Solutions to Assignment 10


 Susan Conley
 1 years ago
 Views:
Transcription
1 Solutions to Assignment 10 Math 412, Winter Prove that each element of a finite group G appears exactly once in each row and exactly once in each column of the operation table. It is not difficult to show that each element appears at least once in each row and column. In fact, this is equivalent to the equations ax = b and xa = b always having a solution, which is easy to demonstrate. The required solutions are x = gb and x = bg where g G is such that ga = ag = e. So suppose an element a G appears twice in some column of the operation table. Then for some element b G, the equation xb = a has at least two solutions, which we will call c and d. Note that cb = a = db. Because there is an element f G such that bf = e, we get that c = ce = cbf = dbf = de = d, a contradiction. The argument is roughly the same if an element a G appears twice in some row of the operation table. It must then be the case that for some element b G, the equation bx = a has at least two solutions, which we will call c and d. Note that bc = a = bd. Then because there is an element f G such that fb = e, we get that c = ec = fbc = fbd = ed = d, a contradiction A partial operation table for a group G = {e, a, b, c, d, f} is shown below. Complete the table. a a b e d d d f f This is going to be a little painful! We proceed with the following steps: (a) We know by the one per row and column rule that ad = f and af = c. 1
2 d d f f (b) We know that ca = f. So da = (ac)a = aca = af = c. d d c f f (c) We know that cf = a. Thus df = (ac)f = acf = a 2 = b. d d c b f f (d) We know that da = c so db = d(aa) = daa = ca = f. f f (e) We know that df = b, so ff = (ad)f = adf = ab = e. f f e (f) By the one in each row and column trick we know that bf = d, and hence that ba = e and fa = d. 2
3 e d f f d e (g) We know that db = f, so fb = (ad)b = adb = af = c. e d f f d c e (h) We know that fb = c, so cb = (af)b = afb = ac = d. e d c c f d a f f d c e (i) Again by the one per row and column trick, we see that bb = a, and then bc = f and bd = c. e a f c d c c f d a f f d c e (j) We know that cf = a, so cc = c(fb) = cfb = ab = e. e a f c d c c f d e a f f d c e (k) Again by the one per row and column trick, we see that cd = b, dc = a, dd = e, fc = b, and fd = a. 3
4 e a f c d c c f d e b a d d c f a e b f f d c b a e Let G = {0, 1, 2, 3, 4, 5, 6, 7} and assume G is a group under an operation with these properties: (a) a b a + b for all a, b G; (b) a a = 0 for all a G. Write out the operation table for G. Again, fun to do, painful to type up! (a) We get for free that all of the entries on the main diagonal are (b) Now note that because , it must be that 0 0 = 0. Then due to the one per row and column rule, and the fact that , it must be that 0 1 = 1. Continuing in this manner allows us to fill in the first row and column of our table (and note that this makes zero the identity) (c) Now = 3, and we have already used 0, 1, and 2 in the second row and third column, so 2 1 = 3. Similarly, we may say that 1 2 = 3. At this point I ll note be so diligent about saying each time that we have already used such and so numbers in the column and row we are interested in. 4
5 (d) Now = 4, so 3 1 is either 2 or 4. If 3 1 = 4, then 3 = 3 (1 1) = = 4 1. This cannot be the case because we have already used 3 in the second column. It must be that 3 1 = 2. A similar argument demonstrates that 1 3 = (e) Note that = 5, and we have already used 4 in the fifth row, so 4 1 = 5. A similar argument demonstrates that 1 4 = (f) Now = 6, so that 5 1 is either 4 or 6. If 5 1 = 6, however, we get that 5 = 5 (1 1) = = 6 1, which cannot happen (we have already used 5 in the second column), so we conclude that 5 1 = 4. A similar argument show that 1 5 =
6 (g) It is now clear that 6 1 = 7, 7 1 = 6, 1 6 = 7, and 1 7 = (h) Note that 3 2 5, so 3 2 can equal 5, 4, or 1. But if 3 2 = 5, then 3 = 3 (2 2) = = 5 2, a contradiction as we have already used 3 in the third column. If 3 2 = 4, we get the same contradiction. Thus 3 2 = 1. A similar argument shows that 2 3 = (i) Note that 4 2 6, and that we have used 0, 1, 2, 3, 4, and 5 already in the corresponding column and row. So 4 2 = 6. A similar argument works for (j) Now 5 2 7, so 5 2 = 7 (by the one per column and row trick yet again!). The same can be said for
7 (k) Now 6 2 can equal 4 or 5. If 6 2 = 4, then 6 = 6 (2 2) = = 4 2 = 5, a contradiction. So 6 2 = 5, and we may then conclude that 7 2 = 4. The same goes for 2 6 and (l) We can now see that 4 3 = 7 = 3 4, 5 3 = 6 = 3 5, 6 3 = 4 = 3 6, and 7 3 = 5 = 3 7 (all using the one per column/row trick) (m) Note that 5 6 is either 1, 2, or 3. If 5 6 = 3, then 5 = 5 (6 6) = = 3 6 = 4, a contradiction. Similarly, if 5 6 = 1, then 5 = 5 (6 6) = = 1 6 = 7, a contradiction. Thus 5 5 = 2. We may conclude the same for (n) Note that 5 7 is either 1 or 3. If 5 7 = 3, then 5 = 5 (7 7) = = 7
