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.

Thanks in advance!

## Best Answer

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:

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.