Abelianization of free group is the free abelian group

Question

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 $X$?

Answer 1

Since both the free group and abelianization are left adjoint (and hence preserve coproduts) we shall have up to canonical isomorphism $(FX)^{\mathrm{ab}}=(F\coprod_{x\in X}\left\lbrace x\right\rbrace)^\mathrm{ab}=(\ast_{x\in X}F_1)^{\mathrm{ab}}=\bigoplus_{x\in X}\mathbf{Z}$, since $F_1=\mathbf{Z}$.

Answer 2

By means of universal property:

• For a set $X$, for every set map $X \to G$ to a group, there exists a unique morphism $FX \to G$ extending the map.
• For a group $H$, for every morphism $H \to A$ to an abelian group, there exists a unique morphism $H_{ab} \to A$ that factors the map from $H$.

Combining these two, you see that for any set map $X \to A$ to an abelian group, there exists a unique morphism $(FX)_{ab} \to A$ that factors the map. This is precisely the "universal property" definition of the free abelian group.

---

Another proof: $FX$ is presented as $\langle X : \emptyset \rangle$, so its abelianization is presented by $\langle X : xy = yx \forall x,y \in X \rangle$ (standard fact about abelianization). This clearly coincide with the free abelian group on $X$.

Answer 3

Another approach:

Try the following cute

Lemma: An element in $\;F(X)\;$, written as a word $\;w(x)=x_i^{a_i}\cdot\ldots\cdot x_j^{a_j}\;,\;\;x_k\in X\;,\;\;a_k\in\Bbb Z$ belongs to the commutator group $\;F(X)'\;$ iff the exponent sum of each element in $\;X\;$ is zero, meaning: if we denote by

$$\;E_x^w:=\{n\in\Bbb Z\;;\;n\;\;\text{appears as exponent of the letter}\;\;x_i\;\;\text{in the word}\;\;w\}\;$$

then

$$\sum_{n\in E^w_x}n=0\;\;,\;\;\forall\,x\;\;\text{in the word}\;\;w$$

For example, the element $\;x^2yx^{-1}zx^{-1}y^2z^{-1}y^{-3}\;$ belongs to $\;F(x,y,z)'\;$ , but $\;xzy^{-2}x^{-3}\;$ doesn't.

Letting $\;F(X)\,,\,A(X)\;$ be the free (free abelian) group on the set $\;X\;$ , resp. , define

$$f:X\to A(X)\;,\;\;f(x):=x$$

By the universal property of $\;F(X)\;$ there exists a unique homomorphism $\;\phi: F(X)\to A(X)\;$ extending $\;f\;$ , and if $\;w(x)\;$ is a word in the "letters" (elements) of $\;X\;$ , we get that (writing all multiplicatively)

$$1=\phi(w(x)) = w(f(x)) \iff \;\text{ the exponent sum of each letter in the word $\;w\;$ is zero}$$

Answer 4

On a categorical tour:

There is the inclusion-functor $U:\mathbf{Ab}\rightarrow\mathbf{Grp}$ and forgetful functor $V:\mathbf{Grp}\rightarrow\mathbf{Set}$ and you are interested in the left-adjoint of composition $V\circ U:\mathbf{Ab}\rightarrow\mathbf{Set}$. This composition makes it possible to construct the free abelian group over set $X$ in two steps. Firstly, construct a group free over set $X$. Secondly, construct an abelian group free over the group constructed in the first step. In fact if $F:\mathbf{Set}\rightarrow\mathbf{Grp}$ can be marked as left-adjoint of $V$ and $K:\mathbf{Grp}\rightarrow\mathbf{Ab}$ as left-adjoint of $U$ then $K\circ F:\mathbf{Set}\rightarrow\mathbf{Ab}$ can be marked as left-adjoint of $V\circ U$. This is a general statement and is nice to know. If $G$ is an object in $\mathbf{Grp}$ then the $K\left(G\right):=G/\left[G,G\right]$ is an object in $\mathbf{Ab}$ free over $G$. For $G=F\left(X\right)$ we will get $F\left(X\right)/\left[F\left(X\right),F\left(X\right)\right]$ as free abelian group over set $X$.

Answer 5

Here is an algebraico-topological proof, using :

Hurewicz's theorem. For a topological space $X$, the natural morphism $$ \pi_1(X)^{\rm ab} \to H_1(X) $$ is an isomorphism.

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.