8 3 6 = 4, a contradiction, so 5 7 = 1. The same may be concluded for 7 5. We can then see that 4 5 = 3, 4 6 = 1, 4 7 = 2, 5 4 = 3, 6 4 = 1, 6 7 = 3, 7 4 = 2, and 7 6 = (a) If a G and a 12 = e, find the orders of each of the elements a, a 2,..., a 11. According to theorem 7.8, the order of a i where i is a divisor of 12 is 12/i. Thus a 2 = 6, a 3 = 4, a 4 = 3, and a 6 = 2. We also know that a k = e if and only if 12 k. In this case we want to find the smallest t such that (a k ) t = a kt = e where k = 5, 7, 8, 9, 10, 11, so we need t such that 12 tk. If k is relatively prime to 12, then this is easy enough. In that case it must be that t = 12. Thus the orders of a 5, a 7, and a 11 are 60, 84, and 131. On the other hand, if k and 12 are not relatively prime, we can use a smaller t such that 12 will divide tk. For a 8, for instance, we note that if we take t = 3, then (a 8 ) 3 = (a 12 ) 2 = e. Similarly, for a 9, taking t = 4 will suffice (because this is the smallest integer such that 12 9t). Finally, for a 10, taking t = 6 works. So the orders are: a i order (b) Based on the evidence in part (a), make a conjecture about the order of a k when a = n. 8
9 It seems that we are always looking for a t such that 12 tk (note that this holds even when k 12). So the answer seems to be that the order of a k is 12/(k, 12) Prove that every nonabelian group G has order at least 6; hence, every group of order 2,3,4, or 5, is abelian. Suppose that G is a nonabelian group of order at least 2. We will show that G contains at least six elements. If G is of order exactly two, then because one of the elements must be the identity, all products are forced to commute (if we call the two elements of G e and a, then the only possible multiplications are e e, a a, e a, and a e, and these all commute). Thus we may suppose that G contains at least three elements, which we call e, a, and b. We can choose a and b in G such that ab ba because G is nonabelian. Of course, ab, ba G, and by cancellation, ab a ba and ab b ba. Also, because ab ba, we have that ab e ba. We have now identified five distinct elements in G: e, a, b, ab, and ba. So it is enough to show that {e, a, b, ab, ba} is not closed (because then we will have demonstrated that G contains at least six elements). First we show that if one of a 2 or b 2 is equal to e, then we are done. Suppose that a 2 = e. Then we know that aba ba, and aba ab (by cancellation), aba a (because ab e), aba e (because ba a), and aba b (because ab ba), and thus aba {e, a, b, ab, ba} so that G has at least six elements. This argument is symmetric in a and b, so we get the same result if b 2 = e. If neither a 2 = e or b 2 = e, then we will demonstrate that either a 2 {e, a, b, ab, ba} or a 1 {e, a, b, ab, ba}, again completing the proof. It is easy to see, again using cancellation, that a 2 a, a 2 ab, and a 2 ba. Suppose therefore that a 2 = b. We note that unless a 1 {e, a, b, ab, ba}, it must be the case that a(ab) = e or a(ba) = e. The first case, however, implies that b 2 = e, a contradiction. In the second case, we may assume that ba = a 1 and thus that baa = b 2 = e, again a contradiction. Thus if G is abelian, we have shown that it must have at least six elements Let G be a nonempty set equipped with an associative operation with these properties: (a) There is an element e G such that ea = a for every a G. This is a little tricky to show, until you see the trick, of course! We will let d be such that da = e and q be such that qd = e (these things exist by part (b) of the question). Now simply note that because ea = a for all a G, we have that ae = (ea)e = eae = (qd)ae = qdae = q(da)e = qee = q(ee) = qe = qda = ea as required. (b) For each a G, there exists d G such that da = e. 9
10 Now that we know part (a), this is not difficult. Using the notation from above we have that a = ea = qda = qe = eq = q so that ad = qd = e = da as required If a is the only element of order 2 in a group G, prove that a Z(G). We need to show that for all g G, ag = ga. We do know that for all g G, (gag 1 ) 2 = gag 1 gag 1 = gaeag 1 = ga 2 g 1 = geg 1 = gg 1 = e. But a is the only order two element of G. So if gag 1 e (the only order 1 element), it must be that gag 1 = a, and then we will be done because multiplying both sides on the right by g we then obtain gag 1 g = ag, that is, ga = ag. Suppose then that gag 1 = e. This implies that ga = g, or that a = g 1 ga = g 1 (ga) = g 1 g = e. But this is a contradiction because a has order two and hence a e Let G be an abelian group and n a fixed positive integer. (a) Prove that H = {a G a n = e} is a subgroup of G. As we proved in class, it is enough to show that H is nonempty, and for all a, b H, ab 1 H. The first condition is quite simple, because e n = e, so e H. The second is similarly straight forward. If a, b H, that is, if a n = e and b n = e, then (ab 1 ) n = a n (b n ) 1 = ee = e, so that ab 1 H. (b) Show by example that part (a) may be false if G is nonabelian. Consider the nonabelian group S 3, and let n = 2. Then it is not difficult to see that the elements ρ S 3 such that ρ 2 = e are H = {( ), ( ), ( ), ( )}. But this set is not closed under composition, for example ( ) ( ) ( ) = H, and hence H is not a subgroup Let G = a be a cyclic group of order n. (a) Prove that the cyclic subgroup generated by a m is the same as the cyclic subgroup generated by a d where d = (m, n). It is enough to show that a m a d and a d a m. In fact, because a m and a d are the respective generators of a m and a d, it is enough to show that a m a d and a d a m. Note that because d = (m, n) is is clear that there is a t Z such that dt = m. Thus a m = (a d ) t a d. On the other hand, we know that there exist u, v Z such that un + vm = d. Thus a d = a un+vm = a un a vm = (a n ) u a vm = ea vm = a vm a m. (b) Prove that a m is a generator of G if and only if (m, n) = 1. Suppose that (m, n) = 1, then again there are u, v Z such that un + vm = 1. Thus for any a i a we have that i = iun + ivm so that 10
