Solutions to Assignment 10


 Susan Conley
 11 months 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
Basic Properties of Rings
LECTURE 15 Basic Properties of Rings Theorem 15.1. For any element a in a ring R, the equation a + x 0 R has a unique solution. We know that a + x 0 R has at least one solution u R by Axiom (5) in the
More informationVisa Smart Debit/Credit Certificate Authority Public Keys
CHIP AND NEW TECHNOLOGIES Visa Smart Debit/Credit Certificate Authority Public Keys Overview The EMV standard calls for the use of Public Key technology for offline authentication, for aspects of online
More informationFAN DECK FEDERAL STANDARD 595B COLORS (July 1994) 2002 Rob Graham. Special thanks to Pete Chalmers.
7690011622210 FAN DECK FEDERAL STANDARD 595B COLORS (July 1994) Introduction 2002 Rob Graham. Special thanks to Pete Chalmers. The Federal Standard 595B specification is a U.S. government paint color
More information3.3 Constructing Examples
33 JABeachy 1 33 Constructing Examples from A Study Guide for Beginner s by JABeachy, a supplement to Abstract Algebra by Beachy / Blair From this point on, if the modulus n is fixed throughout the problem,
More informationElements 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 information1. The set of all nilpotent elements in a commutative ring forms an ideal.
Hungerford: Algebra III.2. Ideals 1. The set of all nilpotent elements in a commutative ring forms an ideal. Proof: Let R be a commutative ring and let N denote the set of all nilpotent elements in R.
More informationSolutions to Assignment 4
Solutions to Assignment 4 Math 412, Winter 2003 3.1.18 Define a new addition and multiplication on Z y a a + 1 and a a + a, where the operations on the righthand side off the equal signs are ordinary
More informationAlgebraic structures: from groups to fields
EE 387, Notes 6, Handout #8 Algebraic structures: from groups to fields This course concentrates on linear block codes. Codeword vectors are linear transforms of message vectors: c = mg. codeword c is
More informationSboxes generated using Affine Transformation giving Maximum Avalanche Effect
Sboxes generated using Affine Transformation giving Maximum Avalanche Effect Chandrasekharappa T.G.S., Prema K.V. and Kumara Shama Department of Electronics and Communication Engineering Manipal Institute
More informationHW3, Math 307. CSUF. Spring 2007.
HW, Math 07. CSUF. Spring 2007. Nasser M. Abbasi Contents Section.6, problem 2 Section.6, problem 2 Section.6, problem 5 4 Section.6, problem 7 6 5 Section.6, problem 25 8 6 Section.6, Problem 28 0 7 Section.6,
More informationA NOTE ON DISTORTED CYCLIC SUBSETS
Mathematica Pannonica 20/1 (2009), 1 5 A NOTE ON DISTORTED CYCLIC SUBSETS A. D. Sands Department of Mathematics, Dundee University, Dundee DD1 4HN, Scotland Received: February 2009?? MSC 2000: 20 K 01
More informationHW2 Solutions. Let d be the distance from the point P to the origin. Then the orbit of P is the circle of radius d centered at the origin.
HW2 Solutions Section 16 13) Let G be the additive group of real numbers Let the action of θ G in the real plane R 2 be given by rotating the plane counterclockwise about the origin through θ radians Let
More informationINTRODUCTION TO PROOFS: HOMEWORK SOLUTIONS
INTRODUCTION TO PROOFS: HOMEWORK SOLUTIONS STEVEN HEILMAN Contents 1. Homework 1 1 2. Homework 2 6 3. Homework 3 10 4. Homework 4 16 5. Homework 5 19 6. Homework 6 21 7. Homework 7 25 8. Homework 8 28
More informationEquivalence relations
Equivalence relations A motivating example for equivalence relations is the problem of constructing the rational numbers. A rational number is the same thing as a fraction a/b, a, b Z and b 0, and hence
