An isomorphism and functions

  • Thread starter Artusartos
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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.
 

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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).
 
247
0
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.
 

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