Proving the $\mathbb{Z}$-module structure on an abelian group is unique

Question

I'm trying to prove that given any abelian group $G$, we can define on $G$ the structure of a module over $\mathbb{Z}$ and that this structure is unique.

What I tried to do is define an addition and scalar multiplication. Addition is just addition in $G$. Scalar multiplication I defined by $ng$ is $g + \ldots + g$ $n$ times if $n > 0$, $-g + \ldots - g$ if $n < 0$, and $0$ if $n = 0$.

With this definition, I didn't have trouble proving this is a module over $\mathbb{Z}$, but I cannot figure out how to show this structure is unique. I think what this means is that given a group $G$, I can only define addition and scalar multiplication in this way. The addition is basically fixed because that's the addition defined on $G$, but I'm not sure how to prove that no other definition would work.

Think about, for instance, what $2=1+1$ tells you about $2g$. — Eric Wofsey, Oct 17 '23 at 18:20
@EricWofsey Does it tell me something about its order? If $2g = 0$, then $g = -g$. — Valor Vaporeon, Oct 17 '23 at 18:23
See this duplicate: https://math.stackexchange.com/q/4220167/104041 — Shaun, Oct 17 '23 at 18:54
Does this answer your question? Every Abelian group is canonically a $\mathbb{Z}$-module. Is this just a coincidence? — user10354138, Oct 17 '23 at 18:54
I closed then reopened, as I'd got the wrong duplicate. Sorry. — Shaun, Oct 17 '23 at 18:54
It's unique if you require modules to be unital (that is, if you require $1x=x$ for all $x\in M$). If you don't require modules to be unital, then it need not be unique. — Arturo Magidin, Oct 17 '23 at 18:55
Another solution in fancy language: to extend an abelian group structure on $G$ to an $R$-module structure is equivalent to giving a ring homomorphism $R \to \operatorname{End}{\mathsf{Ab}}(G)$. And in the case of $\mathbb{Z}$, that's an initial object of the category of rings, so there is always a unique ring homomorphism $\mathbb{Z} \to \operatorname{End}{\mathsf{Ab}}(G)$. — Daniel Schepler, Oct 17 '23 at 19:14

score 3 · Answer 1 · answered Oct 17 '23 at 19:02

A module $M$ over a ring $R$ with unity is "unital" if and only if $1m=m$ for all $m\in M$.

If you require $\mathbb{Z}$-modules to be unital, then the definition you give is indeed the unique module structure you can put on $G$ (writing $G$ additively).

To see this, we prove by induction on $n$ that $ng$ must be equal to the sum of $n$ copies of $g$, for all $g\in G$. Indeed, we know that $0g=0_G$, and that because the module is unital, $1g=g$. Assuming that we know that $$ng = \underbrace{g+\cdots+g}_{n\text{ summands}}$$ then we have $$\begin{align*} (n+1)g &= ng + g &\text{(module property)}\\ &= \underbrace{g+\cdots+g}_{n\text{ summands}} + g\\ &= \underbrace{g+\cdots+g+g}_{n+1\text{ summands}}. \end{align*}$$ And $(-n)g + ng = (-n+n)g = 0g = 0$, so $(-n)g = -(ng) = n(-g)$ holds.

Thus, the $\mathbb{Z}$-module structure is unique.

If you do not require modules to be unital, then the structure is not unique: define $\cdot\colon \mathbb{Z}\times G\to G$ by $m\cdot g = 0_G$; then this defines a non-unital module structure, and if $G\neq\{0\}$, then it is not isomorphic to standard structure. More generally, if we can express $G$ as a direct sum, $G=\oplus_{i\in I}A_i$, then for each subset $I_0\subseteq I$ we can define a module structure by letting $n$ act on $(g_i)_{i\in I}$ by letting $n\cdot (g_i) = (x_i)$ where $$x_i = \left\{\begin{array}{ll} ng_i &\text{if }i\in I_0\\ 0&\text{if }i\notin I_0, \end{array}\right.$$ This will be a (non-unital if $I_0\neq I$) $\mathbb{Z}$-module structure on $G$. You get the "standard" structure when $I_0=I$, and the zero structure when $I_0=\varnothing$.

Proving the $\mathbb{Z}$-module structure on an abelian group is unique

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