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# Comprehensive Study Guide up to Test 1 4193

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This 10 page Study Guide was uploaded by Arnold E on Wednesday January 6, 2016. The Study Guide belongs to 4193 at Fort Valley State University taught by Dr. Zhu in Winter 2016. Since its upload, it has received 44 views. For similar materials see Abstract Algebra in Math at Fort Valley State University.

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Date Created: 01/06/16

Study Guide Test 1 Monday, September 14, 2011:09 AM No section 1 material on test. Section 2: Definition of Binary Operation Whether or not an operation is a binary operation (check it using definition). Know whether or not a binary operation is commutative or associative There may be tables on the test True or False questions from page 27 Section 3: Know definition of iso- and homo- morphisms Be able to find isomorphism such that the homomorphism property is satisfied (it is one to one and onto,etc.) If there is a given binary structure, verify that it is homomorphism Prove uniqueness of identity element in a binary structure for ANY binary structure (which is a theorem 3.13 in the text, be able to prove it). Be able to prove theorem 3.14 which is if the S and S' are isomorphic then the identity elements image is an identity elements pre-image (kinda? Did I explain that correctly? Check text). Section 4: Definition of Group (Definition 4.1 in notes, must be complete not a "word" like associative, entire thing) Page 47 True or False (answers F T T F F T T T F T) 4.15 and 4.17 Theorem Proofs Section 5: Definition of Subgroup Verify if it is a subgroup True or False page 57 #39 (T F T F F F F F F T T) Section 3 Isomorphic Binary Structure Monday, August 24, 211:29 AM Proof. Assume that e and e' are identity elements of <S,*> Since e = e * e' = e' * e given by definition 3.12 And since e' = e' * e = e * e' Then e = e' Therefore theorem 3.13 is true Proof. Prove that Phi ( e ) *' S' = S' *' Phi ( e ) = S' where s' belongs to S' Since Phi is an isomorphism, then for s' belongs to S' there is an element s that bleongs to S such that Phi (s) = S' . Since e is the identity element in S, then e * s = s * e = s And Phi ( e * s ) = Phi ( s * e ) = Phi ( s ) Notes Page 1 Section 5 Subgroups Wednesday, September 9, 11:12 AM You will use the same operation in the subgroup as the group, it is called the induced operation: Proper Trivial Trivial and Non Trivial Subgroupts & Proper and Improper Subgroups Notes Page 1 Section 4 Groups Monday, August 31, 201511:03 AM The identity element must EXIST and be UNIQUE. Notes Page 1 "Proof by Construction" is when you find a solution to prove that a solution does in fact exist. Notes Page 2 IF there IS an identity element in a binary structure, THEN it is unique. For every GROUP there is an identity element and it is unique. This is guaranteed in every group. A group is a binary structure but a binary structure is not necessarily a group. To prove that two items are unique, you supose there at at least two and then prove that they are the same one. Notes Page 3 Notes Page 4 Section 2 Binary Operations Wednesday, August 19,11:25 AM Audio recording started: 11:28 AM Wednesday, August 19, 2015 Definitions of Sets SxS is a product R is all real numbers, one dimensional space R^2 is two dimensional space, or RxR SxS is a set in two dimensional space represented as an ordered pair (x,y) such that SxS = R^2, (x,y) e R^2 If S = {a,b,c} then SxS = {(a,a),(a,b),(a,c),(b,a),(b,a)…(c,b)}. Notice that order matters so (a,b) and (a,c) are two different elements of the set SxS read as "S cross S". Binary Operations If the result of an operation is still in the original set, then it is called a binary operation. One example of a binary operation on R is addition. 1 belongs to R, 2 belongs to R, 1+2 =3 which also belongs to R. Multiplication is also a binary operation in R. Addition is a binary operation in R^- because a negative number plus a negative number is a negative number Multiplication is NOT a binary operation in R^- because a negative times negative is positive Division is NOT a binary operation on R because a number divided by 0 is undefined. Note that… a*b does not mean multiplication necessarily in this context, it stands for any binary operation. a*b = a+b **Will continue section 1.2 on Friday Audio recording started: 11:03 AM Friday, August 21, 2015 M( R ) is the set of all matrices with real entries. To perform a binary operation, you need an operation and a set. Question: is the following a binary operation: M( R ) + No, because a binary operation must be defined for all elements of the set, and matrices of different sizes cannot be added together. If you have a set with elements in a subset, a binary operation on those elements is still in the subset as long as the binary operation is always true in the subset. Question: If you have a binary operation defined in a set S, and a subset H with two elements a and b, is a * b in H? Is it in S? It is not necessarily in H, we do not know if the binary operation is defined in H It is definitely in S because the binary operation is defined in S and the elements both belong to S. Problem 18, p26 Yes, this is a binary operation. Any positive integer raised to the power of another positive integer is still in the set of positive integers. Notes Page 1 No, a positive integer minus a positive integer is not necessarily a positive integer because if the second number is larger than the first, you get a negative integer which is not in the set of positive integers. If you have a binary operation, you can switch the order and they will be the same. NOT ALL binary operations are commutative. If a * b is equal to b * a for all elements in the set, then it is commutative. Where there is ONE binary operation done more than once, if you can change the order in which the binary operation is done then it is associative. Not all binary operations are associative. Notes Page 2 The values in the table are given. Not something you have to figure out. The table reads, for example a * a = b 1. Is it a binary operation? Because all results of the binary operation are in the given set (a, b, c) it IS a binary operation. 2. Is it commutative? It is NOT commutative because if you look at a * b = c and b * a = a we see that the outcome is not the same. 3. Is it associative? No it is not associative because (a * b) * c = c * c = a, and a * ( b * c ) = a * b = b and these are not the same. HW: p25 1, 2, 5, 7, 9, 14, 16, 17-20, 24, 29, 36 #24 answers F T F F F T T T T F Answer for #36 (Scratch work, informal proof) Need to show that H is closed under star For every two elements a, b in a set H, you need to show that a * b belongs to set H See the definition of H = in problem 36 Therefor to prove a and b are elements in H, you must show each belongs to S and that they are commutative within S We know that * is an associative operation given by the problem, so show this verbatim using three variables such as a, b, and x Proof. Let c and b be in H, Then by the definition of H we know that c * x = x * c for x that belong to S d * x = x * d for x that belong to S Since (c * d) * x = c * (d * x) = c * (x * d) = (c * x) * d = x * (c * d) Then c * d belongs to H Therefor H is closed under * Answer for #5 Then these must be symmetric. d * b = b * d and so on Notes Page 3

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