# Show that the free abelian group is a group.

Let S be a set and let $$F\langle S\rangle = \{\phi : S \to \mathbb{Z}\mid \phi(x) = 0\ \text{ for all but finitely many } x \in S\}$$.

Show that $$F\langle S\rangle$$ is an abelian group w.r.t. the usual addition of maps (i.e., $$(f +g)(x) = f(x)+g(x)$$). We call $$F\langle S\rangle$$ the free abelian group generated by $$S$$.

So I have to show that the addition is associative, closed/well-defined. Also $$F\langle S\rangle$$ must contain a neutral element and inverses. So for the neutral element I thought that its the kernel since we have some elements which are mapped to 0. For the inverse element/function I tried to say that φ is bijective so an inverse exists. But nearly all $$x$$ are mapped to 0, so φ is not injektive maybe I miss here something:/ it would be nice to give me some help regarding the group operation (how to show it) and the inverse.

• Your comment "for the neutral element I thought that its the kernel" doesn't make sense. The kernel of what? The neutral element has to be a specific map $\phi$. The maps $\phi$ are not normally bijective. $S$ is an arbitrary set so how could they be? Remember that the group operation is addition, not composition. Commented Apr 30, 2022 at 12:26
• Maybe for the addition: since we only have finite many x which are non-zero there addition is also non-zero. For the other x the addition is Zero. Still there is the case when in g x is mapped to a non-zero value and f maps x to 0 then f+g (x) =g(x) Commented Apr 30, 2022 at 14:54
• Yes you can show closure that way. But you have to find the neutral element. Commented Apr 30, 2022 at 15:05
• Can you give me some advice? Commented Apr 30, 2022 at 18:41
• None of them. But you still haven't proved that they hold under the given assumptions. Commented May 3, 2022 at 21:45

For any set $$S$$ and group $$G$$, the set $$\operatorname{Maps}(S,G):=\{f\mid f\colon S\to G\}$$, endowed with the pointwise multiplication "$$*$$": $$(f*g)(x):=f(x)g(x)\tag 1$$ is a group. In fact:
1. closure: for every $$f,g\in\operatorname{Maps}(S,G)$$, $$f*g\in\operatorname{Maps}(S,G)$$ by definition $$(1)$$;
2. associativity: for every $$x\in S$$, $$(f*(g*h))(x)=$$ $$f(x)((g*h)(x))=$$ $$f(x)(g(x)h(x))=$$ $$(f(x)g(x))h(x)=$$ $$((f*g)(x))h(x)=$$ $$((f*g)*h)(x)$$, whence $$f*(g*h)=(f*g)*h$$;
3. identity: $$\exists e\in \operatorname{Maps}(S,G)$$, defined by $$e(x)=1_G$$ for every $$x\in S$$; then, for every $$f\in\operatorname{Maps}(S,G)$$ and $$x\in S$$, $$(f*e)(x)=$$ $$f(x)e(x)=$$ $$f(x)1_G=$$ $$f(x)$$, whence $$f*e=f$$ for every $$f\in\operatorname{Maps}(S,G)$$ and hence $$e$$ plays as identity of $$\operatorname{Maps}(S,G)$$;
4. inverse element: for every $$f\in\operatorname{Maps}(S,G)$$, $$\exists \tilde f\in \operatorname{Maps}(S,G)$$, defined by $$\tilde f(x)=f(x)^{-1}$$ for every $$x\in S$$; then, for every $$f\in\operatorname{Maps}(S,G)$$ and $$x\in S$$, $$(f*\tilde f)(x)=$$ $$f(x)\tilde f(x)=$$ $$f(x)f(x)^{-1}=$$ $$1_G$$, whence $$f*\tilde f=e$$ for every $$f\in\operatorname{Maps}(S,G)$$ and hence $$\tilde f$$ plays as inverse of $$f$$.
Now, the subset $${\rm FS}\subseteq\operatorname{Maps}(S,G)$$, which comprises all and only the maps $$f$$ "with finite support" (i.e. such that $$X_f:=\{x\in S\mid f(x)\ne 1_G\}$$ is finite) is indeed a subgroup of $$\operatorname{Maps}(S,G)$$. In fact, for every $$f,g\in {\rm FS}$$ (one-step subgroup test; note that $${\rm FS}\ne\emptyset$$, since $$e\in {\rm FS}$$ because $$X_e=\emptyset$$ is finite): \begin{alignat}{1} X_{f*\tilde g} &= \{x\in S\mid (f*\tilde g)(x)\ne 1_G\} \\ &= \{x\in S\mid f(x)\tilde g(x)\ne 1_G\} \\ &= \{x\in S\mid f(x)g(x)^{-1}\ne 1_G\} \\ &= \{x\in S\mid f(x)\ne g(x)\} \\ &\subseteq X_f\cup X_g \end{alignat} which is finite because $$X_f$$ and $$X_g$$ are finite. Therefore $$f*\tilde g\in {\rm FS}$$ for every $$f,g\in {\rm FS}$$, and hence $${\rm FS}\le\operatorname{Maps}(S,G)$$.
Your case is just the special one $$G=(\Bbb Z,+)$$ and "$$*=+$$", because then $${\rm FS}$$ boils down to your $$F\langle S\rangle$$. Finally, $${\rm FS}$$ is Abelian if $$G$$ is such, as in your case.