11 a i = a iun+ivm = a iun a ivm = (a n ) iu a ivm = e(a m ) iv = (a m ) iv, so that a m is a generator of a. On the other hand, if a m is a generator of a, then a a m, that is, there is a v such that (a m ) v = a. By theorem 7.8, this implies that mv 1 (mod n), and thus that there is an u Z such that mv 1 = un. Thus mv + ( u)n = 1 and we know that this implies that (m, n) = 1. 11
Elements of Abstract Group Theory
Chapter 2 Elements of Abstract Group Theory Mathematics is a game played according to certain simple rules with meaningless marks on paper. David Hilbert The importance of symmetry in physics, and for
More informationA Hajós type result on factoring finite abelian groups by subsets II
Comment.Math.Univ.Carolin. 51,1(2010) 1 8 1 A Hajós type result on factoring finite abelian groups by subsets II Keresztély Corrádi, Sándor Szabó Abstract. It is proved that if a finite abelian group is
More information4. FIRST STEPS IN THE THEORY 4.1. A
4. FIRST STEPS IN THE THEORY 4.1. A Catalogue of All Groups: The Impossible Dream The fundamental problem of group theory is to systematically explore the landscape and to chart what lies out there. We
More informationGroups in Cryptography
Groups in Cryptography Çetin Kaya Koç http://cs.ucsb.edu/~koc/cs178 koc@cs.ucsb.edu Koç (http://cs.ucsb.edu/~koc) ucsb cs 178 intro to crypto winter 2013 1 / 13 Groups in Cryptography A set S and a binary
More informationAdvanced Encryption Standard by Example. 1.0 Preface. 2.0 Terminology. Written By: Adam Berent V.1.5
Written By: Adam Berent Advanced Encryption Standard by Example V.1.5 1.0 Preface The following document provides a detailed and easy to understand explanation of the implementation of the AES (RIJNDAEL)
More informationSERVER CERTIFICATES OF THE VETUMA SERVICE
Page 1 Version: 3.4, 19.12.2014 SERVER CERTIFICATES OF THE VETUMA SERVICE 1 (18) Page 2 Version: 3.4, 19.12.2014 Table of Contents 1. Introduction... 3 2. Test Environment... 3 2.1 Vetuma test environment...
More informationAdvanced Encryption Standard by Example. 1.0 Preface. 2.0 Terminology. Written By: Adam Berent V.1.7
Written By: Adam Berent Advanced Encryption Standard by Example V.1.7 1.0 Preface The following document provides a detailed and easy to understand explanation of the implementation of the AES (RIJNDAEL)
More informationSERVER CERTIFICATES OF THE VETUMA SERVICE
Page 1 Version: 3.5, 4.11.2015 SERVER CERTIFICATES OF THE VETUMA SERVICE 1 (18) Page 2 Version: 3.5, 4.11.2015 Table of Contents 1. Introduction... 3 2. Test Environment... 3 2.1 Vetuma test environment...
More informationMath 312 Homework 1 Solutions
Math 31 Homework 1 Solutions Last modified: July 15, 01 This homework is due on Thursday, July 1th, 01 at 1:10pm Please turn it in during class, or in my mailbox in the main math office (next to 4W1) Please
More informationAssignment 8: Selected Solutions
Section 4.1 Assignment 8: Selected Solutions 1. and 2. Express each permutation as a product of disjoint cycles, and identify their parity. (1) (1,9,2,3)(1,9,6,5)(1,4,8,7)=(1,4,8,7,2,3)(5,9,6), odd; (2)
More information6 Commutators and the derived series. [x,y] = xyx 1 y 1.
6 Commutators and the derived series Definition. Let G be a group, and let x,y G. The commutator of x and y is [x,y] = xyx 1 y 1. Note that [x,y] = e if and only if xy = yx (since x 1 y 1 = (yx) 1 ). Proposition
More information= 2 + 1 2 2 = 3 4, Now assume that P (k) is true for some fixed k 2. This means that
Instructions. Answer each of the questions on your own paper, and be sure to show your work so that partial credit can be adequately assessed. Credit will not be given for answers (even correct ones) without
More informationSolutions to TOPICS IN ALGEBRA I.N. HERSTEIN. Part II: Group Theory
Solutions to TOPICS IN ALGEBRA I.N. HERSTEIN Part II: Group Theory No rights reserved. Any part of this work can be reproduced or transmitted in any form or by any means. Version: 1.1 Release: Jan 2013
More informationFACTORING CERTAIN INFINITE ABELIAN GROUPS BY DISTORTED CYCLIC SUBSETS
International Electronic Journal of Algebra Volume 6 (2009) 95106 FACTORING CERTAIN INFINITE ABELIAN GROUPS BY DISTORTED CYCLIC SUBSETS Sándor Szabó Received: 11 November 2008; Revised: 13 March 2009
More informationGROUPS ACTING ON A SET
GROUPS ACTING ON A SET MATH 435 SPRING 2012 NOTES FROM FEBRUARY 27TH, 2012 1. Left group actions Definition 1.1. Suppose that G is a group and S is a set. A left (group) action of G on S is a rule for
More informationKALE: A HighDegree AlgebraicResistant Variant of The Advanced Encryption Standard
KALE: A HighDegree AlgebraicResistant Variant of The Advanced Encryption Standard Dr. Gavekort c/o Vakiopaine Bar Kauppakatu 6, 41 Jyväskylä FINLAND mjos@iki.fi Abstract. We have discovered that the
More informationCommunity College of Philadelphia Calling Code 218 Employer Scan Client Approved: November 17, 2005 Region (CIRCLE) City MSA
Community College of Philadelphia Calling Code 218 Employer Scan Client Approved: November 17, 2005 Region (CIRCLE) City MSA Zip V0 V1 V2 Month/ Day/ Year of Contact: Business Name: Address: V3 City: V4
More informationBaltic Way 1995. Västerås (Sweden), November 12, 1995. Problems and solutions
Baltic Way 995 Västerås (Sweden), November, 995 Problems and solutions. Find all triples (x, y, z) of positive integers satisfying the system of equations { x = (y + z) x 6 = y 6 + z 6 + 3(y + z ). Solution.
