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

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

Let $$\circ$$ be the mapping from $$\Z \times G$$ to $$G$$ defined as in Product and Index Laws for Monoids: $$\forall n \in \Z: \forall x \in G: n \circ x = *^n x$$.

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 $$x^n$$.


 * VS 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, $$\left({x * y}\right)^n = x^n * y^n = \left({n \circ x}\right) * \left({n \circ y}\right)$$.


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

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

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


 * VS 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.


 * VS 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.