More informationCONVERSION CHART PROBATE CODE TO ESTATES CODE
Revised January 27, 2014 Prepared by Prof. Gerry W. Beyer CONVERSION CHART PROBATE CODE TO ESTATES CODE 2(a) (part) 21.006 2(c) 21.004 3 (part) 22.001 3(a) 22.002 3(b) 22.004 3(c) 22.005 3(d) 22.006 3(e)
More information!" #" $ % $ & ' % "( ) $* ( + $,  $. /
!" #"$ %$ & '% "( )$* (+$,  $. / 0 & 1&&##+.2 " 2&%%.&, &#& $ & 1 $ + $1 & 31,4!%'"%# 1$4 % 2.& # &2 5 + 670894, ' &" '$+.&2 2$%&##" +..$&&+%1$%114+1 &$ +1&#&+,#%11" &&.1&.$$!$%&'" &:#1;#.!, $& &' +".2
More informationAlgebra. Sample Solutions for Test 1
EPFL  Section de Mathématiques Algebra Fall semester 20082009 Sample Solutions for Test 1 Question 1 (english, 30 points) 1) Let n 11 13 17. Find the number of units of the ring Z/nZ. 2) Consider the
More informationDepartment of Mathematics Exercises G.1: Solutions
Department of Mathematics MT161 Exercises G.1: Solutions 1. We show that a (b c) = (a b) c for all binary strings a, b, c B. So let a = a 1 a 2... a n, b = b 1 b 2... b n and c = c 1 c 2... c n, where
More informationMagic Squares and Modular Arithmetic
Magic Squares and Modular Arithmetic Jim Carlson November 7, 2001 1 Introduction Recall that a magic square is a square array of consecutive distinct numbers such that all row and column sums and are the
More informationMath 430 Problem Set 5 Solutions
Math 430 Problem Set 5 Solutions Due April 1, 2016 13.2. Find all of the abelian groups of order 200 up to isomorphism. Solution. Every abelian group is a direct product of cyclic groups. Using the fact
More informationGroup Theory. Solutions to Problem Set 2 October 26, Suppose that e is the right identity of a group G, e g = g (2) e e = e (3)
Group Theory Solutions to Problem Set 2 October 26, 2001 1. Suppose that e is the right identity of a group G, for all g in G, and that e is the left identity, ge g (1) e g g (2) for all g in G. The choice
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 information25 Integral Domains. Subrings
Arkansas Tech University MATH 4033: Elementary Modern Algebra Dr. Marcel B. Finan 25 Integral Domains. Subrings In Section 24 we defined the terms unitary rings and commutative rings. These terms together
More informationSent sent. lavoro licenziamento marittimo ISSN Pubblicato dal 01/04/2010
ISSN 11278579 Pubblicato dal 01/04/2010 All'indirizzo http://www.diritto.it/docs/29254tribunaledimodicagiudicedellavorodottssasandralevantisentenzan742008intemadiirrogazionedisanzionedisciplinarenellambitodirapportodilavoronelsettoremarittimo
More informationMODULAR ARITHMETIC STEVEN DALE CUTKOSKY
MODULAR ARITHMETIC STEVEN DALE CUTKOSKY 1. Relations, Rational Numbers and Modular Arithmetic 1.1. Relations. Suppose that X is a set. a relation on X is a subset R X X = {(a, b) a, b X}. Write a b if
More informationSMS4 Encryption Algorithm for Wireless Networks
SMS4 Encryption Algorithm for Wireless Networks Translated and typeset by Whitfield Diffie of Sun Microsystems and George Ledin of Sonoma State University 15 May 2008 Version 1.03 SMS4 is a Chinese block
More informationMATH 118 HW 1 LINDSAY CROSS, KELLY DOUGAN, KEVIN LIU
MATH 118 HW 1 LINDSAY CROSS, KELLY DOUGAN, KEVIN LIU 1.2 Problem 2. Find the greatest common divisor g of the numbers 1819 and 3574, and then find integers x and y to satisfy 1819x + 3587y = g. Use the
More informationI. GROUPS: BASIC DEFINITIONS AND EXAMPLES
I GROUPS: BASIC DEFINITIONS AND EXAMPLES Definition 1: An operation on a set G is a function : G G G Definition 2: A group is a set G which is equipped with an operation and a special element e G, called
More informationLAGRANGE S THEOREM. Definition: An operation on a set G is a function : G G G.
