Theorem:
Every infinite cyclic group is Isomorphic to $\left ( \mathbb{Z},+ \right )$
Proof:
We show first that the map $\phi$ is a homomorphism.
Then, show that $\phi$ is a bijection.
Showing $\phi$ is a homomorphism here is trivial so I'll leave that out.
Let $G=\left \langle g \right \rangle$ be an infinite cyclic group.
As a note, recall that $\left | g \right |=\left | \left \langle g \right \rangle \right |$
Define
$\phi:\mathbb{Z}\rightarrow G$
$\space$ $n \mapsto g^{n}$
$\phi\left ( n \right ) =g^{n}$
Since $G=\left \{ g^{n}\mid n \in \mathbb{Z} \right \}$,
Then, $\space$ $\forall x \in G$
$x=g^{n}$ $\space\ \exists n \in \mathbb{Z}$
Evidently, this means that every element in G is the image of at least one element n in $\mathbb{Z}$. This is equivalent to the definition of onto.
Moving on to showing one-to-one,
a function $\phi$ is one-to-one IFF $\phi\left ( n \right )=\phi\left ( m \right )$ implies n=m
$\forall n,m \in \mathbb{Z}$
Here's my reasoning and I'll like to know if it is valid.
The group G is an infinite cyclic group. So its order is infinite.
This is equivalent to stating that there is an element g in G with infinite order. Then, all power of the element g in G is distinct group elements.
If $\phi\left ( m \right )=\phi\left ( n \right )$ then $\space$ $g^{m}=g^{n}$
But the elements in G are distinct so the only solution is for $m=n$.
Hence, $\phi$ is one-to-one.
So, we have an Isomorphism $\phi$.
Best Answer
Your solution has some strange typos in it, but to answer the actual question ("is the argument for one-to-one valid?"): yes it is, but could be made a bit clearer.
As you note, we want to show that $\phi(m) = \phi(n)$ implies $m = n$. But if $\phi(m) = \phi(n)$ then $g^m = g^n$, so $g^{m-n} = 1$. Since $g$ has infinite order, the only value of $m-n$ for which this is true is $m-n = 0$, i.e. $m = n$.