Proving a Cyclic group $G$ of order n is isomorphic to $\mathbb{Z_n}$

Question

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?

Answer 1

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 $j\in [i]_n\Rightarrow [j]_n=[i]_n$.