Definition: LAGRANGE S THEOREM An operation on a set G is a function : G G G. Definition: A group is a set G which is equipped with an operation and a special element e G, called the identity, such that
More informationClassifying All Groups of Order 16 David Clausen MATH 434 University of Puget Sound Spring 2012
Classifying All Groups of Order 16 David Clausen MATH 434 University of Puget Sound Spring 2012 c 2012, David Clausen. Permission is granted to copy, distribute, and/or modify this document under the terms
More informationMatrices: 2.2 Properties of Matrices
Goals We will describe properties of matrices with respect to addition, scalar multiplications and matrix multiplication and others. Among what we will see 1. Matrix multiplication does not commute. That
More informationSolutions For Homework 3. Ch. 2
Solutions For Homework 3 Ch. 2 2. Show that the set {5, 15, 25, 35} is a group under multiplication modulo 40. What is the identity element of this group? Can you see any relationship between this group
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 informationEECS Homework 9 Solutions
EECS 2031 Homework 9 Solutions Total Points: 50 Page 413: 10) Let R be the relation on the set of ordered pairs of positive integers such that ((a, b), (c, d)) R if and only if ad = bc. Show that R is
More informationAdvanced Encryption Standard. Z. Jerry Shi Department of Computer Science and Engineering University of Connecticut
Advanced Encryption Standard Z. Jerry Shi Department of Computer Science and Engineering University of Connecticut February 2012 Classes of cryptographic algorithms Symmetrickey algorithms Use the same
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 information9. For each of the following values of n, find all generators of the cyclic group Z n under addition.
Math 110A HW 34 Solutions 6 Find the order of the following elements in GL 4 (R) (a) (b) A A 2 Thus A 2 A A 2 A 3 A 4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
More informationDefinition For a set F, a polynomial over F with variable x is of the form
*7. Polynomials Definition For a set F, a polynomial over F with variable x is of the form a n x n + a n 1 x n 1 + a n 2 x n 2 +... + a 1 x + a 0, where a n, a n 1,..., a 1, a 0 F. The a i, 0 i n are the
More informationHomework 2 Solutions
Homework 2 Solutions Enrique Treviño September 23, 2014 1 Chapter 3 Problem 1. (Exercise 2) Which of the following multiplication tables defined on the set G = {a, b, c, d} form a group? Support your answer
More information2011 Canadian Senior Mathematics Contest
The CENTRE for EDUCATION in MATHEMATICS and COMPUTING www.cemc.uwaterloo.ca 011 Canadian Senior Mathematics Contest Tuesday, November, 011 (in North America and South America) Wednesday, November 3, 011
More informationQuick Installation Guide TBKSER1001
Quick Installation Guide TBKSER1001 This manual has been designed to guide you to basic settings of your IP devices such as installation and configuration for using them. 1. Connect cables to IP device
More informationMATH Algebra for High School Teachers Integral Domains
MATH 57091  Algebra for High School Teachers Integral Domains Professor Donald L. White Department of Mathematical Sciences Kent State University D.L. White (Kent State University) 1 / 8 Cancellation
More informationProblem Set 1 Solutions Math 109
Problem Set 1 Solutions Math 109 Exercise 1.6 Show that a regular tetrahedron has a total of twentyfour symmetries if reflections and products of reflections are allowed. Identify a symmetry which is
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 informationGroup Theory E d i t h L a w
Group Theory E d i t h L a w 7.. 7 Puzzle Group Theory in the Bedroom Reference: Scientific American, 9(5)95 What is a Group? A Familiar Group To solve the equation + x =  + (+x) =  + (+) + x = 6 +
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 informationType: Universal Relais Modbus protocol description. EANr.: 1451 Replace for: Page: 1 von 9
EANr.: 1451 Replace for: Page: 1 von 9 MODBUS TCP/IP TCP Port: 502 Max. TCP connections): 5 MODBUS RTU Supported function codes Function code Name Utilization 3 (03H) Read Holding Registers Read data
More information1 Symmetries of regular polyhedra
1230, notes 5 1 Symmetries of regular polyhedra Symmetry groups Recall: Group axioms: Suppose that (G, ) is a group and a, b, c are elements of G. Then (i) a b G (ii) (a b) c = a (b c) (iii) There is an
More informationNew CUSD 712 Secondary High School Proposed Boundary Recommendation. CUSD Board Study Session November 5, 2014
New CUSD 712 Secondary High School Proposed Boundary Recommendation CUSD Board Study Session November 5, 2014 1 Goals Develop school boundaries for the new 712 secondary high school Balance projected
More informationLet A be a square matrix of size n. A matrix B is called inverse of A if AB = BA = I.
