4 definitions by qsqazxcvfrew

abelian

In abstract algebra, an abelian group is a group that is commutative.

The group of intergers modulo p excluding zero where p is a prime is said to be an abelian group under multiplication.

by qsqazxcvfrew March 29, 2018

injective

An injection is a map f: A -> B such that
f(a)=f(b) => a=b for all a,b in A

The map f: R -> R given by f(x)=ax+b is injective.

Proof:

f(c)=f(d) <=> ac+b = ad+b <=> c=d

Q. E. D.

by qsqazxcvfrew March 29, 2018

Group

Abstract Algebra: If G is a nonmpty set that is equipped with a binary relation * that satisfies the following axioms:

1. Closure
2. Associativity
3. There is an element e in G s.t. a*e=a=e*a for all a in G
4. For each a in G, there is an element d in G such that a*d=e and d*a=e

Then G is a group under the operation *

The set of integers is a group under addition.

The set of permutations in S_n is a group under composition.
The subset from the complex numbers {1, -1, i, -1} is a group under multiplication.

by qsqazxcvfrew March 29, 2018

isomorphism

Abstract algebra: Take a group G, and a group H, with a group operation *. G is isomorphic to H if there exist a map f: G->H such that:

1: f is injective
2: f is surjective
3: f(a*b)=f(a)*f(b) for all a,b in G

If f satisfies these three properties, f is called an isomorphism.

The map f: Z -> E given by f(a)=2a where Z is the integers and E is the even integers is an isomorphism.

Proof:
Showing injectivity

f(b)=f(a) => 2a=2b (from the given function) <=> a=b

Showing surjectivity
Suppose n is in E. n is an even integer hence n=2k for some integer k.

f(k)=2k=n, hence f is surjective.

Homomorphism:
f(a+b)=2(a+b)=2a+2b=f(a)+f(b)

Hence f is an isomorphism. Q. E. D.

by qsqazxcvfrew March 29, 2018