How does one prove that if $X$ is a set, then the abelianization of the free group $FX$ on $X$ is the free abelian group on $X$?
Abstract Algebra – Abelianization of Free Group is the Free Abelian Group
abstract-algebracategory-theoryfree-groupsgroup-theory
Related Solutions
An abelian group is the same thing as a module over the ring $\mathbb{Z}$ (think about it). The ring $\mathbb{Z}$ is a PID, thus submodules of free $\mathbb{Z}$-modules are free. Reformulated in the context of abelian groups, submodules of free abelian groups are free abelian. You can use this fact to show that every abelian group has a length 2 free resolution (this works over any PID $R$, in particular $R = \mathbb{Z}$):
Let $P_0 = \bigoplus_{m \in M} R_m$ be a direct sum of copies of $R$, one for each element of $M$ (the index is just here for bookkeeping reasons). This is a free $R$-module. This maps to $M$ through $\varepsilon : P_0 \to M$ by defining $\varepsilon_m : R_m \to M$, $x \mapsto x \cdot m$ and extending to the direct sum (coproduct).
The kernel $P_1 = \ker(P_0 \to M)$ is a submodule of the free module $P_0$, hence it is free as $R$ is a PID. Thus you get a free resolution (exact sequence): $$0 \to P_1 \to P_0 \to M \to 0.$$
In the above proof, notice that "free $\mathbb{Z}$-module" is also the same thing as "free abelian group", so everything works out. You can also choose a set of generators for your module $M$, but I feel it's cleaner by just taking every element of $M$ and be done with it.
As Bernard mentions in the comments, for finitely generated abelian groups it's easy to see from the structure theorem (e.g. $\mathbb{Z}/n\mathbb{Z}$ has the free resolution $0 \to \mathbb{Z} \xrightarrow{\cdot n} \mathbb{Z} \to \mathbb{Z}/n\mathbb{Z} \to 0$).
Other people have already commented on the difference between resolutions of abelian groups and groups in general. Surprisingly enough, subgroups of free groups are free too by the Nielsen–Schreier theorem, and the standard proof even uses algebraic topology! It all comes around. So you can directly adapt the above argument to show that every (not necessarily abelian) group has a free resolution of length at most two:
Let $G$ be a group. Let $P_0 = \bigstar_{g \in G} \mathbb{Z}_g$ be the free product of copies of $\mathbb{Z}$, one for each element of $g$. This is a free group. By the universal property of free groups, this maps to $G$ by sending $1 \in \mathbb{Z}_g$ to $g \in G$. The kernel $P_1 = \ker(P_0 \to G)$ is a subgroup of a free group, thus it is free itself, and you get a free resolution of $G$ of length at most 2: $$0 \to P_1 \to P_0 \to G \to 0.$$
In the above proof, $\mathbb{Z}$ appears too, but for different reasons: it is the free group on one generator. It also happens to be the free abelian group on one generator, but that's not its role in the proof above. The groups $P_0$ and $P_1$ that appear in the proof are, in general, not free abelian.
As in the comments: free (abelian group) is not the same as abelian (free group). The order/grouping of words cannot be interchanged, as it happens, (though in a different universe it might have been otherwise), because most free groups are not abelian (that is "abelian (free group)" refers to only one non-trivial thing, $\mathbb Z$), while there are many free (abelian groups).
In both cases the constructions given are typical, but do not clearly explain the function of the things constructed. A simple and standard categorical characterization succeeds in characterizing both: given a category $C$ (of either all groups, or just abelian groups, or possibly some other types of things), so that every object $X$ in $C$ has an underlying set ${\mathrm set}X$ (for example, $X$ is a set plus structure), a free object $F(S)$ in $C$ on a set $S$ is an object in $C$ such that every map of sets $f:S\to {\mathrm set}X$ gives a unique $C$-morphism $F(S)\to X$ (restricting to $f$ on $S$...) We can check that the two constructions you mention succeed in exhibiting free objects in the two categories.
The free group $F(S)$ on a set $S$ does map to the free abelian group $FA(S)$, induced from the identity map $f:S\to S$... so the free abelian group is a quotient of the free group. You can check that the kernel is generated by commutators, unsurprisingly.
EDIT: it might be useful to add another operation here, "abelianization" of a group $G$, which creates the largest quotient ${\mathrm ab}G$ of $G$, with a quotient map $G\to {\mathrm ab}G$. (No, it's not a-priori clear that there is a unique such, etc...) The map from free group on $S$ to free abelian group on $S$ is exactly the abelianization map.
Best Answer
Here is an algebraico-topological proof, using :
The fundamental group of $\bigvee_{s \in S} \mathbb S^1$ is the free group on the set $S$ (using Van Kampen for example). The $1$-homology group of $\bigvee_{s \in S} \mathbb S^1$ is the free $\mathbb Z$-module on $S$ (using Mayer-Vietoris, or another long exact sequence-wise proof). So Hurewicz's theorem concludes.