Z-Module Associated with Abelian Group is Unitary Z-Module

Theorem
Let $\left({G, *}\right)$ be an abelian group.

Let $\left({\Z, +, \times}\right)$ be the ring of integers.

Let $\circ$ be the mapping from $\Z \times G$ to $G$ defined as:
 * $\forall n \in \Z: \forall x \in G: n \circ x = *^n x$

where $*^n x$ is defined as in Index Laws for Monoids:
 * $*^n x = x * x * \ldots (n) \ldots * x$

Then $\left({G, *, \circ}\right)_\Z$ is a unitary $\Z$-module.

This is called the $\Z$-module associated with $G$.

Proof
The notation $*^n x$ can be written as $x^n$.


 * Module (1): We need to show that $n \circ \left({x * y}\right) = \left({n \circ x}\right) * \left({n \circ y}\right)$.

From the definition, $n \circ x = x^n$ and so $n \circ \left({x * y}\right) = \left({x * y}\right)^n$

From Powers of Elements in Abelian Groups, $\left({x * y}\right)^n = x^n * y^n = \left({n \circ x}\right) * \left({n \circ y}\right)$.


 * Module (2): We need to show that $\left({n + m}\right) \circ x = \left({n \circ x}\right) * \left({m \circ x}\right)$.

That is, that $x^{n + m} = x^n * x^m$.

This follows directly from Powers of Group Elements; no further comment required.


 * Module (3): We need to show that $\left({n \times m}\right) \circ x = n \circ \left({m \circ x}\right)$.

That is, that $x^{nm} = \left({x^m}\right)^n$.

This also follows directly from Powers of Group Elements; no further comment required.


 * Module (4): We need to show that $\forall x \in G: 1 \circ x = x$.

That is, that $x^1 = x$, and of course by Powers of Group Elements it does.