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Published byBasil James Modified over 8 years ago
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Group A set G is called a group if it satisfies the following axioms. G 1 G is closed under a binary operation. G 2 The operation is associative. G 3 There is an identity element in G. G 4 Every element of G has an inverse in G.
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Theorem Every cyclic group is abelian. Every subgroup H of a cyclic group G = g is cyclic. If H = {1}, then H = 1 is cyclic. Let m be the smallest positive integer such that g m ∈ H, then H = g m is cyclic.
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Fundamental Theorem of Finite Cyclic Groups. Let G = g be a cyclic group of order n. If H is any subgroup of G, then (1) H = g d = {1, g d, g 2d, …, g n-d } for some d|n. (2) |H| = n/d. (3) H is the unique subgroup of G of order n/d.
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Find all subgroups of C 12 and draw the lattice diagram. C 12 = g , | g | = 12. The divisors of 12 are 1, 2, 3, 4, 6, and 12. {1} = g 12 , g 6 , g 4 , g 3 , g 2 , and g = G. C 12 g3 g3 g6 g6 g2 g2 1 1 g4 g4
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Find all subgroups of C 12 and draw the lattice diagram. C 12 = g , | g | = 12. The divisors of 12 are 1, 2, 3, 4, 6, and 12. {1} = g 12 , g 6 , g 4 , g 3 , g 2 , and g = G. C 12 g3 g3 g6 g6 g2 g2 1 1 g4 g4
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homomorphism What mappings preserve the group multiplication? If G and G 1 are groups, a mapping :G G 1 is called a homomorphism if (ab)= (a)· (b) for a and b in G.
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Example The mapping :Z Z give by (a) = 3a is a homomorphism of additive groups because (a+b) = 3(a+b) = 3a+3b = (a)+ (b) for all a,b ∈ Z.
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Example The identity map 1 G : G G is a homomorphism for any group G because 1 G (ab) = ab = 1 G (a) · 1 G (b) for all a,b in G.
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homomorphism If : G H and : H K are homomorphisms, then : G K is also a homomorphism. Proof: For all a and b in G, (ab) = [ (ab)] = [ (a)· (b)] = [ (a)]· [ (b)] = (a)· (b).
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Theorem Let : G G 1 be a homomorphism. Then: (1) (1) = 1. ( preserves the identity element) (2) (g −1 ) = (g) −1 for all g ∈ G. ( preserves inverses) (3) (g k ) = (g) k for all g ∈ G. ( preserves powers)
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isomorphism Consider the groups G = {1,–1} and Z 4 * = {1,3}. The two Cayley tables are G 1 –1 – 1 1 1 – 1 Z4*Z4* 1 3 3 1 1 3
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isomorphism The mapping : G Z 4 * given by (1) = 1 and (–1) = 3 is a bijection, and we can obtain the entire Cayley table for Z 4 * from that of G by replacing a with (a) for every a in G.
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isomorphism If G and G 1 are groups, a mapping : G G 1 is called an isomorphism if is a bijection (one-to-one and onto) which is also a homomorphism. When an isomorphism exists from G to G 1 we say that G is isomorphic to G 1 and write G G 1.
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Example The set 2 Z = { 2k | k ∈ Z } of even integers is an additive group, in fact a subgroup of Z. Show that Z 2 Z. Proof: : Z 2 Z given by (k) = 2k O Onto is clearly. (k) = (m) implies k = m one to one (k +m) = 2(k +m) = 2k +2m = (k) + (m) homomorphism
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Show that R R +, where R is additive and R + is multiplicative. Proof: DDefine : R R + is by (r) = e r LLet (r) = (s). TThen e r = e s and r = ln(e r ) = ln(e s ) = s. TThus is one-to-one. IIf t R +, then lnt ∈ R and (lnt) = e lnt = t. HHence is onto. (r+s) = e r+s = e r e s = (r)· (s) for all r, s R homomorphism.
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Show that Q is not isomorphic to Q *. Proof: SSuppose that : Q Q * is an isomorphism. LLet q ∈ Q satisfy (q) = 2, and write (q/2) =a. TThen a 2 = (q/2) (q/2) = (q/2 + q/2) = (q) =2. BBut a ∈ Q such that a 2 = 2, so no such isomorphism can exist.
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Let G = a be a cyclic group. Show that: (1) If |G| = n, then G Z n. (2) If |G| = ,then G Z. Proof: If |G| = n, then |a| = n. We define : Z n G by. Then k ≡ m(mod n) a k = a m So is well defined and one-to-one. is clearly onto.
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Proof of Example is an homomorphism. HHence is an isomorphism. TThe proof of (2) is similar and we leave it as exercise.
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Theorem If G H,and G has a structural property, then H also has that structural property. For examples: 1) G has order n. 2) G is finite. 3) G is abelian. 4) G is cyclic. 5) G h as no element of order n. 6) G has exactly m elements of order n.
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Theorem Let G, G 1, and G 2 denote groups. (1) The identity map 1 G : G G is an isomorphism for every group G. (2) If : G G 1 is an isomorphism, the inverse mapping 1 : G 1 G is also an isomorphism. (3) If : G G 1 and : G 1 G 2 are isomorphisms, their composite : G G 2 is also an isomorphism.
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Corollary 1 The isomorphic relation is an equivalence for groups. That is: (1) G G for every group G. (2) If G G 1 then G 1 G. (3) If G G 1 and G 1 G 2 then G G 2.
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Theorem Let : G G 1 be an isomorphism. Then | (g)| = |g| for all g ∈ G. Proof: g k = 1 [ (g)] k = (g k ) = (1) = 1 [ (g)] k = 1 (g k ) = [ (g)] k = 1 k = 1 = (1) g k = 1 g k = 1 [ (g)] k = 1
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|X| = n S X S n Proof: Let h : X {1,2,…,n} be a bijection. Suppose ∈ S X. F: S X S n defined by F( ) = where (k) = h[ [h 1 (k)]] = h h 1 (k)
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Cayley’s Theorem Every group G of order n is isomorphic to a subgroup of S n. Proof: Let a ∈ G, a : G G by r a (g) = ag (1) r a ∈ S G (2) G 1 ={r a | a ∈ G} is a subgroup of S G f : G G 1 a r a is isomorphs G G 1 S G S n G is isomorphic to a subgroup of S n.
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