subgroups of Abelian group are normal subgroups S 3 ={e, 1, 2, 3, 4, 5 } : H 1 ={e, 1 }; H 2 ={e, 2 }; H 3 ={e, 3 }; H 4 ={e, 4, 5 } are subgroups of S 3. H 4 is a normal subgroup
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(1) If H is a normal subgroup of G, then Hg=gH for g G (2)H is a subgroup of G. (3)Hg=gH, it does not imply hg=gh. (4) If Hg=gH, then there exists h' H such that hg=gh' for h H
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Theorem 6.21: Let H be a subgroup of G. H is a normal subgroup of G iff g -1 hg H for g G and h H. Example:Let G ={ (x; y)| x,y R with x 0}, and consider the binary operation ● introduced by (x, y) ● (z,w) = (xz, xw + y) for (x, y), (z, w) G. Let H ={(1, y)| y R}. Is H a normal subgroup of G? Why? 1. H is a subgroup of G 2. normal?
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Let H be a normal subgroup of G, and let G/H={Hg|g G} For Hg 1 and Hg 2 G/H, Let Hg 1 Hg 2 =H(g 1 *g 2 ) 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 Hg 1 =Hg 3 and Hg 2 =Hg 4 G/H, Hg 1 Hg 2 =H(g 1 *g 2 ), Hg 3 Hg 4 =H(g 3 *g 4 ), Hg 1 Hg 2 ?=Hg 3 Hg 4 ? H(g 1 *g 2 )=?H(g 3 *g 4 ) g 3 *g 4 ?H(g 1 *g 2 ), i.e. (g 3 g 4 ) (g 1 *g 2 ) -1 ?H.
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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=H G/H is identity element of G/H Inverse element: For Ha G/H, Ha -1 G/H is inverse element of Ha, where a -1 G is inverse element of a.
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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 Hg 1 Hg 2 = H(g 1 *g 2 ). If G is a finite group, then G/H is also a finite group, and |G/H|=|G|/|H|
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Next: quotient group The fundamental theorem of homomorphism for groups Exercise: P362 21, 22,23, 26,28,33,34
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