Prove that if $G$ is cyclic and infinite then $G$ is isomorphic to $\mathbb{Z}$ Assume $G$ is generated by $a$, so $G = \langle a\rangle$. Since $G$ is infinite for all $m \in \mathbb{Z}$, $a^m \neq e$.
Suppose $a^h = a^k$ then $a^h\cdot a^{-k} = a^{h-k} = e$, but this is a contradiction.
$\rightarrow$ that if $h \neq k$ then $a^h \neq a^k$ where $h,k \in \mathbb{Z}$ and $a^h,a^k \in G$. 
Now if we have the map $\phi : G \rightarrow \mathbb{Z}$ with $\phi(a^i) = i$ we want to show this is an isomorphism.
Homomorphism: $\phi(a^ia^j) = \phi(a^{i+j}) = i + j = \phi(a^i) + \phi(a^j)$.
One-to-One: Suppose $\phi(a^i) = \phi(a^j)$ then $i = j \rightarrow a^i = a^j$.
Question: How can I show this is onto?
I want to say if $b \in \mathbb{Z}$, then $a^b$ is an element in $G$.
 A: Since $G$ is infinite and cyclic, generated by $a$, then we know that $a^{n}\neq a^{m}$ if and only if $n \neq m$, otherwise $a$ would have finite order (thus $G$ would have finite order). Since this is true, $a^{n}$ is distinct for any integer $n$. So for any $n\in\mathbb{Z}$, we have $a^{n}\in G$, with $\phi(a^{n})=n$. So $\phi$ is onto.
Note that the $\phi$ you constructed is defined by $$\phi(e)=0,$$
$$\phi(a)=1,$$ 
$$\phi(a^{n})=\phi(\underbrace{a\cdots a}_{n\mathrm{-times}})=\underbrace{\phi(a)+\cdots +\phi(a)}_{n\mathrm{-times}}=n\phi(a)=n\cdot 1 = n$$
$$\phi((a^{n})^{-1})=\phi((a^{-1})^{n})=\phi(\underbrace{a^{-1}\cdots a^{-1}}_{n\mathrm{-times}})=\underbrace{\phi(a^{-1})+\cdots +\phi(a^{-1})}_{n\mathrm{-times}}=\underbrace{-\phi(a)+\cdots +(-\phi(a))}_{n\mathrm{-times}}=\underbrace{-(\phi(a)+\cdots +\phi(a))}_{n\mathrm{-times}}=-n\phi(a)=-n\cdot 1 = -n.$$
A: Suppose $m \in \mathbb{Z}$, then $a^m \in G$ and $\phi(a^m)=m$.
You still have to prove $\phi$ is well-defined i.e. to show $\forall x \in G$ there is an unique $k \in \mathbb{Z}$ so that $a^k=x$