Matrix Inverses Let A be a square matrix of size n A matrix B is called inverse of A if AB = BA = I If A has an inverse, then we say that A is invertible The inverse of A is unique and we denote it by
More informationMath 122, Solution Set No. 2
Math 122, Solution Set No. 2 As a general note, all elements of S n will be written in disjoint cycle notation unless otherwise specified. Also, as a notational convention, H G means H is a subgroup of
More information2215 Solution Problem Sheet 4 1. Irreducibility and Factorization of Polynomials
2215 Solution Problem Sheet 4 1 Irreducibility and Factorization of Polynomials 1. Show that p(x) is irreducible in R[X], R any ring, if and only if p(x a) is irreducible in R[x] for any a R. Solution:
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 informationNumber Theory Vocabulary (For Middle School Teachers)
Number Theory Vocabulary (For Middle School Teachers) A Absolute value the absolute value of a real number is its distance from zero on the number line. The absolute value of any real number, a, written
More informationCHAPTER 1 INTRODUCTION
CHAPTER 1 INTRODUCTION 1.1 Motivations Vision is the process of discovering from images what is present in the world and where it is (Marr [88], p. 3). Due to the apparent simplicity of the action of seeing,
More informationSolutions to 2015 CMO (DRAFT as of April 6, 2015)
Solutions to 015 CMO Solutions to 015 CMO (DRAFT as of April 6, 015) Problem 1. Let N = {1,, 3,...} be the set of positive integers. Find all functions f, defined on N and taking values in N, such that
More informationOn the generation of elliptic curves with 16 rational torsion points by Pythagorean triples
On the generation of elliptic curves with 16 rational torsion points by Pythagorean triples Brian Hilley Boston College MT695 Honors Seminar March 3, 2006 1 Introduction 1.1 Mazur s Theorem Let C be a
More informationTheorem 2. If x Q and y R \ Q, then. (a) x + y R \ Q, and. (b) xy Q.
Math 305 Fall 011 The Density of Q in R The following two theorems tell us what happens when we add and multiply by rational numbers. For the first one, we see that if we add or multiply two rational numbers
More informationQuick Installation Guide
Quick Installation Guide This manual has been designed to guide you through basic settings of your IP devices, such as installation and configuration for using them. Step1. Assemble and install IP device
More information3ab, a, b R, Find the identity element in R under. Ans: a*e= a e 1 and e*a=e a 1 i.e a*e e*a Therefore identity does not exist GROUPS PART A
[ 1 MARK ] 1. Define a binary operation GROUPS PART A Ans: Let S be a nonempty set.* is said to be binary operation on S if a, b S, a* b S and is unique 2. a * b = ab, Where a and b are the elements of
More informationMath 31 Show Your Work! Page 1 of (10) Complete the following definitions.
Math 1 Show Your Work! Page 1 of 4 1. (10) Complete the following definitions. (a) A nonempty set G together with a binary operation (a,b) ab from G G to G is called a group if... ANS: if (1) the operation
More information1.4 More Matrix Operations and Properties
8 CHAPTER. SYSTEMS OF LINEAR EQUATIONS AND MATRICES. More Matrix Operations Properties In this section, we look at the properties of the various operations on matrices. As we do so, we try to draw a parallel
More informationUSB HID to PS/2 Scan Code Translation Table
Key Name HID Usage Page HID Usage ID PS/2 Set 1 Make* PS/2 Set 1 Break* PS/2 Set 2 Make PS/2 Set 2 Break System Power 01 81 E0 5E E0 DE E0 37 E0 F0 37 System Sleep 01 82 E0 5F E0 DF E0 3F E0 F0 3F System
More informationAlgebraic Structures II
MAS 305 Algebraic Structures II Notes 10 Autumn 2006 Ring Theory A ring is a set R with two binary operations + and satisfying (a) (R,+) is an Abelian group; (b) R is closed under ; (c) is associative;
More informationQuick Start Guide GCode/EEC2400
Quick Start Guide GCode/EEC2400 This manual provides instructions for quick installation and basic configuration of your IP device. 1. Part names and positions Please check names and positions of each
More information3.2 Subgroups. 3.2 J.A.Beachy 1. from A Study Guide for Beginner s by J.A.Beachy, a supplement to Abstract Algebra by Beachy / Blair
3.2 J.A.Beachy 1 3.2 Subgroups from A Study Guide for Beginner s by J.A.Beachy, a supplement to Abstract Algebry Beachy / Blair 28. In Z n, show that if gcd(a,n) = d, then a n = d n. Note: This result
More informationType: Universal Relais Modbus protocol description. EANr.: 1451 Replace for: Page: 1 von 10
EANr.: 1451 Replace for: 12280160500 Page: 1 von 10 MODBUS TCP/IP TCP Port: 502 Max. TCP connections): 5 MODBUS RTU Connection diagram RS 485 Connection name Modbus Ziehl EIA/TIA485  wire D0 B (B
More informationELEMENTS OF NUMBER THEORY & CONGRUENCES. Lagrange, Legendre and Gauss. Mth Mathematicst
ELEMENTS OF NUMBER THEORY & CONGRUENCES Lagrange, Legendre and Gauss ELEMENTS OF NUMBER THEORY & CONGRUENCES 1) If a 0, b 0 Z and a/b, b/a then 1) a=b 2) a=1 3) b=1 4) a=±b 2) 0and 1 ae are 1) primes 2)
More informationSummer High School 2009 Aaron Bertram
Summer High School 2009 Aaron Bertram 2. Modular Arithmetic and Algebra. Notation: The expression k n means k divides n. Now fix a natural number k > 1. Definition 2.1. Integers a and b are congruent modulo
More informationMATH 304 Linear Algebra Lecture 8: Inverse matrix (continued). Elementary matrices. Transpose of a matrix.
