GROUPS Compiled by: Nyasha P. Tarakino (Trockers) +263772978155/+263717267175 ntarakino@gmail.com 30 JANUARY 2020 Tarakino N.P. (Trockers) ~ 0772978155/ 0717267175 Page 1 SYLLABUS (6042) REQUIREMENTS define a binary operation define closure, commutation, association, distribution, identity and inverse element define a group use the basic properties to show that a given structure is, or is not, a group solve problems involving binary operations and properties of a group Tarakino N.P. (Trockers) ~ 0772978155/ 0717267175 Page 2 BINARY OPERATION Definition of a Binary Operation Binary operations on the set are calculations that combine two elements of the set (called operands) to produce another element of the same set. Algebraic Definition If G is a nonempty set, a binary operation on is a function NOTES A binary operation is represented by Binary operations are: . The properties of a binary operation are: Closure, Commutative, Associative, Identity Inverse and Distributive. A binary operation can be represented on a table called Cayley table/ Latin square Modular Arithmetic Definition For any integers, by and , we write to denote the remainder when is divided . So in this case, the answer is the remainder when it is divided by the modulo number. Tarakino N.P. (Trockers) ~ 0772978155/ 0717267175 Page 3 Hence it is a set of remainders/residues when dividing two Integers together. Example remainder Example Find the value of: (i) (ii) (iii) (iv) Solution (i) (ii) (iii) Tarakino N.P. (Trockers) ~ 0772978155/ 0717267175 Page 4 (iv) Properties of a Binary Operation A binary operation has the following properties: Closure, Associativity, Distribution, Commutative, Identity and Inverse. Property 1: Closure A set is said to be closed under a binary operation the result of the binary operation if for any two members from the set, returns a member of the same set. Algebraically we define closure property as follows: If is a set with elements the Example Show that addition is closed under the set of real numbers. Solution Let and . This is true for all real numbers, addition is closed under the set of real numbers. Example If and . Is closed under ? Solution Let Now and one would say it is closed. NOTE: When attempting these particular questions on a given binary definition we have to be very careful before we give a conclusion. To circumvent the problem of giving a false Tarakino N.P. (Trockers) ~ 0772978155/ 0717267175 Page 5 conclusion, we should construct the Cayley table. If there are no new values in the table then the set is said to be closed under a binary ( ) operation, otherwise it is not closed. How to find the elements? Remember that Let . Now Members in blue are not contained in is not closed under Example If and . Is closed under ? Solution Tarakino N.P. (Trockers) ~ 0772978155/ 0717267175 Page 6 How to find the elements? Remember that Let . . Now All values in the table are contained in is closed under Property 2: Commutative A binary operation defined on set of real numbers is commutative if for all Note Common binary operations which are commutative are addition and multiplication . Example (Addition is commutative) (Multiplication is commutative) (Subtraction is not commutative) (Division is not commutative) Example Is commutative for all ? Solution Since addition and multiplication Since are commutative then: is commutative. Note If a binary operation is commutative and when given a table, the elements must reflect each other along the leading diagonal. Example Is the following binary operation commutative? Tarakino N.P. (Trockers) ~ 0772978155/ 0717267175 Page 7 Solution The elements are reflecting along the leading diagonal therefore it is commutative. Example Is the following binary operation commutative? Solution The elements are not reflecting along the leading diagonal therefore it is not commutative. Example Complete the table so that it is commutative? 1 Solution Members should reflect along the leading diagonal. Therefore the table is given by: 1 Property 3: Associative A closed binary operation defined on set of real numbers is associative if for all Tarakino N.P. (Trockers) ~ 0772978155/ 0717267175 Page 8 Note Common binary operations which are associative are addition and multiplication . Example and (Addition is associative) and (Multiplication is associative) and (Subtraction is not associative) and (Division is not associative) Example If , is associative. Solution and . Since is not associative. Example If , is associative. Solution . Since is associative. Property 4: Distributive A closed binary operation and are defined on set of real numbers if : Left Distributive (LD) and : Right Distributive (RD). Then the operator is said to be distributive over the operator . Note Common binary operation multiplication is distributive over addition . Example and (LD). and (RD). Also Tarakino N.P. (Trockers) ~ 0772978155/ 0717267175 Page 9 Since both LD and RD hold the operation of multiplication is distributive over addition. Example Given the binary operation and where and . Is the operation distributive over ? Solution We have