More information(0, 0) : order 1; (0, 1) : order 4; (0, 2) : order 2; (0, 3) : order 4; (1, 0) : order 2; (1, 1) : order 4; (1, 2) : order 2; (1, 3) : order 4.
11.01 List the elements of Z 2 Z 4. Find the order of each of the elements is this group cyclic? Solution: The elements of Z 2 Z 4 are: (0, 0) : order 1; (0, 1) : order 4; (0, 2) : order 2; (0, 3) : order
More information1 = (a 0 + b 0 α) 2 + + (a m 1 + b m 1 α) 2. for certain elements a 0,..., a m 1, b 0,..., b m 1 of F. Multiplying out, we obtain
Notes on realclosed fields These notes develop the algebraic background needed to understand the model theory of realclosed fields. To understand these notes, a standard graduate course in algebra is
More informationFuture Trends in Airline Pricing, Yield. March 13, 2013
Future Trends in Airline Pricing, Yield Management, &AncillaryFees March 13, 2013 THE OPPORTUNITY IS NOW FOR CORPORATE TRAVEL MANAGEMENT BUT FIRST: YOU HAVE TO KNOCK DOWN BARRIERS! but it won t hurt much!
More informationCROSS REFERENCE. Cross Reference Index 110122. Cast ID Number 110111 Connector ID Number 111 Engine ID Number 112122. 2015 Ford Motor Company 109
CROSS REFERENCE Cross Reference Index 110122 Cast ID Number 110111 Connector ID Number 111 112122 2015 Ford Motor Company 109 CROSS REFERENCE Cast ID Number Cast ID Ford Service # MC Part # Part Type
More informationINCIDENCEBETWEENNESS GEOMETRY
INCIDENCEBETWEENNESS GEOMETRY MATH 410, CSUSM. SPRING 2008. PROFESSOR AITKEN This document covers the geometry that can be developed with just the axioms related to incidence and betweenness. The full
More informationSL8800 HDCP 2.2 and HDCP 1.x Protocol Analyzer for HDMI User Guide
SL8800 HDCP 2.2 and HDCP 1.x Protocol Analyzer for HDMI SimplayUG02003A July 2015 Contents 1. Overview... 4 1.1. SL8800 HDCP Protocol Analyzer Test Equipment... 4 1.2. HDCP 2.2/HDCP 1.x Protocol Analyzer
More informationOnline EFFECTIVE AS OF JANUARY 2013
2013 A and C Session Start Dates (AB Quarter Sequence*) 2013 B and D Session Start Dates (BA Quarter Sequence*) Quarter 5 2012 1205A&C Begins November 5, 2012 1205A Ends December 9, 2012 Session Break
More informationRijndael Encryption implementation on different platforms, with emphasis on performance
Rijndael Encryption implementation on different platforms, with emphasis on performance KAFUUMA JOHN SSENYONJO Bsc (Hons) Computer Software Theory University of Bath May 2005 Rijndael Encryption implementation
More informationChapter 4, Arithmetic in F [x] Polynomial arithmetic and the division algorithm.