MATH 304 Linear Algebra Lecture 8: Inverse matrix (continued). Elementary matrices. Transpose of a matrix. Inverse matrix Definition. Let A be an n n matrix. The inverse of A is an n n matrix, denoted
More informationi I Thus H is the smallest subgroup that contains S.
4. Cyclic groups Lemma 4.1. Let G be a group and let H i, i I be a collection of subgroups of G. Then the intersection H = H i, is a subgroup of G i I Proof. First note that H is nonempty, as the identity
More informationMA 274: EQUIVALENCE RELATIONS
MA 274: EQUIVALENCE RELATIONS 1. EQUIVALENCE RELATIONS Definition 1.1. Let A be a set. An equivalence relation on A is a relation such that: (1) is reflexive. That is, (2) is symmetric. That is, x A,x
More informationORDERS OF ELEMENTS IN A GROUP
ORDERS OF ELEMENTS IN A GROUP KEITH CONRAD 1. Introduction Let G be a group and g G. We say g has finite order if g n = e for some positive integer n. For example, 1 and i have finite order in C, since
More informationFermat s Little Theorem
Fermat s Little Theorem Theorem (Fermat s Little Theorem): Let p be a prime. Then p n p n (1) for any integer n 1. Proof: We distinguish two cases. Case A: Let p n, then, obviously, p n p n, and we are
More informationAbstract Algebra. Notes by R.J. Buehler Based on J.A. Gallian s Contemporary Abstract Algebra. April 26, 2011
Abstract Algebra Notes by R.J. Buehler Based on J.A. Gallian s Contemporary Abstract Algebra April 26, 2011 0 Preliminaries Theorem 0.1 (The Division Algorithm). Let a and b be integers with b > 0. Then
More informationCOMMUTATIVE RINGS. Definition: A domain is a commutative ring R that satisfies the cancellation law for multiplication:
COMMUTATIVE RINGS Definition: A commutative ring R is a set with two operations, addition and multiplication, such that: (i) R is an abelian group under addition; (ii) ab = ba for all a, b R (commutative
More information18. If x = 2 + 3, find an integer or fraction equal to x x 4.
What is the sum of the digits of (000) 2? 2 Find two numbers whose sum is 52 such that one is three times as large as the other 3 What is the area of a circle inscribed in a square of sidelength 3? 4
More informationApplications of Fermat s Little Theorem and Congruences
Applications of Fermat s Little Theorem and Congruences Definition: Let m be a positive integer. Then integers a and b are congruent modulo m, denoted by a b mod m, if m (a b). Example: 3 1 mod 2, 6 4
More informationSolutions to Homework Problems from Chapter 3
Solutions to Homework Problems from Chapter 3 31 311 The following subsets of Z (with ordinary addition and multiplication satisfy all but one of the axioms for a ring In each case, which axiom fails (a
More informationArts et politique sous Louis XIV
Arts et politique sous Louis XIV Marie Tourneur To cite this version: Marie Tourneur. Arts et politique sous Louis XIV. Éducation. 2013. HAL Id: dumas00834060 http://dumas.ccsd.cnrs.fr/dumas00834060
More information3. Arithmetic of Rational Numbers We now define addition and multiplication on the rational numbers. For addition we should have a b + c d = ad+bc
RATIONAL NUMBERS Abstract. These are notes on the construction of the rational numbers from the integers. We go through equivalence relations, the definition of rationals, addition and multiplication of
More informationMath 541 Solutions to HW #6
Math 541 Solutions to HW #6 The following are from Gallian, Chapters 4 and 5 (6th edition). # 4.8 : Let a be an element of a group and let a = 15. Compute the orders of the following elements of G: a 3,
More informationMatrices Worksheet. Adding the results together, using the matrices, gives
Matrices Worksheet This worksheet is designed to help you increase your confidence in handling MATRICES. This worksheet contains both theory and exercises which cover. Introduction. Order, Addition and
More informationCanadian Aero Manufacturing Applicability list for CAM Lycoming Oil Pump replacement parts
Issue Date August 7, 2002 Print Date 27/02/09 Revision: 5 Page 1 of 5 This applicability list is approved by PDA0017, and its eligibility list CAL01 issued October 30 2001, or later approved revision.
