Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

An isomorphism and functions

  1. Mar 17, 2013 #1
    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.
     

    Attached Files:

    Last edited: Mar 17, 2013
  2. jcsd
  3. Mar 20, 2013 #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).
     
  4. Mar 23, 2013 #3
    Thank you.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: An isomorphism and functions
  1. Are they isomorphic? (Replies: 4)

  2. Isomorphic Subfields (Replies: 1)

  3. Isomorphic groups (Replies: 12)

Loading...