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Math 251W: Foundations of Advanced Mathematics, Spring 2011Portfolio Assignment 10: 5.3, 7.2-3

Name: Sean Fogerty

Problem 5.3.6

proposition: Let A be a non-empty set, and let be an equivalence relation on A. Let x, y A.Show that ifx y, then [x] = [y] and that ifx y, then [x] = [y].

proof (Direct, Contrapositive) Cases

Let a be an arbitrary element of [x] and let x y. By the definition of relation class,x a. By the symmetric definition of a equivalence relation, we have y x. By thetransitive definition of equivalence relations, we have y a. Thus, as a was arbitrarilychosen, [x] [y].

Let a be an arbitrary element of [y] and let x y. By the definition of relation class,y a. By the transitive definition of equivalence relations, we have x a. Thus, as a wasarbitrarily chosen, [y] [x].

Because [y] [x] and [x] [y] they must be equal.

Suppose that the intersection of [x] and [y] is not equal to the null set. Therefore, thereexists an element a in [x] [y]. By the definition of intersection, we have a [x] and a [y].By the definition of relation class, we have x a and y a. By the symmetric property ofequivalence relations, we have a y. By the transitive property of equivalence relations, wehave x y. Thus, by the contrapositive, ifx y, then [x] = [y].

Problem 5.3.14

proposition: Let A be a non-empty set, and let E1 and E2 be equivalence relations on A. LetD1 and D2 denote the partitions ofA that correspond to E1 and E2 respectively. Let E= E1E2.Then E is an equivalence relation on A by Exercise 5.3.7(i). Let D denote the partition ofA thatcorrespond to E. What is the relation between D1,D2 and D? Prove your result.

proof (Direct) Cases

Let the relation between D, D1 and D2 be D = D1 D2. I will prove this by showing that Dand D1 D2 are equal.

Let E = E1 E2 and suppose D = D1 D2. Let a be an arbitrary element of D. By thedefinition of relation, there exists some b A such that (a, b) E. By substitution, we have(a, b) E1 E2. By the definition of intersection, we have (a, b) E1 and (a, b) E2. Bythe definition of relations and because E1 corresponds to D1 and E2 corresponds to D2, we havea D1 and a D2. Therefore, by the definition of intersection, a D1 D2. Because a wasarbitrarily chosen, it follows that D D1 D2.

Let E= E1E2 and suppose D = D1D2. Let a be an arbitrary element ofD1D2. By thedefinition of relations and because E1 corresponds to D1 and E2 corresponds to D2, there existssome b A such that (a, b) E1 E2. By substitution, we have (a, b) E. By the definition ofrelations, we have a D . Because a was arbitrarily chosen, it follows that D D1 D2.

Because D D1 D2 and D D1 D2, they must be equal.

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Problem 7.2.3

proposition: Let A be a set. Define the binary operation on P(A) by XY = (X Y)

(Y X) for all X,Y P(A). Show that (P(A),) is an abelian group.

proof (Direct)

Because the power set contains all subsets ofA, the identity element is the because (X) ( X) = X.

The inverse is the element itself because (X X) (X X) =

We assume the operation is associative.

The operation is trivially commutative, because flipping the variables simply flips the setdifferences across the union.

Because (P(A),) satisfies the identity, inverse, commutative, and associative laws, it must be anabelian group.

Problem 7.2.6

proposition: Let (G, ) be a group. Ifg G, then g has a unique inverse.

proof (Direct) Let g G and let (G, ) be a group. By the definition of a group, there exists

some g( 1) G such that g g( 1) = e(the identity element). Suppose g has two inverses x andy. Therefor, by the definition of the identity element, xg = e and yg = e. Thus, by substitution,we have x g = y g. Additionally, x = x e and y = y e, by the definition of the identity

element. By substitution, we have x = x (g g

1) and x = (x g) g

1, by the associativity ofgroups. By substitution, we have x = (y g) g1. By association and the definition of inverses,we have x = y e, which means that x = y and g has only one unique inverse.

Problem 7.2.12.1

proposition: Let n N. Recall the definition of the set Zn and the operations + and . on Zngiven in section 5.2. Show that (Zn,+) is an abelian group.

proof (Direct) Cases

By the definition of relation classes and modular addition, every element of Zn has an

additive element inZn.

[] is the additive identity element and is included in Zn.

By the definition of set addition, it is associative. Therefore, (Zn,+) is a group.

By the definition of set addition, it is commutative and thus Abelian.

Because (Zn,+) satisfies the identity, inverse, commutative, and associative laws, it must be anabelian group.

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Problem 7.2.12.3

proposition: Recall the definition of the set Zn and the operations + and . on Zn given in section

5.2. Is (Zn

{[0]}, .) a group? If not, can you find any conditions ofn that would guarantee that(Zn {[0]}, .) is a group?

proof (Direct)

No, it is not a group. n would have to be a prime integer.

Problem 7.3.7

proposition: Let G,Hand Kbe groups, and let f : G Hand j : H Kbe homomorphisms.Then j f is a homomorphism.

proof (Direct) Let (G, ), (H,@), and (K, ?) be groups. Let a, b G. By the definition

of homomorphism, we have f(a b) = f(a)@f(b). Thus f(a) H and f(b) H. By thedefinition of homomorphism, we have j(f(a)@f(b)) = j(f(a))?j(f(b)). By substitution, we havej(f(a b)) = j(f(a))?j(f(b)), which is a homomorphism. Thus, j f is a homomorphism.

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