### Create a StudySoup account

#### Be part of our community, it's free to join!

Already have a StudySoup account? Login here

# Logic and Set Theory MAT 301

EKU

GPA 3.58

### View Full Document

## 20

## 0

## Popular in Course

## Popular in Mat Mathematics

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.

## Similar to MAT 301 at EKU

## Popular in Mat Mathematics

## Reviews for Logic and Set Theory

### What is Karma?

#### Karma is the currency of StudySoup.

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

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

### BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.

### You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

#### "There's no way I would have passed my Organic Chemistry class this semester without the notes and study guides I got from StudySoup."

#### "When you're taking detailed notes and trying to help everyone else out in the class, it really helps you learn and understand the material...plus I made $280 on my first study guide!"

#### "I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"

#### "It's a great way for students to improve their educational experience and it seemed like a product that everybody wants, so all the people participating are winning."

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.