Exploring the Isomorphism Between Z_k and Aut(Z_n)

In summary, the conversation discusses the isomorphism \alpha : Z_k \rightarrow Aut(Z_n) and how it functions as a map from elements in Z_k to automorphisms of Z_n. The formula for this isomorphism is given, and it is noted that keeping track of where elements are going is important.
  • #1
Artusartos
247
0
I was a bit confused the last paragraph before "Corollary 4.6.4". It says that we have the isomorphism [itex]\alpha : Z_k \rightarrow Aut(Z_n)[/itex] but then says that [itex]\alpha(a^j)(b^i)=b^{m^ji}[/itex].

In a regular function [itex]f: X \rightarrow Y[/itex], we take one element from X and end up with an element in Y, right? But in this isomorphism, we take two elements [itex]a^j[/itex] and [itex]b^i[/itex] (and b^i is not even necessarily in Z_k). and end up with an element in [itex]Z^x_n[/itex]. So how can that happen with a function?

Thanks in advance.
 

Attachments

  • Steinberger 112.jpg
    Steinberger 112.jpg
    59.4 KB · Views: 403
  • Steinberger 112 (2).jpg
    Steinberger 112 (2).jpg
    58.5 KB · Views: 425
Last edited:
Physics news on Phys.org
  • #2
Hello,

Your alpha function sends elements from Z_k to the automorphisms of Z_n. This means that for every element a of Z_k alpha applied to a is a function from Z_n to Z_n. So alpha is an isomorphism and the formula gives you a description of the image of a^j under this isomorphism namely exactly that function that sends b^i to B^(bladiebla). The only obstacle here is to keep track of where elements are going alpha is a perfectly fine map and so is alpha(a).
 
  • #3
conquest said:
Hello,

Your alpha function sends elements from Z_k to the automorphisms of Z_n. This means that for every element a of Z_k alpha applied to a is a function from Z_n to Z_n. So alpha is an isomorphism and the formula gives you a description of the image of a^j under this isomorphism namely exactly that function that sends b^i to B^(bladiebla).


The only obstacle here is to keep track of where elements are going alpha is a perfectly fine map and so is alpha(a).

Thank you.
 

1. What is an isomorphism?

An isomorphism is a mathematical concept that describes a one-to-one correspondence between two mathematical structures. It essentially means that the two structures are structurally identical and can be transformed into each other without losing any information.

2. What is the difference between an isomorphism and a homomorphism?

While both isomorphisms and homomorphisms describe relationships between mathematical structures, an isomorphism is a one-to-one correspondence that preserves all mathematical operations, while a homomorphism may only preserve some operations.

3. How do you determine if two functions are isomorphic?

To determine if two functions are isomorphic, you must check if they are both one-to-one and onto (or bijective). This means that every element in the domain of one function has a corresponding element in the codomain, and vice versa.

4. What is the significance of an isomorphism in mathematics?

Isomorphisms are important in mathematics because they allow us to identify and study the structural similarities between different mathematical structures. This can lead to a deeper understanding of these structures and the relationships between them.

5. Can you give an example of an isomorphism?

One example of an isomorphism is the relationship between the group of rotations in 3D space and the group of complex numbers with absolute value 1. Both structures have the same algebraic properties and can be transformed into each other without losing any information.

Similar threads

  • Linear and Abstract Algebra
Replies
1
Views
933
  • Linear and Abstract Algebra
Replies
15
Views
3K
  • Linear and Abstract Algebra
2
Replies
52
Views
2K
  • Linear and Abstract Algebra
Replies
8
Views
1K
  • Linear and Abstract Algebra
Replies
2
Views
2K
  • Linear and Abstract Algebra
Replies
11
Views
2K
  • Linear and Abstract Algebra
Replies
7
Views
3K
Replies
3
Views
1K
  • Linear and Abstract Algebra
Replies
1
Views
1K
  • Linear and Abstract Algebra
Replies
5
Views
2K
Back
Top