to show that and if they agree then it is distributive. and Since it is LD Also and Since it is RD Since both LD and RD hold the operation distributive over . Property 5: Identity Element For a binary operation , if there exist just one unique element then such that: is called an identity element for the operation Common Identity Elements For a binary operation, addition multiplication , the identity element, , the identity element, and for a binary operation, . Example For addition, For multiplication, Tarakino N.P. (Trockers) ~ 0772978155/ 0717267175 Page 10 Identifying/finding the identity element in a Cayley Table Look for where the elements match the outer elements i.e. the row that matches the top upper row in green and the column that matches the outside column in green. The intersection of the two lines (red) gives the identity element. The identity element, . Finding the identity when given a binary definition Example If , where , find the identity element . Solution We know that Now and replacing we have: Also The identity element is . Property 6: Inverse Element An element is called an inverse of under the binary operation if: Tarakino N.P. (Trockers) ~ 0772978155/ 0717267175 Page 11 where is the identity element under the operation Common Inverse Elements For a binary operation, addition Thus the inverse of is , where the identity element, under addition. If we take any real number, say , the inverse is the negative of that particular number e.g. The inverse of is – under addition. ALSO For a binary operation, multiplication Thus the inverse of , where the identity element, is under multiplication. If we take any real number, say , the inverse is the reciprocal of that particular number e.g. The inverse of NB: is . under multiplication does not have an inverse under multiplication. Identifying/finding the identity element in a Cayley Table We have to find the identity element first (as illustrated under identity element section above) Tarakino N.P. (Trockers) ~ 0772978155/ 0717267175 Page 12 Therefore, in this case the identity element . To find the inverse of each element you identify the position where the identity element ( in red) is located in that particular row/column, and that is the inverse. Since we know that thus in our case: , where . THEREFORE: (i) (ii) (iii) Finding the inverse when given a binary definition Example If , where , find , the inverse of . Solution NB: We have to find the identity element first. We know that Now and replacing we have: Tarakino N.P. (Trockers) ~ 0772978155/ 0717267175 Page 13 The identity element is . Now let the inverse be and we know that and in our case GROUP Definition of a Group A group is a non-empty set with a binary operation which satisfies the following properties: Closure, Associativity, Identity and Inverse. Properties of a Group A group has 4 properties which are: Closure, Associativity, Identity and Inverse. The other special property is called Commutative. Property 1: Closure For every elements and where in , the result must also lie in i.e . Example Given that then suppose , . Show that is closed under multiplication modulo . Solution Since therefore that is closed under multiplication modulo . Tarakino N.P. (Trockers) ~ 0772978155/ 0717267175 Page 14 Property 2: Associative For every Example Show that , under multiplication modulo is associative. Solution The Cayley table for is: For Associativity, Let LHS ; RHS Since LHS RHS the associative axiom/property holds. Property 3: Identity element An element such that , for all , is called an identity element for . Example Find the identity element for , under multiplication modulo Solution The Cayley table for is: If you are asked to find the identity element in an exam you must look for the column which is identical to the 1st column (in blue). The number or element on the 1st row of that column is the identity element. OR Look for where the elements match the outer elements i.e. the row Tarakino N.P. (Trockers) ~ 0772978155/ 0717267175 Page 15 that matches the top upper row and the column that matches the outside column. The intersection of that column and the row gives the identity element. The identity element is . Property 4: Inverse Every element has its own inverse. For every in there exists such that . Example Find the inverse element for , under multiplication modulo Solution The Cayley table for is: If you are asked to find the inverse element in an exam you must look for the identity element (in blue) for each element. The number or element on the 1st row of that column or 1st column of that row is the inverse of that element. The inverse of: is (self inverse) is (self inverse) is (self inverse) is (self inverse). HOW TO SHOW THAT A GIVEN SET IS A GROUP? If you are asked to show that something is a group in an exam you must tick off each of the above criteria one-by-one i.e. it has to satisfy all the above properties. If one of the properties is not met then it is not a group. Tarakino N.P. (Trockers) ~ 0772978155/ 0717267175 Page 16 Special Property: Commutative If a group is commutative, then it’s called an Abelian