# 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: $G \cong \struct {\Z, +}$.

### Corollary

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

## Proof

From the definition of an infinite cyclic group, we have:

$G = \gen a = \set {a^k: k \in \Z}$

Let us define the mapping:

$\phi: \Z \to G: \map \phi k = a^k$.

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

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

As $G$ is cyclic, every element of $G$ is apower of $a$ for some $a \in G$ such that $G = \gen a$.

Thus:

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

By the definition of $\phi$:

$\map \phi k = a^k = x$

Thus $\phi$ is surjective.

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

This follows directly from Powers of Infinite Order Element, where:

$\forall m, n \in \Z: m \ne n \implies a^m \ne a^n$

Thus $\phi$ is an injective, surjective homomorphism, thus:

$G \cong \struct {\Z, +}$

as required.

$\blacksquare$

## 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 $\struct {\Z, +}$ as an "archetypal" infinite cyclic group.