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My impression was that $\mathbb{Z_n} = \{0,1,...,n-1\}$ and so to prove this result, you would only have to define an isomorphism from the set $\{0,1,...,n-1\}$ to $G$. However in the proof I was given, the author defines isomorphism $\theta: \mathbb{Z_n} \rightarrow G$ s.t $\theta([x]) = g^x$ where $g$ is generator for cyclic group $G$. He then goes on to show initially that the map is well defined by showing that for any $x' \in [x]$, (Where the equivalence relation must be $x \sim y \iff x \equiv y$ mod $n$) the function is well defined. This confused me because my impression was that $\mathbb{Z_n}$ was only the set $\{0,1,...,n-1\}$ so I can't see why you have to worry about equivalence classes here. Any integer that is not contained in set $\{0,1,...,n-1\}$ is surely irrelevant right?

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    $\begingroup$ Well, the idea is that every integer has some remainder when divided by $n$, and $\mathbb{Z}_n$ is the set of all remainders, which forms a group under addition. This is where the equivalence classes come from, we put all those integers that have the same remainder together in a single class. For example, 1, $n+1$, $2n+1$, ... are all represented by the same element of $\mathbb{Z}_n$, what we call 1 in there. $\endgroup$ – Debmalya Bandyopadhyay Nov 16 '20 at 13:10
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    $\begingroup$ Implicitly they use the map as $\theta_0:\Bbb Z\to G$, then what is happening here is the verification that $\ker\theta_0=n\Bbb Z$. $\endgroup$ – Berci Nov 16 '20 at 13:16
  • $\begingroup$ @DebmalyaBandyopadhyay This confuses me. So are the elements of $\mathbb{Z_n}$ equivalence classes then? $\endgroup$ – Henry Brown Nov 16 '20 at 13:25
  • $\begingroup$ @HenryBrown for all practical purposes you can use $\mathbb{Z}_n$ as the set of integers from 0 to $n-1$. What I said is the motivation where it comes from, and that would help you in understanding the proof you have mentioned. It is interpreted as the set of remainder classes mostly, so that is why the author has mentioned $[x]$. $\endgroup$ – Debmalya Bandyopadhyay Nov 16 '20 at 13:32
  • $\begingroup$ So would it not be sufficient to show that the map given is well defined for just integers 0,1,..., n-1. Would you have to show the map is well defined for all integers by considering equivalence classes? $\endgroup$ – Henry Brown Nov 16 '20 at 13:35
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If $G$ is cyclic of order $n$, then the underlying set is $G=\{g,\dots,g^n(=e)\}$. Consider the map $\varphi\colon G\to \Bbb Z_n$, defined by $g^i\mapsto [i]_n$.

  • Injectivity: $[i]_n=[j]_n\Rightarrow j-i\equiv 0\pmod n\Rightarrow j=kn+i\Rightarrow g^j=g^{kn+i}=(g^n)^kg^i=e^kg^i=g^i$.
  • Surjectivity: by construction.
  • Homomorphism: $\varphi(g^ig^j)=\varphi(g^{i+j\pmod n})=[i+j]_n=[i]_n+[j]_n=\varphi(g^i)+\varphi(g^j)$.

You have to firstly take care about good-definition in case you consider the other way around, namely $\psi\colon \Bbb Z_n\to G$ defined by $[i]_n\mapsto g^i$. In fact, here we are using a class representative, $i$ (yes, the elements of $\Bbb Z_n$ are equivalence classes) to define a map whose argument is an equivalence class as such; so, we want $j\in [i]_n\Rightarrow \psi([j]_n)=\psi([i]_n)$; but this is the case, because $j\in [i]_n\Rightarrow [j]_n=[i]_n$.

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  • $\begingroup$ This is correct, but it doesn't answer the question, which was concerned with confusion over the definition of ${\mathbb Z}_n$. $\endgroup$ – Derek Holt Nov 16 '20 at 14:16

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