group. Definition A group is abelian if for all elements . NOTE: When using a Latin Square/ Cayley table, a group is Abelian when it is symmetrical along the leading diagonal or when the elements are reflected in the leading diagonal. How to observe symmetry along the leading diagonal? The elements on the leading diagonal are the same OR The elements on the leading diagonal are the symmetric either or How to identify commutation using the leading diagonal? The elements are reflecting along the leading diagonal Tarakino N.P. (Trockers) ~ 0772978155/ 0717267175 Page 17 The order of a group is the number of elements the group contains. If a group contains an infinite number of elements it is said to be of infinite order. SUBGROUPS Definition of a Subgroup If is a nonempty subset of a group , then same operation as . is a subgroup of if is a group under the Properties of a Subgroup If the following three points are met then we say 1. is a nonempty subset of a group of 2. is a group itself 3. and is subgroup use the same binary operation. Types of Subgroups 1. Trivial subgroup - A subgroup includes the subset containing just the identity set 2. Improper subgroup – the group itself since is a subset itself. 3. A proper subgroup is any subgroup with order not one or the same as the original group. The order of an element, say is the smallest positive integer ( ) in which we raise the element ( ) to get the identity element i.e. If there is no such an element then and it is denoted: . has an infinite order. Example Find all the subgroups for the set , under multiplication modulo . Tarakino N.P. (Trockers) ~ 0772978155/ 0717267175 Page 18 Solution The Cayley table for is: NB: A subgroup has to satisfy the following properties: 1. is a nonempty subset of a group of 2. is a group itself 3. and use the same binary operation. For the set (i) the identity (ii) the group , the subgroups are: - trivial subgroup itself - improper subgroup (iii) The proper subgroups are: a) b) c) NB: For each Subgroup we have to check if the closure property holds since it is also a Group. Tarakino N.P. (Trockers) ~ 0772978155/ 0717267175 Page 19 ASSIGNMENT-ZIMSEC QUESTIONS ZIMSEC SPECIMEN QUESTION NOVEMBER 2018 PAPER 2 Let be the set of matrices of the form where . Show that (i) does not form a group under matrix addition. [4] (ii) forms an Abelian group under matrix multiplication. [6] [Assume associativity.] ZIMSEC NOVEMBER 2018 PAPER 2 a)(i) Show that the set of integers under multiplication forms a group. (ii) Write down any three subgroups. b) The set [7] [3] under the operation of composition of functions forms a group H, where Show the operation table for . Tarakino N.P. (Trockers) ~ 0772978155/ 0717267175 [6] Page 20 ZIMSEC JUNE 2019 PAPER 2 Show that the set forms a group under multiplication. [6] ZIMSEC NOVEMBER 2019 PAPER 2 (i) If is a set of all matrices of the form where a group under addition of matrices. (ii) Verify if the set forms a group under addition modulo . Tarakino N.P. (Trockers) ~ 0772978155/ 0717267175 , show that forms [7] [9] Page 21 SOLVED QUESTIONS & ANSWERS Question 1 The set of matrices under matrix multiplication. Construct its Cayley table/Latin square. Suggested Solution The Cayley table for this group is: NB: Question 2 The set of complex numbers forms a group under complex number multiplication. Construct its Cayley table/Latin square. Hence show that is a group and write down all subgroups of . Suggested Solution 1. The Cayley table for this group is: NB: Tarakino N.P. (Trockers) ~ 0772978155/ 0717267175 Page 22 2. a) Checking for closure: Checking in the Cayley table we see that there are no new values i.e all values in set are there. Hence it is closed. b) For Associativity, Let LHS ; RHS Since LHS RHS the associative axiom holds. c) The identity element is . d) The inverse of: is is (self inverse) is (self inverse) is Since all properties hold is a group. 3. For the set (i) the identity , the subgroups are: - trivial subgroup (ii) the group itself - improper subgroup (iii) The proper subgroups are: a) NB This is not a proper subgroup because it does not satisfy closure b) NB This is the only proper subgroup c) NB This is not a proper subgroup because it does not satisfy closure Tarakino N.P. (Trockers) ~ 0772978155/ 0717267175 Page 23 Question 3 (i)Show that the set of numbers , under multiplication , does not form a group. (ii) The set of numbers , under multiplication , forms a group. Write down the value of . Suggested Solution (i) The Cayley table for this set is: NB: then write down the remainder. (Check the cell containing a letter with different colour) The leading diagonal contains 1’s which are new elements not found in the original set. Hence the closure property does not hold and therefore it does not form a group. (ii) The Cayley table for this set is: If , then closure property holds and the set forms a group. Question 4 The set S consists of all non-singular real numbers such that , where . (i) Prove that each matrix A must be of the form . Tarakino N.P. (Trockers) ~ 0772978155/ 0717267175 Page 24 (ii) State clearly