Chapter 4, Arithmetic in F [x] Polynomial arithmetic and the division algorithm. We begin by defining the ring of polynomials with coefficients in a ring R. After some preliminary results, we specialize
More informationService Instruction. 1.0 SUBJECT: ECi Accessory Cases for Lycoming 4Cylinder engines with single magneto configurations and TITAN 361 Engines
Title: Service Instruction ECi Accessory Cases Installed on Engines S.I. No.: 031 Page: 1 of 7 Issued: 2/28/2003 Revision: 2 (4/13/2009) Technical Portions are FAA DER Approved. 1.0 SUBJECT: ECi Accessory
More informationMathematical Induction
Mathematical Induction (Handout March 8, 01) The Principle of Mathematical Induction provides a means to prove infinitely many statements all at once The principle is logical rather than strictly mathematical,
More informationHow to create OpenDocument URL s with SAP BusinessObjects BI 4.0
How to create OpenDocument URL s with SAP BusinessObjects BI 4.0 Creator: Twitter: Blog: Pieter Verstraeten http://www.twitter.com/pverstraeten http://www.pieterverstraeten.com/blog Hi, Thanks for downloading
More informationASCII CODES WITH GREEK CHARACTERS
ASCII CODES WITH GREEK CHARACTERS Dec Hex Char Description 0 0 NUL (Null) 1 1 SOH (Start of Header) 2 2 STX (Start of Text) 3 3 ETX (End of Text) 4 4 EOT (End of Transmission) 5 5 ENQ (Enquiry) 6 6 ACK
More informationNumber Theory Hungarian Style. Cameron Byerley s interpretation of Csaba Szabó s lectures
Number Theory Hungarian Style Cameron Byerley s interpretation of Csaba Szabó s lectures August 20, 2005 2 0.1 introduction Number theory is a beautiful subject and even cooler when you learn about it
More informationA Second Course in Mathematics Concepts for Elementary Teachers: Theory, Problems, and Solutions
A Second Course in Mathematics Concepts for Elementary Teachers: Theory, Problems, and Solutions Marcel B. Finan Arkansas Tech University c All Rights Reserved First Draft February 8, 2006 1 Contents 25
More informationChapter 13. Fractional Factorials. 13.1 Fractional replicates
244 Chapter 13 Fractional Factorials 13.1 Fractional replicates A factorial design is a fractional replicate if not all possible combinations of the treatment factors occur. A fractional replicate can
More information3. April 2013 IT ZERTIFIKATE. Zertifizierungsstellen / Certification Center. IT Sicherheit UNTERNEHMENSBEREICH IT
IT Sicherheit UNTERNEHMENSBEREICH IT IT ZERTIFIKATE 3. April 2013 Zertifizierungsstellen / Certification Center DTRUST DTrust Root Class 2 CA2007 Aussteller/Issuer: DTRUST Root Class 2 CA 2007 Gültig
More informationInner products on R n, and more
Inner products on R n, and more Peyam Ryan Tabrizian Friday, April 12th, 2013 1 Introduction You might be wondering: Are there inner products on R n that are not the usual dot product x y = x 1 y 1 + +
More informationTHE FUNDAMENTAL THEOREM OF ALGEBRA VIA PROPER MAPS
THE FUNDAMENTAL THEOREM OF ALGEBRA VIA PROPER MAPS KEITH CONRAD 1. Introduction The Fundamental Theorem of Algebra says every nonconstant polynomial with complex coefficients can be factored into linear
More informationFactoring Polynomials
Factoring Polynomials Sue Geller June 19, 2006 Factoring polynomials over the rational numbers, real numbers, and complex numbers has long been a standard topic of high school algebra. With the advent
More information2. Let H and K be subgroups of a group G. Show that H K G if and only if H K or K H.
Math 307 Abstract Algebra Sample final examination questions with solutions 1. Suppose that H is a proper subgroup of Z under addition and H contains 18, 30 and 40, Determine H. Solution. Since gcd(18,
More information02421. HSR TRAINING COURSE REQUIREMENTS HSR Training Course Guidance Booklet 2
02421 HSR TRAINING COURSE REQUIREMENTS HSR Training Course Guidance Booklet 2 SafeWork SA 2 Contents Introduction... 4 Learning resources... 4 PART 1 UNDERPINNING PRINCIPLES FOR THE DEVELOPMENT OF A SAFEWORK
More informationChapter 2 Remodulization of Congruences Proceedings NCUR VI. è1992è, Vol. II, pp. 1036í1041. Jeærey F. Gold Department of Mathematics, Department of Physics University of Utah Don H. Tucker Department
More informationLuxembourg (Luxembourg): Trusted List
Luxembourg (Luxembourg): Trusted List Institut Luxembourgeois de la Normalisation, de l'accréditation de la Sécurité et qualité des produits et services Scheme Information TSL Version 4 TSL Sequence Number
More informationGENERATING SETS KEITH CONRAD
GENERATING SETS KEITH CONRAD 1 Introduction In R n, every vector can be written as a unique linear combination of the standard basis e 1,, e n A notion weaker than a basis is a spanning set: a set of vectors
More informationINTRODUCTORY SET THEORY
M.Sc. program in mathematics INTRODUCTORY SET THEORY Katalin Károlyi Department of Applied Analysis, Eötvös Loránd University H1088 Budapest, Múzeum krt. 68. CONTENTS 1. SETS Set, equal sets, subset,
More informationCOUNCIL OF INTERNATIONAL SCHOOLS (CIS) SCHOOL IMPROVEMENT THROUGH ACCREDITATION
COUNCIL OF INTERNATIONAL SCHOOLS (CIS) SCHOOL IMPROVEMENT THROUGH ACCREDITATION Guide to School Evaluation and Accreditation  8 th Edition (Version 8.2) STANDARDS AND INDICATORS SECTION A SCHOOL GUIDING
More informationMATH 289 PROBLEM SET 4: NUMBER THEORY
MATH 289 PROBLEM SET 4: NUMBER THEORY 1. The greatest common divisor If d and n are integers, then we say that d divides n if and only if there exists an integer q such that n = qd. Notice that if d divides
More informationPUTNAM TRAINING POLYNOMIALS. Exercises 1. Find a polynomial with integral coefficients whose zeros include 2 + 5.
