Every Abelian Group is an Integer Module
Theorem
Let \((G, \ast)\) be an abelian group. Then \(G\) is a module over \(\mathbb{Z}\) with scalar multiplication \(\mathbb{N} \times G \to G\) defined by
\[ (n, g) \mapsto \begin{cases}
\underbrace{g \ast \dots \ast g}_{n \ \text{terms}} & n > 0 \\
\mathrm{id} & n = 0 \\
\underbrace{g^{-1} \ast \dots \ast g^{-1}}_{n \ \text{terms}} & n < 0 \\
\end{cases}\]
where \(g^{-1}\) is the inverse of \(g\) under \(\ast\).
In the context of an additive group, this is consistent with the notation \((n, g) \mapsto ng\) and in a multiplicative group, this is consistent with the notation \((n, g) \mapsto g^n\).
While this result is relatively simple, actually proving it can get rather ugly because of the piecewise nature of the multiplication map.
Proof
Clearly if \(n = 1\) then we have \((1, g) \mapsto g\).
If \(n\) is \(0\), then
\[ n.(m.g) = 0.(m.g) = (0m).g = 0.g = m.(0.g)\]
noting that \(0.g = (1 - 1).g = 1.g - 1.g = 0\) and hence \(m.(0.g) = m.0 = m.(1 - 1) = m.1 - m.1 = 0\), thus the rightmost equality holds. Likewise if \(m = 0\). As such we assume that \(m \neq 0\) and \(n \neq 0\)
Then, letting \(\mathrm{sign}(n) = 1\) for positive integers and \(-1\) for negative integers, we have
\[\begin{align*}
n.(m.g) &= (\underbrace{m.g \ast \dots \ast m.g}_{|n|})^{\mathrm{sign}(n)} \\
&= (\underbrace{(\underbrace{g \ast \dots \ast g}_{|m|})^{\mathrm{sign}(m)} \ast \dots \ast (\underbrace{g \ast \dots \ast g}_{|m|})^{\mathrm{sign}(m)}}_{|n|})^{\mathrm{sign}(n)} \\
&= (\underbrace{g \ast \dots \ast g}_{|mn|})^{\mathrm{sign}(n)\mathrm{sign}(m)} \\
&= (\underbrace{g \ast \dots \ast g}_{|mn|})^{\mathrm{sign}(mn)} \\
&= (nm).g \\
\end{align*}\]
Now, for distributivity consider that
\[\begin{align*}
(n + m).g &= (\underbrace{g \ast \dots \ast g}_{|n + m|})^{\mathrm{sign}(n + m)} \\
\end{align*}\]
If either \(m\) or \(n\) is zero, the result is again trivial. If \(m, n > 0\) or \(m, n < 0\), then the result is clear because \(|n + m| = |n| + |m|\) and \(\mathrm{sign}(n + m) = \mathrm{sign}(n) = \mathrm{sign}(m)\), therefore
\[\begin{align*}
(n + m).g &= (\underbrace{g \ast \dots \ast g}_{|n + m|})^{\mathrm{sign}(n + m)} \\
&= (\underbrace{g \ast \dots \ast g}_{|n|})^{\mathrm{sign}(n)} \ast (\underbrace{g \ast \dots \ast g}_{|m|})^{\mathrm{sign}(m)} \\
&= n.g \ast m.g.
\end{align*}\]
In the case where the signs differ, say \(m > 0\) and \(n < 0\), we still have that
\[ (n + m).g = (\underbrace{g \ast \dots \ast g}_{|n + m|})^{\mathrm{sign}(n + m)} = (\underbrace{g \ast \dots \ast g}_{|n|})^{\mathrm{sign}(n)} \ast (\underbrace{g \ast \dots \ast g}_{|m|})^{\mathrm{sign}(m)}\]
because exactly \(|m + n| = ||m| - |n||\) terms remain after the cancellation of terms from each of the two products, given that \(\mathrm{sign}(m) \neq \mathrm{sign}(n)\).
Likewise, for distributivity across \(\ast\) we have
\[\begin{align*}
n.(g \ast h) &= (\underbrace{(g \ast h) \ast \dots \ast (g \ast h)}_{|n|})^{\mathrm{sign}(n)} \\
&= (\underbrace{(g \ast \dots \ast g)}_{|n|} \ast \underbrace{(h \ast \dots \ast h)}_{|n|})^{\mathrm{sign}(n)} & G\ \text{is abelian} \\
&= (\underbrace{(g \ast \dots \ast g)}_{|n|})^{\mathrm{sign}(n)} \ast (\underbrace{(h \ast \dots \ast h)}_{|n|})^{\mathrm{sign}(n)} \\
&= n.g \ast n.h.
\end{align*}\]