the restrictions on the value of such that is in Suggested Solution (i) Let . and Now Comparing corresponding elements (a) (b) (c) (d) (Matrix (ii) Since is non-singular, conforms to the required form). . Question 5 The set S consists of the numbers , where . ( denotes the set of integers ). Prove that the element of S, under multiplication forms a commutative group G. (You may assume that addition of integers is associative and commutative.) Suggested Solution (i) hence it is closed under multiplication. (ii) For Associativity, LHS RHS Since LHS = RHS it is associative. (iii) The identity element is (iv) The inverse element is Tarakino N.P. (Trockers) ~ 0772978155/ 0717267175 Page 25 (v) For commutative property to hold, Let . LHS RHS Since LHS RHS element of S, under multiplication forms a commutative group G. or simply it is commutative. NB: A group which is commutative is called an Abelian group. Question 6 A group G of order 6 has the combination table shown below. (i) State, with a reason, whether or not is commutative. (ii) Write down the inverse of each element Suggested Solution (i) It is not commutative because: (a) the leading diagonal elements (in yellow) are not symmetric, i.e they are instead of . or (b) For commutative property to hold, Tarakino N.P. (Trockers) ~ 0772978155/ 0717267175 Page 26 Let : LHS (in grey) RHS (in green) Since LHS RHS is not commutative (ii) The identity element is NB: To find the identity element you look for a column which is identical to the column outside the table The inverse of: is is is is (self inverse) is (self inverse) is (self inverse). NB: To find the inverse you look for a cell in every row which contains the identity element and in this case the identity is . Question 7 For the group of matrices of the form under matrix addition, where , state the identity element and the inverse of Suggested Solution (i) The identity element is (ii) The inverse of Tarakino N.P. (Trockers) ~ 0772978155/ 0717267175 Page 27 Question 8 consists of the set with the operation of multiplication Write down the operation table and, assuming Associativity, show that . is a group. Suggested Solution (i) The Cayley table for is: a) Checking for closure: Checking in the Cayley table we see that there are no new values i.e all values in set are there. Hence it is closed. b) The associative property holds (from the assumption) OR For Associativity, Let LHS ; RHS Since LHS RHS the associative axiom holds. c) The identity element is . d) The inverse of: is (self inverse) is (self inverse) is (self inverse) is (self inverse). Since all properties hold is a group. Tarakino N.P. (Trockers) ~ 0772978155/ 0717267175 Page 28 Question 9 The group G consists of the set combined under multiplication . Find the inverse of each element. Suggested Solution The Cayley table for set G is given by: a) The identity element is . b) The inverse of: is (self inverse) is is is (self inverse) is (self inverse) is is is (self inverse). Question 10 Elements of the set are combined according to the operation table shown below. Tarakino N.P. (Trockers) ~ 0772978155/ 0717267175 Page 29 (i) Verify that . (ii) Assuming that the associative property holds for all elements, prove that the set , with the operation table shown, forms a group . Suggested Solution (i) LHS . NB: in purple and in green RHS . NB: in yellow and Since LHS = RHS in grey it is associative. (ii) Checking for basic axioms of a group. a) Checking for closure: Checking in the Cayley table we see that there are no new values i.e all values in the given set are there. Hence it is closed. b) The associative property holds (check part (i)) c) The identity element is . d) The inverse of: is is is (self inverse) is (self inverse). is (self inverse). Since all properties hold is a group. Question 10 If is the set of all matrices of the form (i) Prove that :{ and }, forms a group under matrix multiplication. [Assume matrix multiplication is associative] (ii) Determine whether is 1. Abelian, Tarakino N.P. (Trockers) ~ 0772978155/ 0717267175 Page 30 2. closed under matrix addition. Suggested Solution If is the set of all matrices of the form (i) Proving that :{ and }, forms a group under matrix multiplication. [Assuming matrix multiplication is associative] Testing for Closure Let :{ { and and } } is closed under matrix multiplication Testing for Association is associative (from the assumption) Testing for Identity Let the identity element be where { and } has an identity Testing for Inverse Let the inverse element be where { and } Tarakino N.P. (Trockers) ~ 0772978155/ 0717267175 Page 31 has an identity Alternatively, since the identity is we can find the inverse using the normal procedure of finding an inverse of a matrix. Since all four properties/ Axioms hold forms a group under matrix multiplication. (ii) Determining whether is 1. Abelian, Let :{ and { } and } { and } Since and is Abelian. Also 2. closed under matrix addition. Testing for Closure Let :{ and } Tarakino N.P. (Trockers) ~ 0772978155/ 0717267175 Page 32 { and } and also conforms to is closed under matrix addition Question 