Every Abelian Group is an Integer Module

Theorem

Let $(G, *)$ be an abelian group. Then $G$ is a module over $Z$ with scalar multiplication $N \times G \to G$ defined by

(n, g) \mapsto {\begin{cases} \underset{n terms}{\underset{⏟}{g * \dots * g}} & n > 0 \\ id & n = 0 \\ \underset{n terms}{\underset{⏟}{g^{- 1} * \dots * g^{- 1}}} & n < 0 \end{cases}

where $g^{- 1}$ is the inverse of $g$ under $*$ .

In the context of an additive group, this is consistent with the notation $(n, g) \mapsto n g$ 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) = (0 m) . 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 $sign (n) = 1$ for positive integers and $- 1$ for negative integers, we have

\begin{aligned} n . (m . g) & = (\underset{| n |}{\underset{⏟}{m . g * \dots * m . g}})^{sign (n)} \\ = (\underset{| n |}{\underset{⏟}{(\underset{| m |}{\underset{⏟}{g * \dots * g}})^{sign (m)} * \dots * (\underset{| m |}{\underset{⏟}{g * \dots * g}})^{sign (m)}}})^{sign (n)} \\ = (\underset{| m n |}{\underset{⏟}{g * \dots * g}})^{sign (n) sign (m)} \\ = (\underset{| m n |}{\underset{⏟}{g * \dots * g}})^{sign (m n)} \\ = (n m) . g \end{aligned}

Now, for distributivity consider that

\begin{aligned} (n + m) . g & = (\underset{| n + m |}{\underset{⏟}{g * \dots * g}})^{sign (n + m)} \end{aligned}

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 $sign (n + m) = sign (n) = sign (m)$ , therefore

\begin{aligned} (n + m) . g & = (\underset{| n + m |}{\underset{⏟}{g * \dots * g}})^{sign (n + m)} \\ = (\underset{| n |}{\underset{⏟}{g * \dots * g}})^{sign (n)} * (\underset{| m |}{\underset{⏟}{g * \dots * g}})^{sign (m)} \\ = n . g * m . g . \end{aligned}

In the case where the signs differ, say $m > 0$ and $n < 0$ , we still have that

(n + m) . g = (\underset{| n + m |}{\underset{⏟}{g * \dots * g}})^{sign (n + m)} = (\underset{| n |}{\underset{⏟}{g * \dots * g}})^{sign (n)} * (\underset{| m |}{\underset{⏟}{g * \dots * g}})^{sign (m)}

because exactly $| m + n | = | | m | - | n | |$ terms remain after the cancellation of terms from each of the two products, given that $sign (m) \neq sign (n)$ .

Likewise, for distributivity across $*$ we have

\begin{aligned} n . (g * h) & = (\underset{| n |}{\underset{⏟}{(g * h) * \dots * (g * h)}})^{sign (n)} \\ = (\underset{| n |}{\underset{⏟}{(g * \dots * g)}} * \underset{| n |}{\underset{⏟}{(h * \dots * h)}})^{sign (n)} & G is abelian \\ = (\underset{| n |}{\underset{⏟}{(g * \dots * g)}})^{sign (n)} * (\underset{| n |}{\underset{⏟}{(h * \dots * h)}})^{sign (n)} \\ = n . g * n . h . \end{aligned}