Solving Abstract Algebra Problem: Proving Isomorphism & Listing Generators

In summary, the conversation is about a subgroup H of GL(2, R) defined by a function phi(n) and how to prove that phi is an isomorphism and list its generators. The person has tried adding two matrices but is not getting the correct results. There is also confusion about whether [1 n] can be a matrix in GL(2, R) since H is a subgroup of it.
  • #1
WM07
3
0
Can some one help me, how to solve this problem?. Please explain me how is done, been having problem with the subject


Let H be the subgroup of GL(2, R) under Matrix multiplication defined by
H = {[ 1 n ]}| n E Z }
0 1


Let 0: Z à H be the function defined by

phi(n) = [ 1 n ]
0 1

How do I prove phi is an isomorphism and how I list the generators

I tried to add the two matrix, but I am getting 0's, I just need explanation on the problem
 
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  • #2
[1 n] is not a matrix in GL(2,R) so H clearly can not be a subgroup of GL(2,R).

Either you're missing something or I'm missing something completely.
 
  • #3


To prove that phi is an isomorphism, we need to show that it is a bijective homomorphism. This means that phi is both one-to-one and onto, and that it preserves the group operation of multiplication.

To show that phi is one-to-one, we can assume that phi(n1) = phi(n2) for some n1, n2 in Z. This means that [1 n1] = [1 n2] and [0 1] = [0 1]. From this, we can see that n1 = n2, which proves that phi is one-to-one.

To show that phi is onto, we need to show that for every element [1 n] in H, there exists an element n' in Z such that phi(n') = [1 n]. This is true since we can simply choose n' = n. Therefore, phi is onto.

To show that phi preserves the group operation, we need to show that phi(n1 + n2) = phi(n1) * phi(n2) for all n1, n2 in Z. This means that [1 n1 + n2] = [1 n1] * [1 n2] and [0 1] = [0 1]. From this, we can see that n1 + n2 = n1 + n2, which proves that phi preserves the group operation.

Now, to list the generators of H, we need to find the elements in H that can generate all other elements by multiplication. This means that we need to find the elements [1 n] in H such that for any [1 m] in H, there exists an integer k such that [1 m] = [1 n]^k. We can see that [1 1] and [1 -1] are generators of H, since any element [1 n] can be written as [1 1]^n or [1 -1]^(-n). Therefore, the generators of H are [1 1] and [1 -1].
 

1. What is abstract algebra?

Abstract algebra is a branch of mathematics that studies algebraic structures such as groups, rings, fields, and vector spaces. It focuses on the general properties of these structures and their relationships rather than specific equations or problems.

2. What is an isomorphism in abstract algebra?

An isomorphism is a mathematical function that preserves the algebraic structure between two objects. In abstract algebra, it is used to show that two algebraic structures are essentially the same, even if they may look different at first glance.

3. How can I prove an isomorphism?

To prove an isomorphism between two algebraic structures, you need to show that there exists a bijective function between them that preserves the algebraic operations. This can be done by showing that the function is both injective (one-to-one) and surjective (onto).

4. What are generators in abstract algebra?

Generators are elements of an algebraic structure that can generate all the other elements in that structure through repeated application of the algebraic operations. In other words, they are the building blocks of the structure and can create all the other elements.

5. How do I list generators for an algebraic structure?

To list generators for an algebraic structure, you need to identify the elements that can generate all the other elements in the structure. This can often be done by trial and error or by understanding the structure's properties and patterns. Some structures may also have specific rules for determining generators.

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