I am trying to prove that for a fixed $n$, all cyclic groups of order $n$ are isomorphic. Here is my attempt.
Let $G_1 = \langle x \rangle$ and $G_2 = \langle y \rangle$ be cyclic groups of order $n$. Define the map \begin{align*} f: \underset{x^k}{G_1} \underset{\longmapsto}{\to} \underset{y^k}{G_2} \end{align*} for any $k \in \mathbb{Z}$. We first show that $f$ is a well-defined map. Given $x^i = x^j$, we have \begin{align*} f(x^i) = y^i, \; f(x^j) = y^j. \end{align*} But since $x^i = x^j$, with $i > j$ without loss of generality, we have $x^{i-j} = e$, so $n \mid i - j$, so we can write $i - j = kn$ for some $k \in \mathbb{Z}$, i.e., $i = j + kn$, so we have \begin{align*} y^i = y^{j + kn} = y^j (y^n)^k = y^j e^k = y^j, \end{align*} so if $x^i = x^j$, then $f(x^i) = f(x^j)$, so $f$ is well-defined. Next, we prove that $f$ is a bijection. Given $g \in G_2$, there exists $m \in \mathbb{Z}$ such that $g = y^m$. But then $f(x^m) = y^m = g$, so $f$ is surjective. Furthermore, if $f(x^i) = f(x^j)$, then $y^i = y^j$. But then $i \equiv j \text{ (mod $n$)}$, so $x^i = x^j$, so $f$ is injective, hence bijective. Finally, we have \begin{align*} f(x^i x^j) = f(x^{i+j}) = y^{i+j} = y^i y^j = f(x^i) f(x^j), \end{align*} so $f$ is a homomorphism, hence $G_1 \cong G_2$.