PUTNAM TRAINING POLYNOMIALS (Last updated: November 17, 2015) Remark. This is a list of exercises on polynomials. Miguel A. Lerma Exercises 1. Find a polynomial with integral coefficients whose zeros include
More informationminimal polyonomial Example
Minimal Polynomials Definition Let α be an element in GF(p e ). We call the monic polynomial of smallest degree which has coefficients in GF(p) and α as a root, the minimal polyonomial of α. Example: We
More informationAMBIGUOUS CLASSES IN QUADRATIC FIELDS
MATHEMATICS OF COMPUTATION VOLUME, NUMBER 0 JULY 99, PAGES 0 AMBIGUOUS CLASSES IN QUADRATIC FIELDS R. A. MOLLIN Dedicated to the memory ofd. H. Lehmer Abstract. We provide sufficient conditions for the
More informationSUBGROUPS OF CYCLIC GROUPS. 1. Introduction In a group G, we denote the (cyclic) group of powers of some g G by
SUBGROUPS OF CYCLIC GROUPS KEITH CONRAD 1. Introduction In a group G, we denote the (cyclic) group of powers of some g G by g = {g k : k Z}. If G = g, then G itself is cyclic, with g as a generator. Examples
More information"HIGHER EDUCATION VALUES AND OPINIONS SURVEY" ADVANCED PLACEMENT TEACHERS and GUIDANCE COUNSELORS MayJune 1994
"HIGHER EDUCATION VALUES AND OPINIONS SURVEY" ADVANCED PLACEMENT TEACHERS and GUIDANCE COUNSELORS MayJune 1994 VARIABLE SURVEY ANSWER NAME QUESTION CATEGORIES Facsimile the original mail questionnaire
More informationBEGINNING ALGEBRA ACKNOWLEDMENTS
BEGINNING ALGEBRA The Nursing Department of Labouré College requested the Department of Academic Planning and Support Services to help with mathematics preparatory materials for its Bachelor of Science
More informationMath 34560 Abstract Algebra I Questions for Section 23: Factoring Polynomials over a Field
Math 34560 Abstract Algebra I Questions for Section 23: Factoring Polynomials over a Field 1. Throughout this section, F is a field and F [x] is the ring of polynomials with coefficients in F. We will
More informationAttachment "A"  List of HP Inkjet Printers
HP Deskjet 350c Printer HP Deskjet 350cbi Printer HP Deskjet 350cbi Printer w/rollercase HP Deskjet 420 Printer HP Deskjet 420c Printer HP Deskjet 610c Printer HP Deskjet 610cl Printer HP Deskjet 612c
More informationPutnam Notes Polynomials and palindromes
Putnam Notes Polynomials and palindromes Polynomials show up one way or another in just about every area of math. You will hardly ever see any math competition without at least one problem explicitly concerning
More informationLinear Algebra I. Ronald van Luijk, 2012
Linear Algebra I Ronald van Luijk, 2012 With many parts from Linear Algebra I by Michael Stoll, 2007 Contents 1. Vector spaces 3 1.1. Examples 3 1.2. Fields 4 1.3. The field of complex numbers. 6 1.4.
More informationNEOSHO COUNTY COMMUNITY COLLEGE MASTER COURSE SYLLABUS. Medical Administrative Aspects
NEOSHO COUNTY COMMUNITY COLLEGE MASTER COURSE SYLLABUS COURSE IDENTIFICATION Course Code/Number: ALMA 120 Course Title: Medical Administrative Aspects Division: Applied Science (AS) Liberal Arts (LA) Workforce
More information3 1. Note that all cubes solve it; therefore, there are no more
Math 13 Problem set 5 Artin 11.4.7 Factor the following polynomials into irreducible factors in Q[x]: (a) x 3 3x (b) x 3 3x + (c) x 9 6x 6 + 9x 3 3 Solution: The first two polynomials are cubics, so if
More informationHandout NUMBER THEORY
Handout of NUMBER THEORY by Kus Prihantoso Krisnawan MATHEMATICS DEPARTMENT FACULTY OF MATHEMATICS AND NATURAL SCIENCES YOGYAKARTA STATE UNIVERSITY 2012 Contents Contents i 1 Some Preliminary Considerations
More informationGroup Fundamentals. Chapter 1. 1.1 Groups and Subgroups. 1.1.1 Definition
Chapter 1 Group Fundamentals 1.1 Groups and Subgroups 1.1.1 Definition A group is a nonempty set G on which there is defined a binary operation (a, b) ab satisfying the following properties. Closure: If
More informationChapter 13: Basic ring theory
Chapter 3: Basic ring theory Matthew Macauley Department of Mathematical Sciences Clemson University http://www.math.clemson.edu/~macaule/ Math 42, Spring 24 M. Macauley (Clemson) Chapter 3: Basic ring
More informationIdentifying second degree equations
Chapter 7 Identifing second degree equations 7.1 The eigenvalue method In this section we appl eigenvalue methods to determine the geometrical nature of the second degree equation a 2 + 2h + b 2 + 2g +
More informationSOLUTIONS FOR PROBLEM SET 2
SOLUTIONS FOR PROBLEM SET 2 A: There exist primes p such that p+6k is also prime for k = 1,2 and 3. One such prime is p = 11. Another such prime is p = 41. Prove that there exists exactly one prime p such
More informationDatabases Normalization III. (N Spadaccini 2010 and W Liu 2012) Databases  Normalization III 1 / 31
Databases Normalization III (N Spadaccini 2010 and W Liu 2012) Databases  Normalization III 1 / 31 This lecture This lecture describes 3rd normal form. (N Spadaccini 2010 and W Liu 2012) Databases 
More informationcalculating the result modulo 3, as follows: p(0) = 0 3 + 0 + 1 = 1 0,
Homework #02, due 1/27/10 = 9.4.1, 9.4.2, 9.4.5, 9.4.6, 9.4.7. Additional problems recommended for study: (9.4.3), 9.4.4, 9.4.9, 9.4.11, 9.4.13, (9.4.14), 9.4.17 9.4.1 Determine whether the following polynomials
More informationEMV (ChipandPIN) Protocol
EMV (ChipandPIN) Protocol Märt Bakhoff December 15, 2014 Abstract The objective of this report is to observe and describe a real world online transaction made between a debit card issued by an Estonian
More informationMath 115A HW4 Solutions University of California, Los Angeles. 5 2i 6 + 4i. (5 2i)7i (6 + 4i)( 3 + i) = 35i + 14 ( 22 6i) = 36 + 41i.
