Could someone please help me correct my following proof?
Prove that if $G$ is a cyclic group of order $n$, then $G$ is isomorphic to $\mathbb{Z}_{n}$.
Proof: Let $G=\left \langle a \right \rangle$ with $|G|=n$. Define the relation $\phi:\mathbb{Z}_{n}\to G$ by $\phi(k)=a^{k}$ with $0\leq k<n$.
Well-defined: For $0\leq k_{1}<n$ and $0\leq k_{2}<n$, if $k_{1}=k_{2}$, then $\phi(k_{1})=a^{k_{1}}=a^{k_{2}}=\phi(k_{2})$.
Operation preserving: Since $\phi(k_{1}+k_{2})=a^{k_{1}+k_{2}}=a^{k_{1}}a^{k_{2}}=\phi(k_{1})\phi(k_{2})$, the operation is preserved.
One-to-one: For $m,n\in \mathbb{Z}_{n}$, let $m>n$ and assume $\phi(m)=\phi(n)$. Then $a^{m}=a^{n}$ or $a^{m-n}=e$. Then $m-n>0$.
I am not sure how to prove that $m-n<0$. Is it due to $G$ being abelian?
Onto: $\phi$ is onto iff for all $a^{k}\in G$, there exists a $k\in \mathbb{Z}_{n}$ such that $\phi(k)=a^{k}$. By definition, $\phi$ is onto.
When you write $m,n\in\mathbb{Z}_n$, the symbol $n$ has two distinct meanings. In order to prove that $\phi$ is ono-to-one, all it take is to prove that $\ker\phi=\{0\}$. But$$\phi(k)=e\iff a^k=e\iff k=0,$$since $\operatorname{ord}a=n$.
And since $\phi$ is one-to-one and both groups have $n$ elements, $\phi$ is surjective too.