More informationHonors Algebra 2 Learning Targets. Unit 1  Equations and Inequalities Learning Target
Honors Algebra 2 s I can Unit 1  Equations and Inequalities 1. Describe the subsets of the real numbers and show how they are related to one another. 2. Use the properties of real numbers to justify the
More informationMODULAR ARITHMETIC. a smallest member. It is equivalent to the Principle of Mathematical Induction.
MODULAR ARITHMETIC 1 Working With Integers The usual arithmetic operations of addition, subtraction and multiplication can be performed on integers, and the result is always another integer Division, on
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 informationGeometric Transformations
Geometric Transformations Definitions Def: f is a mapping (function) of a set A into a set B if for every element a of A there exists a unique element b of B that is paired with a; this pairing is denoted
More informationEmbargos de Declaração no Processo do Trabalho
ISSN 11278579 Pubblicato dal 12/03/2013 All'indirizzo http://www.diritto.it/docs/34770embargosdedeclaraonoprocessodotrabalho Autore: José Geraldo da Fonseca Embargos de Declaração no Processo do
More informationSection I.7. Generating Sets and Cayley Digraphs
I.7 Generating Sets 1 Section I.7. Generating Sets and Cayley Digraphs Note. In this section, we generalize the idea of a single generator of a group to a whole set of generators of a group. Remember,
More informationCHAPTER 6 Space Trusses
CHAPTER 6 Space Trusses INTRODUCTION A space truss consists of members joined together at their ends to form a stable threedimensional structures A stable simple space truss can be built from the basic
More informationH/wk 9, Solutions to selected problems
H/wk 9, Solutions to selected problems Ch. 2.8, Problem 7 If H G and H = 2, show that H Z(G). Is this true when H = 3? Since H = 2, we have H = {1, a}, where a G is such that a = 2. Let g G be arbitrary.
More information11. SOLUBLE GROUPS The Derived Series nth derived subgroup derived series Example 1: soluble group soluble of length n Theorem 1: Proof:
11. SOLUBLE GROUPS 11.1. The Derived Series Recall that the derived (commutators) subgroup is generated by all the commutators [a, b] = a 1 b 1 ab for all a, b. The derived subgroup of G is denoted by
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 informationNORMAL SUBGROUPS AND HOMOMORPHISMS
NORMAL SUBGROUPS AND HOMOMORPHISMS f ( xy) = f ( x) f ( y) Previously, we talked about what it means for two groups to be isomorphic, and we explained that this means that they have the same structure
More informationAbstract Algebra Cheat Sheet
Abstract Algebra Cheat Sheet 16 December 2002 By Brendan Kidwell, based on Dr. Ward Heilman s notes for his Abstract Algebra class. Notes: Where applicable, page numbers are listed in parentheses at the
More informationProp and Def: Let G be a group and H, K be subgroups of G. Then the function H K G : (h, k) hk is an isomorphism iff
Section 14: Finite Abelian Groups On the third exam, you were asked to prove the following: Prop and Def: Let G be a group and H, K be subgroups of G. Then the function H K G : (h, k) hk is an isomorphism
More informationPattern Co. Monkey Trouble Wall Quilt. Size: 48" x 58"
.............................................................................................................................................. Pattern Co..........................................................................................
More informationPractice Exam 1. Proof.
Practice Exam 1 1. Let f : A B and g : B C be maps. (a) If f and g are both onetoone functions, show that g f is onetoone. (b) If g f is onto, show that g is onto. (c) If g f is onetoone, show that
More informationLes identités féminines dans le conte traditionnel occidental et ses réécritures : une perception actualisée
Les identités féminines dans le conte traditionnel occidental et ses réécritures : une perception actualisée des élèves Sarah Brasseur To cite this version: Sarah Brasseur. Les identités féminines dans
More information