Math 5A HW4 Solutions September 5, 202 University of California, Los Angeles Problem 4..3b Calculate the determinant, 5 2i 6 + 4i 3 + i 7i Solution: The textbook s instructions give us, (5 2i)7i (6 + 4i)(
More informationDecember 4, 2013 MATH 171 BASIC LINEAR ALGEBRA B. KITCHENS
December 4, 2013 MATH 171 BASIC LINEAR ALGEBRA B KITCHENS The equation 1 Lines in twodimensional space (1) 2x y = 3 describes a line in twodimensional space The coefficients of x and y in the equation
More informationBoolean Algebra Part 1
Boolean Algebra Part 1 Page 1 Boolean Algebra Objectives Understand Basic Boolean Algebra Relate Boolean Algebra to Logic Networks Prove Laws using Truth Tables Understand and Use First Basic Theorems
More informationThe Dirichlet Unit Theorem
Chapter 6 The Dirichlet Unit Theorem As usual, we will be working in the ring B of algebraic integers of a number field L. Two factorizations of an element of B are regarded as essentially the same if
More informationOverview of Number Theory Basics. Divisibility
Overview of Number Theory Basics Murat Kantarcioglu Based on Prof. Ninghui Li s Slides Divisibility Definition Given integers a and b, b 0, b divides a (denoted b a) if integer c, s.t. a = cb. b is called
More informationNEOSHO COUNTY COMMUNITY COLLEGE MASTER COURSE SYLLABUS. Medical Professional Issues
NEOSHO COUNTY COMMUNITY COLLEGE MASTER COURSE SYLLABUS COURSE IDENTIFICATION Course Code/Number: ALMA 110 Course Title: Medical Professional Issues Division: Applied Science (AS) Liberal Arts (LA) Workforce
More informationAlgebraic Structures II
MAS 305 Algebraic Structures II Notes 12 Autumn 2006 Factorization in integral domains Lemma If a, b, c are elements of an integral domain R and ab = ac then either a = 0 R or b = c. Proof ab = ac a(b
More informationALGEBRAIC APPROACH TO COMPOSITE INTEGER FACTORIZATION
ALGEBRAIC APPROACH TO COMPOSITE INTEGER FACTORIZATION Aldrin W. Wanambisi 1* School of Pure and Applied Science, Mount Kenya University, P.O box 55350100, Kakamega, Kenya. Shem Aywa 2 Department of Mathematics,
More informationMOP 2007 Black Group Integer Polynomials Yufei Zhao. Integer Polynomials. June 29, 2007 Yufei Zhao yufeiz@mit.edu
Integer Polynomials June 9, 007 Yufei Zhao yufeiz@mit.edu We will use Z[x] to denote the ring of polynomials with integer coefficients. We begin by summarizing some of the common approaches used in dealing
More informationTHE AVERAGE DEGREE OF AN IRREDUCIBLE CHARACTER OF A FINITE GROUP
THE AVERAGE DEGREE OF AN IRREDUCIBLE CHARACTER OF A FINITE GROUP by I. M. Isaacs Mathematics Department University of Wisconsin 480 Lincoln Dr. Madison, WI 53706 USA EMail: isaacs@math.wisc.edu Maria
More informationChapter 20. Vector Spaces and Bases
Chapter 20. Vector Spaces and Bases In this course, we have proceeded stepbystep through lowdimensional Linear Algebra. We have looked at lines, planes, hyperplanes, and have seen that there is no limit
More informationCryptography and Network Security. Prof. D. Mukhopadhyay. Department of Computer Science and Engineering. Indian Institute of Technology, Kharagpur
Cryptography and Network Security Prof. D. Mukhopadhyay Department of Computer Science and Engineering Indian Institute of Technology, Kharagpur Module No. # 01 Lecture No. # 12 Block Cipher Standards
More informationCopy in your notebook: Add an example of each term with the symbols used in algebra 2 if there are any.