11 consists of the set of matrices of the form where and are real and , combined under operation of matrix multiplication. (i) Prove that is a group. You may assume that matrix multiplication is associative. (ii) Determine whether (i) is commutative. consists of the set of matrices of the form Checking for closure Let and Now: Also: Hence is closed under matrix multiplication Checking for association is associative under matrix multiplication (Assumption) Checking for identity Let be the identity element of . Since , is invertible Checking for inverse Since is invertible therefore its inverse is given by: Tarakino N.P. (Trockers) ~ 0772978155/ 0717267175 Page 33 Since all the four properties hold is a group. (ii) Checking for commutation Let and Now: Also: Since is commutative under matrix multiplication Tarakino N.P. (Trockers) ~ 0772978155/ 0717267175 Page 34 PRACTISE QUESTIONS Question 1 The function is defined by is defined by for . The function g . (i) Show that and that It is given that . are elements of a group under the operation of composition of functions. The element is the identity, where e : (iii) The inverse of the element for . is denoted by h. Find (iv) Construct the operation table for the elements . of the group . Question 2 Given that under addition . a) Show that i. forms a group, ii. is an Abelian group. b) Find all subgroups of Question 3 (i) The operation is defined by , where and are real numbers and is a real constant. (a) Prove that the set of real numbers, together with the operation , forms a group. (b) State with a reason, whether the group is commutative. (ii) The operation by , where and are positive real numbers. By giving a numerical example in each case, show that two of the basic groups properties are not necessarily satisfied. Tarakino N.P. (Trockers) ~ 0772978155/ 0717267175 Page 35 Question 4 (i) on (ii) If ,a set of rational numbers. Is commutative? and (iii) Find the values of , find such that . if . Question 5 The set M consists of the six matrices forms a group , where . It is given that under matrix multiplication, with numerical addition and multiplication both being carried out (i) Determine whether . is a commutative group, justifying your answer. (ii) Write down the identity element of the group and find the inverse of Question 6 Find all Latin squares of side in standard form with respect to the sequence . For each square found determine whether or not it is the multiplication table of a group. Question 7 The set of polynomials , where and , is denoted by . Assuming the Associativity property holds, prove that , under addition, is a group. Question 8 Show that the Latin Square Tarakino N.P. (Trockers) ~ 0772978155/ 0717267175 Page 36 is not the multiplication table of a group. Question 9 Groups A, B, C and D are defined as follows: A: The set under multiplication modulo B: The set under multiplication modulo C: The set . . under multiplication modulo D: The set . under multiplication. (i) Write the identity element for each of groups A, B, C and D (ii) Prove the closure property for D (iii) Elements of the set are combined under addition. State which of the four basics group properties are not satisfied. (Justification is not required.) Question 10 consist of the set of matrices of the form , where and are real and , combined under the operation of matrix multiplication. (i) Prove that is a group. You may assume that the matrix is associative. (ii) Determine whether is commutative. Question 11 Given that . (a) Show that (i) is a group under multiplication modulo . (ii) is an abelian group. (b) Find all subgroups of . Tarakino N.P. (Trockers) ~ 0772978155/ 0717267175 Page 37 Question 12 (i) If , where , addition . Determine whether forms a group or not. (ii) Find the value of (a) . (b) (iii) If , evaluate . Question 13 (a) Is the binary operation closed on (b) If , find the identity element . where (c) Show that . does not form a group under multiplication . Question 14 Consider the set of all matrices of the form , where is a non-zero rational number. Show that , under operation of matrix multiplication, forms a group, (You may assume that matrix multiplication is associative). Question 15 A group has distinct elements , where is the identity element and is the binary operation. (i) Prove that if and , the set of elements forms a subgroup of . (ii) Prove that if , and , the set elements does not form a subgroup of . Tarakino N.P. (Trockers) ~ 0772978155/ 0717267175 Page 38 Question 16 The group consists of the set multiplication modulo combined under . Find the inverse of each element. Question 17 If , where , multiplication . Show that is not a group. Tarakino N.P. (Trockers) ~ 0772978155/ 0717267175 Page 39 ASANTE SANA Tarakino N.P. (Trockers) ~ 0772978155/ 0717267175 Page 40 *******DONT BE SATISFIED BY MEDIOCRE WHILST EXCELLENCE IS THERE******* CONSTRUCTIVE COMMENTS ON THE FORM OF THE PRESENTATION, INCLUDING ANY OMISSIONS OR ERRORS, ARE WELCOME. ***ENJOY*** Nyasha P. Tarakino (Trockers) +263772978155/+263717267175 ntarakino@gmail.com Tarakino N.P. (Trockers) ~ 0772978155/ 0717267175 Page 41