I have to prove that the groups $(\mathbb{Z_n}, +)$ and $(U_\mathbb{n}, \cdot)$ are isomorphic, where $\mathbb{Z}_n$ is the set of residue classes modulo $n$:
$$\mathbb{Z_n} = \{\hat{0}, \hat{1}, …, \widehat{n – 1} \}$$
and $(U_\mathbb{n}, \cdot)$ is the set of $n$-th roots of unity:
$$U_\mathbb{n} = \bigg { \{ } \cos \frac{2 \pi k}{n} + i \sin \frac{2 \pi k}{n} \bigg{|} \hspace{.1cm} k = 0, 1, …, n – 1 \bigg {\} }$$
Correct me if I'm wrong, but I think we could also declare the elements of $U_n$ to be
$$e^{\frac{2 \pi k i}{n}}$$
for $k = 0, 1, …, n – 1$.
Anyway, so that's what I have to prove. I know that in order to prove that these two groups are isomorphic, I have to find a bijective function $f: \mathbb{Z_n} \rightarrow U_\mathbb{n}$ that has the property:
$$f(x + y) = f(x) \cdot f(y)$$
$\forall \hspace{.1cm} x, y \in \mathbb{Z_n}$. But I cannot come up with any such function. These sets look rather complicated and not only do I have to find a function that has the above property, but I also have to make sure it's bijective. I don't know in which direction to look and how I should search such a function.
I know that this exact same problem has been asked before (e.g., here) but I really didn't understand that answer, either the function or how it was found.
Best Answer
Hint: It suffices to prove that they are both cyclic of order $n$, as there's only one cyclic group of given order (up to isomorphism).