Show that a group of order $n$ is isomorphic to $\mathbb{Z}/{n \mathbb{Z} }$ if and only if it contains an element of order $n.$
How should I define a function from $G$ to$\mathbb{Z}/{n \mathbb{Z} }$ to show that it's a bijective homomorphism. We know that elements of $\mathbb{Z}/{n \mathbb{Z} }$ are $[0]_n,[1]_n, \cdots , [n-1]_n$ and it's an (abelian) group with operation $+:~[a]_n + [b]_n=[a+b]_n.$ Any help is much appreciated.
On
I think what you're looking for is:
Given $(G,\cdot)$ a cyclic group of order $n$, we can take a primitive $g$ such that $G = (g, ... , g^n)$ and then define the following isomorphism:
$I: G \rightarrow \mathbb{Z}/n\mathbb{Z} \\ g^i \mapsto [i]_n$.
This way, it is quite clear that if $g^j = g^i \cdot g^k = g^{i+j}$, $I(g^i) + I(g^k) = [i]_n + [k]_n = [j]_n = I(g^j) = I(g^{i+k}) = I(g^i\cdot g^k)$
Also, $|G| = |\mathbb{Z}/n\mathbb{Z}|$ so it's enough to see that the mapping is injective: if $1 \leq i,k \leq n,\\ I(g^i) = I(g^k) \iff [i]_n = [k]_n \iff i = k \iff g^i = g^k$
Let $f$ send $x$, the element of order $n$ to $[1]$. Then $f(x^n) = [n] = [0] = f(0)$, so it's well-defined; $f(x^m)\neq [0]$ for all $m< n$, so it is injective; surjectivity is just as clear.