Algebra 2  Chapter Prerequisites Vocabulary Copy in your notebook: Add an example of each term with the symbols used in algebra 2 if there are any. P1 p. 1 1. counting(natural) numbers  {1,2,3,4,...}
More informationRESULTANT AND DISCRIMINANT OF POLYNOMIALS
RESULTANT AND DISCRIMINANT OF POLYNOMIALS SVANTE JANSON Abstract. This is a collection of classical results about resultants and discriminants for polynomials, compiled mainly for my own use. All results
More informationApplication Note RMF Magic 5.1.0: EMC Array Group and EMC SRDF/A Reporting. July 2009
Application Note RMF Magic 5.1.0: EMC Array Group and EMC SRDF/A Reporting July 2009 Summary: This Application Note describes the new functionality in RMF Magic 5.1 that enables more effective monitoring
More informationMAT188H1S Lec0101 Burbulla
Winter 206 Linear Transformations A linear transformation T : R m R n is a function that takes vectors in R m to vectors in R n such that and T (u + v) T (u) + T (v) T (k v) k T (v), for all vectors u
More informationBX in ( u, v) basis in two ways. On the one hand, AN = u+
1. Let f(x) = 1 x +1. Find f (6) () (the value of the sixth derivative of the function f(x) at zero). Answer: 7. We expand the given function into a Taylor series at the point x = : f(x) = 1 x + x 4 x
More informationSPECIAL PRODUCTS AND FACTORS
CHAPTER 442 11 CHAPTER TABLE OF CONTENTS 111 Factors and Factoring 112 Common Monomial Factors 113 The Square of a Monomial 114 Multiplying the Sum and the Difference of Two Terms 115 Factoring the
More informationRow Ideals and Fibers of Morphisms
Michigan Math. J. 57 (2008) Row Ideals and Fibers of Morphisms David Eisenbud & Bernd Ulrich Affectionately dedicated to Mel Hochster, who has been an inspiration to us for many years, on the occasion
More informationit is easy to see that α = a
21. Polynomial rings Let us now turn out attention to determining the prime elements of a polynomial ring, where the coefficient ring is a field. We already know that such a polynomial ring is a UF. Therefore
More informationFactoring, Solving. Equations, and Problem Solving REVISED PAGES
05W4801AM1.qxd 8/19/08 8:45 PM Page 241 Factoring, Solving Equations, and Problem Solving 5 5.1 Factoring by Using the Distributive Property 5.2 Factoring the Difference of Two Squares 5.3 Factoring
More informationFACTORING POLYNOMIALS IN THE RING OF FORMAL POWER SERIES OVER Z
FACTORING POLYNOMIALS IN THE RING OF FORMAL POWER SERIES OVER Z DANIEL BIRMAJER, JUAN B GIL, AND MICHAEL WEINER Abstract We consider polynomials with integer coefficients and discuss their factorization
More information51 NUMBER THEORY: DIVISIBILITY; PRIME & COMPOSITE NUMBERS 210 f8
51 NUMBER THEORY: DIVISIBILITY; PRIME & COMPOSITE NUMBERS 210 f8 Note: Integers are the w hole numbers and their negatives (additive inverses). While our text discusses only whole numbers, all these ideas
More informationFIRST YEAR CALCULUS. Chapter 7 CONTINUITY. It is a parabola, and we can draw this parabola without lifting our pencil from the paper.
FIRST YEAR CALCULUS WWLCHENW L c WWWL W L Chen, 1982, 2008. 2006. This chapter originates from material used by the author at Imperial College, University of London, between 1981 and 1990. It It is is
More informationRolle s Theorem. q( x) = 1
Lecture 1 :The Mean Value Theorem We know that constant functions have derivative zero. Is it possible for a more complicated function to have derivative zero? In this section we will answer this question
More informationFACTORING IN QUADRATIC FIELDS. 1. Introduction. This is called a quadratic field and it has degree 2 over Q. Similarly, set
FACTORING IN QUADRATIC FIELDS KEITH CONRAD For a squarefree integer d other than 1, let 1. Introduction K = Q[ d] = {x + y d : x, y Q}. This is called a quadratic field and it has degree 2 over Q. Similarly,
More informationSOLICITATION/CONTRACT/ORDER FOR COMMERICAL ITEMS
SOLICITATION/CONTRACT/ORDER FOR COMMERICAL ITEMS 1. REQUISITION NUMBER PAGE 1 OF OFFEROR TO COMPLETE BLOCKS 12, 17, 23, 24 & 30 2. CONTRACT NO. 3. AWARD/EFFECTIVE DATE 4. ORDER NUMBER 5. RFQ NUMBER 6.
More informationThe Structure of Galois Algebras
The Structure of Galois Algebras George Szeto Department of Mathematics, Bradley University Peoria, Illinois 61625 { U.S.A. Email: szeto@hilltop.bradley.edu and Lianyong Xue Department of Mathematics,
More informationFactoring Polynomials and Solving Quadratic Equations
Factoring Polynomials and Solving Quadratic Equations Math Tutorial Lab Special Topic Factoring Factoring Binomials Remember that a binomial is just a polynomial with two terms. Some examples include 2x+3
More informationSecure Network Communication Part II II Public Key Cryptography. Public Key Cryptography
Kommunikationssysteme (KSy)  Block 8 Secure Network Communication Part II II Public Key Cryptography Dr. Andreas Steffen 20002001 A. Steffen, 28.03.2001, KSy_RSA.ppt 1 Secure Key Distribution Problem
More informationUnique Factorization
Unique Factorization Waffle Mathcamp 2010 Throughout these notes, all rings will be assumed to be commutative. 1 Factorization in domains: definitions and examples In this class, we will study the phenomenon
More informationBoolean Algebra (cont d) UNIT 3 BOOLEAN ALGEBRA (CONT D) Guidelines for Multiplying Out and Factoring. Objectives. Iris HuiRu Jiang Spring 2010
Boolean Algebra (cont d) 2 Contents Multiplying out and factoring expressions ExclusiveOR and ExclusiveNOR operations The consensus theorem Summary of algebraic simplification Proving validity of an
More informationCM2202: Scientific Computing and Multimedia Applications General Maths: 2. Algebra  Factorisation
CM2202: Scientific Computing and Multimedia Applications General Maths: 2. Algebra  Factorisation Prof. David Marshall School of Computer Science & Informatics Factorisation Factorisation is a way of
More informationJUST THE MATHS UNIT NUMBER 1.8. ALGEBRA 8 (Polynomials) A.J.Hobson
JUST THE MATHS UNIT NUMBER 1.8 ALGEBRA 8 (Polynomials) by A.J.Hobson 1.8.1 The factor theorem 1.8.2 Application to quadratic and cubic expressions 1.8.3 Cubic equations 1.8.4 Long division of polynomials
More information