Infinite Cyclic Group is Isomorphic to Integers

Theorem
Let $G$ be an infinite cyclic group.

Then $G$ is isomorphic to the additive group of integers: $G \cong \left({\Z, +}\right)$.

Corollary
All infinite cyclic groups are isomorphic.

Or, up to isomorphism, there is only one infinite cyclic group.

Proof
From infinite cyclic group, we have $G = \left \langle {a} \right \rangle = \left\{{a^k: k \in \Z}\right\}$.

Let us define $\phi: \Z \to G: \phi \left({k}\right) = a^k$.

We now show that $\phi$ is an isomorphism.


 * First we show that $\phi$ is a homomorphism.

Let $k, l \in \Z$.

thus proving that $\phi$ is a homomorphism as required.


 * Now we show that $\phi$ is a surjection.

As $G$ is cyclic, every element of $G$ is a power of $a$ (for some $a \in G$ such that $G = \left \langle {a} \right \rangle$).

Thus, $\forall x \in G: \exists k \in \Z: x = a^k$.

By the definition of $\phi$, $\phi \left({k}\right) = a^k = x$.

Thus $\phi$ is surjective.


 * Now we show that $\phi$ is an injection.

This follows directly from Powers of Infinite Order Element, where $\forall m, n \in \Z: m \ne n \Longrightarrow a^m \ne a^n$

Thus $\phi$ is an injective, surjective homomorphism, thus $G \cong \left({\Z, +}\right)$ as required.

Proof of Corollary
Follows directly from the fact that isomorphism is an equivalence relation.

Comment
Now that as we have, in a sense, defined an infinite cyclic group with reference to the additive group of integers that we painstakingly constructed in the definition of integers, it naturally follows that we should use $\left({\Z, +}\right)$ as an "archetypal" infinite cyclic group.