# Abstract Algebra flashcards, diagrams and study guides

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## Top 11 sets of about 2,660

#### MATH 151 Final Exam

Let a and b be integers with b>0. Then there exists unique integers q and r with the property that a=bq+r, where 0<=r<b

A positive integer d such that d is a common divisor of a and b and if d' is any other common divisor of a and b, then d' | d. d = gcd(a,b)

the group of all transformations under which the object is invariant with composition as the group operation.

#### Modern Algebra Test 1 (Chap 1 - 4)

A. For any element y as an element of f(A1UA2), there exists an element x element in A1UA2 such that f(x) = y. By the definition of union, x is in A1 or x is in A2. This implies that y is in A1 or y is in A2. Therefore, y is an element of f(A1) or f(A2). Next, let y be an element of f(A1)Uf(A2). Then, y is an element of f(A1) or y is an element of f(A2), so there is an X1 in A1 such that f(X1) = y or there is X2 in A2 such that f(x2) = y. In either case, there is an element x in A1 U A2 such that f(x) = y. Therefore, f(A1)Uf(A2) is a subset of f(A1UA2) and vice versa.

Let y be an element of f(A1NA2). Then there exists a in A1 N A2 such that f(x) = y. Since x is in A1 and x is in A2, we get y in f(A1) N f(A2). ex: Let f: R -> R. f(x) = x^2. Let A1 = {-2, 4} and A2 = {2, 4}. f(A1NA2) = f({4}) = {16}. f(A1) N f(A2) = {4, 16} N {4, 16} or {4, 16}. Because {16} ≠ {4, 16}, the example fails.

Let x ∈ f^-1(B1UB2). Then f(x) ∈ B1 U B2. So f(x) ∈ B1 or f(x) is in B2. Therefore, x ∈ f^-1(B1) or f^-1(B2), which implies that x ∈ in f^-1(B1) U f^-1(B2). Now, let x ∈ f^-1(B1) U f^-1(B2). Therefore f(x) ∈ B1 or B2, which implies that x ∈ f^-1 (B1 U B2).

#### abstract algebra exam 1

a) multiple of an integer b if a = bq for some integer q b) a=bq

Every nonempty set of natural numbers contains a smallest element

Q={(m/n)|m,n ∈ Z and n≠0} Note: m/n and p/q represent the same ratio if mq=np

#### Inverse Matrices

False

CA = I and AC = I, where I is the identity matrix

I (Identity Matrix)

#### Abstract Algebra - Exam 3

G is a group, H c G For all a in G, define aH = {ah | h E H} and Ha = {ha | h E H}. When H is a subgroup of G, say aH is the left coset of H in G containing a. Ha is the right coset. a is the coset representative of aH or Ha

a E aH AH = H iff a E H AH = bH or aH (Union) bH = the empty set AH = bH iff a^(-1)b E H |aH| = |bH| aH = Ha iff H = aHa^-1 aH is a subgroup of G iff a E H

If G is a finite group and H < G, then |H| divides |G|. Order of a subgroup divides the order of the group. Moreover, the number of distinct left or right cosets of HcG is |G| / |H|

#### Abstract: Exam 1

n|a-b

[a]={b∈Z :b≡a (mod n)}

a nonempty set R with 2 binary operation, + and *, that satisfies: closure, associative, and communative in addition, additive identity, closure for multiplication, associative multiplication, distribuitive

#### Abstract Algebra

addition and multiplication of real numbers is (x+y)+z = x + (y+z)

P => Q is logically equivalent to not Q => not P

A type of proof where it is first assumed that the statement is false. If the assumption leads to an impossibility, than the original statement has to be proven true

#### Abstract Algebra (S1-S9)

A function that assigns to each ordered pair of elements of G an element of G.

ab = ba

(ab)c = a(bc)

#### Linear Algebra Terms Valenza 2018

An op * is called associative if (s*t)*u = s*(t*u)

A (possibly infinite) collection of B or vectors is called a basis for V is 1. B is linearly independent 2. B spans V

A function from f: S → T is called bijective if it is both injective and surjective

#### Number Theory

Axiom: every non-empty subset S of positive integers has a smallest element Proof: Suppose there's a non-empty set S of positive integers that cannot be factored into primes. By axiom, there exist a smallest element n in S. n cannot be a prime because prime numbers can be factored into primes => n = a x b for a,b<n. Since a,b is less than n, they cannot be in S, and can be factored into primes => n can also be factored into primes. This is a contradiction, S must be empty

Theorem: Every positive integer can be uniquely, up to ordering of factors, factored as a product of primes. Lemma: if prime integer p is s.t p|ab, p|a or p|b. Proof: if p does not divide a, by fundamental theorem of arithmetic, ax + py = 1. -> (multiply by b ) axb + pyb = b. Since p|(ab)x, p|abx + pyb -> p|b. so, if p|a1a2a3...ap, p|ai for some i Proof of uniqueness: assume there's 2 prime factorization of integer n. n = p1p2p3...pk = q1q2q3...qm. pi and qi are primes. if pi|n, then pi|q1...qk => pi|qj. Because qj is prime, it is only divisible by 1 or itself. So pi = qj.

Proof: Assume primes are in a finite set S. S = p1p2...pk. See that pi doesn't divide (x = p1p2p3...pk + 1).

#### Crypto Chapter 4

A group is defined as: a set of elements, together with an operation performed on pairs of these elements such that: The operation, when given two elements of the set as arguments, always returns an element of the set as its result. It is thus fully defined, and closed over the set. One element of the set is an identity element. Thus, if we call our operation op, there is some element of the set e such that for any other element of the set x, e op x = x op e = x. Every element of the set has an inverse element. If we take any element of the set p, there is another element q such that p op q = q op p = e. The operation is associative. For any three elements of the set, (a op b) op c always equals a op (b op c).

A ring is a set of elements that is closed under two binary operations, addition and subtractions, with the following: the addition operation is a group that is commutative; the multiplication operation is associative and is distributive over the addition operation.

A field is a ring which the multiplication operation is commutative, has no zero divisors and includes an identity element and a inverse element