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: $$G \cong \left({\mathbb{Z}, +}\right)$$.

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 $$G = \left \langle {a} \right \rangle = \left\{{a^k: k \in \mathbb{Z}}\right\}$$.

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

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


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

Let $$k, l \in \mathbb{Z}$$.

$$ $$ $$

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, $$\forall x \in G: \exists k \in \mathbb{Z}: x = a^k$$.

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 $$\forall m, n \in \mathbb{Z}: m \ne n \Longrightarrow a^m \ne a^n$$

Thus $$\phi$$ is an injective, surjective homomorphism, thus $$G \cong \left({\mathbb{Z}, +}\right)$$ 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 $$\left({\mathbb{Z}, +}\right)$$ as an "archetypal" infinite cyclic group.