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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.

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Presentation on theme: "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."— Presentation transcript:

1 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.

2 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.

3 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.

4 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

5 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

6 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.

7 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.

8 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.

9 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).

10 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)

11 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

12 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.

13 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.

14 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

15 Show that R  R +, where R is additive and R + is multiplicative. Proof: DDefine  : R  R + is by  (r) = e r LLet  (r) =  (s). TThen e r = e s and r = ln(e r ) = ln(e s ) = s. TThus  is one-to-one. IIf t  R +, then lnt ∈ R and  (lnt) = e lnt = t. HHence  is onto.    (r+s) = e r+s = e r e s =  (r)·  (s) for all r, s  R  homomorphism.

16 Show that Q is not isomorphic to Q *. Proof: SSuppose that  : Q  Q * is an isomorphism. LLet q ∈ Q satisfy  (q) = 2, and write  (q/2) =a. TThen a 2 =  (q/2)   (q/2) =  (q/2 + q/2) =  (q) =2. BBut  a ∈ Q such that a 2 = 2, so no such isomorphism  can exist.

17 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.

18 Proof of Example   is an homomorphism. HHence  is an isomorphism. TThe proof of (2) is similar and we leave it as exercise.

19 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.

20 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.

21 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.

22 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

23 |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)

24 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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