How Does the Isomorphism Between Z_k and Aut(Z_n) Work?

[Artusartos]

I was a bit confused the last paragraph before "Corollary 4.6.4". It says that we have the isomorphism \alpha : Z_k \rightarrow Aut(Z_n) but then says that \alpha(a^j)(b^i)=b^{m^ji}.

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

Thanks in advance.

 

[conquest]

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).

 

[Artusartos]

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.