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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 $X$?

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$X \neq \emptyset$ is not necessary. –  Martin Brandenburg Feb 6 at 15:19
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$\;X=\emptyset\;$ and $\;|X|=1\;$ give pretty boring cases... –  DonAntonio Feb 6 at 15:27
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Sure, but it is somewhat misguided to think that the general proofs need extra care in these simple cases. Often people make assumptions excluding trivial cases only because they aren't sure what happens then ... A must-read in that direction is mathoverflow.net/questions/45951/sexy-vacuity –  Martin Brandenburg Feb 6 at 16:12
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5 Answers

up vote 10 down vote accepted
+500

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 $F_1=\mathbf{Z}$.

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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 $X$.

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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.

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Very funny! Actually I am pretty sure that this proof is not circular ... –  Martin Brandenburg Feb 6 at 17:43
    
so how is hurewicz's theorem obvious? –  a more acceptable name Feb 7 at 13:18
    
@amoreacceptablename I didn't say it was obvious, just that it provides an alternative proof. It is elementary though (the idea is to exhibit an inverse $H_1(X) \to \pi_1(X)^{\rm ab}$, mapping any path in $X$ to a loop completing it and showing that this morphism induces a quotient one). –  Pece Feb 7 at 17:04
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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 $X$.

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Are there mistakes in it, or is there another reason for downvoting? Be so kind to explain. –  drhab Feb 6 at 15:35
    
I guess some people don't like categorical explanations, my answer has been downvoted too. –  Najib Idrissi Feb 6 at 15:38
    
It's annoying sometimes to get downvoted but we all must get used to. Many of those downvotes are from frustrated individuals or else from serial downvoters who are bored. Don't you people take that seriously. I'm upvoting both your answers to balance stuff here, but only one question: what is that 'forgetful" functor $\;U\;$ ? $\;V \;$ is clear, though... –  DonAntonio Feb 6 at 15:40
    
@nik Yours is upvoted now and $10-2>8$. :-) –  drhab Feb 6 at 15:40
    
@DonAntonio Thanks. I say 'forgetful' as some custom. In fact it is exactly the inclusion functor here for Ab as a subcategory of Grp. You could say that nothing is 'forgotten', what might explain your question. –  drhab Feb 6 at 15:43
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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)

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

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