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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.
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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