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Logic and Set Theory

by: Dr. Karelle Keeling

Logic and Set Theory MAT 301

Dr. Karelle Keeling
GPA 3.58


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This 6 page Class Notes was uploaded by Dr. Karelle Keeling on Sunday October 11, 2015. The Class Notes belongs to MAT 301 at Eastern Kentucky University taught by Staff in Fall. Since its upload, it has received 20 views. For similar materials see /class/221454/mat-301-eastern-kentucky-university in Mat Mathematics at Eastern Kentucky University.


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Date Created: 10/11/15
Chapter 7 Cosets and Lagrange s Theorem In this unit we consider one of the most important theorem in group theory The result is called LaGrange s Theorem and it relates the order of a subgroup to the order of a super group First some needed background material on cosets De nition 36 LetH be a subgroup ofG and let a b E G l a is right congruent to b modulo H denoted by a E b mod H i a b 391 E H 2 a is left congruent to b modulo H denoted by a E b mod H i 1391 b E H We note that ifGis abelian then a bquot1 E H gt a b39l 1 b a391 a391 b E Hand so right and le congruence modulo H agree Example 49 LetH 12 andG S3 123 132 12 13 23 ThenH s G Now 123 2 13 mod H e 123 131 123 13 12 e H yet 123 as 13 mod H 9 123 1 13 132 13 23 e H Hence there are nonabelian groups G and subgroups H so that right and le congruence fail to coincide Theorem 37 LetH s G 1 Right congruence modulo H is an equivalence relation on G KB 45 2 The equivalence class containing an element a 6 G under right congruence modulo H is the set Ha h a I h e H calledthe right coset oinnG 3 Le congruence modulo H is an equivalence relation on G 4 The equivalence class containing an element a E G under right congruence modulo H is the maHahheHme thmadeG 5 lHa H laHl Proof 1 Let a b c e G Since a aquot e E H itfollowsthata E a mod H and Erisre exive Suppose that a Er b mod H Then ab39leHgtab39139l ba39leH and so I Er a mod H Thus Er is symmetric Now suppose a Er b mod H and b Er 1 mod H Then we have that a b39l b c391 E H and so by closure a b39l b 0391 a c391 E H Thus a Er 6 mod H and Eristransitive Weconcludethat Er is an equivalence relation of G 2 The equivalence class containing an element a e G under right congruence modulo H is given W xeGxEramodhxe xa39leH xeGxa39lheH xeGxhaheH haheH Ha Parts 3 and 4 are done in a similar manner 5 The lnction from H a into H given by dh a h can be shown to be a bijection Thus IH al Inasimilar manner IHI In HI The desired result follows I The next result follows from properties of equivalence classes MAT 301 material and so we omit KB 46 the proof There is a parallel result for le cosets Theorem 38 l G is the union ofthe right cosets ofH in G 2 Two right cosets of H in G are either disjoint or equal 3 ForallabeGHaHbiffab39leH Example 50 1 Let H 12 G S3 123 132 12 13 23 and let a 123 Then we have that a H a a 12 123 23 and Ha a 12 1 123 13 We see here that a H 59 H a and not all le cosets are right cosets 2 Here we let H 123 123 132 and G s3 see above Then H 123 132 H and hence by Theorem 38 it follows that 123 H H 123 and 132 H H 132 Also 12 H 12 12 123 12 132 12 13 23 H 02 12 123 L2 132 12 12 23 13 12 H H 12 It follows by properties of equivalence classes that is Theorem 38 that 13 H H 13 and 23 H H 23 So this particular subgroup H every le coset is also a right coset KB 47 Proposition 39 Let H s G Then the number of distinct le cosets of H in G is the same as the number of distinct right cosets of H in G Sketch of proof The mapping Ma H H a 391 can be shown to be a bijection from the set ofle cosets ofH in G into the set ofright cosets ofH in G I De nition 40 Let H s G Then the number of distinct le or right cosets of H in G is called the index of H in G and is denoted by GH Example 51 1 If H 023 123 132 and G S3 see Example 50 then GHJ 2 2 Let H gt and G Z Then direct computation shows that the distinct le cosets of H inG are given by 01 H 0 3 6 9 11 H 1 4 7 10 219 H 215181111 Hence lelt3gtJ 3 We can state and prove the main result of this section Theorem 41 LaGrange s Theorem If G is a nite group and H s G then the order of H divides evenly into the order of G KB 48 In fact we shall prove that IGI H lG HJ Proof By Theorem 38 G is the union of the disjoint le cosets of H in G and there are precisely G H le cosets ofH in G That is 61 G U akH k1 Because the above is a disjoint union GH IGI U Ink HI kl Since H lakHlfork 1 2 3 GHJ GH GH G U lakHl U H HGH kl kl Thus the order of H divides evenly into the order of G I We note two quick corollaries to LaGrange s Theorem Corollary 42 The order of an element of a group of nite order is a divisor of the order of the group Corollary 43 Every group of prime order p is a cyclic group isomorphic to Zp We observe that in general the converse to LaGrange s Theorem is false KB 49 Example 52 I The alternating group A 4 consists of all the even permutations of S 4 and it has order 12 See Theorem 30 In Example 43 we established that A4 234 243 123 132 124 142 134 143 1324 1423 1234 We prove that A 4 has no subgroup of order 6 On the contrary suppose H is a subgroup of A 4 of order 6 Now let a be any ofthe eight elements of order 3 in A4 Since A4 IHI A4HJ it follows that A4 H J 2 Thus at most two of the following three cosets are distinct H a H a2 H Now 1 ifH a H then a E H 2 ifH a2 H then multiplication ofthe le by a shows that H a H and again a E H 3 ifa H a2 H then multiplication ofthe le bya shows that H a2 H which implies that H a H and yet again a e H Since a is any one ofthe eight elements of order three we have that IH I 2 8 A contradiction We close with a partial converse to LaGrange s Theorem Proposition 44 An abelian group G of nite order has a subgroup of order n for each divisor n of KB 50


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