# Construction of free abelian group from free group

I am reading Fraleigh's Abstract Albebra recently, and I cannot prove a statement about free abelian group:

1. Let $F[A]$ be a free group generated by set $A$ and $C$ is the commutator subgroup of $F[A]$. Then $F[A]/C$ is free abelian with basis $\{aC:a\in A\}$.
2. Let $G$ is a free abelian group with basis $X$. Form the free group $F[X]$ and its commutator $C$. Then $G$ is isomorphic to $F[X]/C$.

Actually I have read the question Abelianization of free group is the free abelian group. However I don't understand with those advanced used of notations. Since Fraleigh mentions that those facts are 'easy to show'. Could any give me a simpler proof of it?

It all boils down to the universal property of free groups.

A free group $F[A]$ with basis $A$ is determined, up to a unique isomorphism, from the property

there is an injection $j_A\colon A\to F[A]$ and, for any group $G$ and map $f\colon A\to G$, there exists a unique group homomorphism $\hat{f}\colon F[A]\to G$ such that $f=\hat{f}\circ j_A$.

Change “group” into “abelian group” and you have the definition of a free abelian group.

Granted the existence of free groups, we can prove statement 1 as follows. The map $j_A$ can be replaced by set inclusion, so it's not restrictive to assume that $A\subseteq F[A]$.

Consider the map $g_A\colon A\to F[A]/C$ defined by $g_A(a)=aC$ and let $f\colon A\to G$ be a map, where $G$ is an abelian group. By the universal property of free groups, there is a unique group homomorphism $\hat{f}\colon F[A]\to G$ such that $\hat{f}(a)=f(a)$, for all $a\in A$. Since $G$ is abelian, it's easy to see that $C\subseteq\ker\hat{f}$, so $\hat{f}$ induces a group homomorphism $\tilde{f}\colon F[A]/C\to G$ such that $\tilde{f}\circ \pi=\hat{f}$, where $\pi\colon F[A]\to F[A]/C$ is the canonical projection.

Now it's obvious that $\tilde{f}\circ g_A=f$; since $A$ generates $F[A]$, also $\{aC:a\in A\}$ generates $F[A]/C$ and so the homomorphism $\tilde{f}$ is unique.

Statement 2 is a variation on the theme.

• I know about the property, but how could I know that is the free abelian group? Since $\{aC:a\in A\}$ generates $F[A]/C$. Suppose $(a_1C)^{k_1}(a_2C)^{k_2}...(a_nC)^{k_n}=C$. Then we need to show all $k_i$'s are zero. That is, by the universal property, we need to show $a_1^{k1}a_2^{k2}...a_n^{kn}\in C$ implies $k_i=0$. But I found a great difficulty on proving this. – Y.H. Chan Jul 5 '15 at 15:06
• @UnemChan Actually you don't need to prove that, that's the sense of my answer. – egreg Jul 5 '15 at 15:28
• @egreg, the free group on $A$ is denoted by $F[A]$. What about a notation for free abelian group on $A$? I never seen such one. – Sigur Jul 16 '19 at 18:27
• @Sigur I just copied $F[A]$ from the OP. I don't think there's a standard notation. – egreg Jul 16 '19 at 18:39
• @egreg, thanks for attention. – Sigur Jul 16 '19 at 18:45

This solution is motivated by DonAntonio's solution to the question Abelianization of free group is the free abelian group (Abelianization of free group is the free abelian group)

Lemma. Consider $$a_1^{k_1}a_2^{k_2}\cdots a_n^{k_n}\in F[A]$$, where $$a_i\in A$$ and $$k_i\in \Bbb{Z}$$. Note that $$a_1, a_2, ..., a_n$$ are not necessarily distinct. Suppose that $$\begin{eqnarray*} b_1 &=& a_{11}=a_{12}=\cdots=a_{1t_1}, \\ b_2 &=& a_{21}=a_{22}=\cdots=a_{2t_2}, \\ \vdots & \vdots& \vdots\\ b_s &=& a_{s1}=a_{s2}=\cdots=a_{st_s}, \\ \end{eqnarray*}$$ where $$a_{jk}\in \{a_1, a_2, ..., a_n\}$$ and $$b_1, b_2, ..., b_s$$ are distinct. If $$a_1^{k_1}a_2^{k_2}\cdots a_n^{k_n}\in C$$, then $$\begin{eqnarray*} 0 &=& r_{11}+r_{12}+\cdots+r_{1t_1} \\ &=& r_{21}+r_{22}+\cdots+r_{2t_2} \\ &\vdots& \vdots \\ &=& r_{s1}+r_{s2}+\cdots+r_{st_s}. \\ \end{eqnarray*}$$

This lemma can be proved by examining the definition of commutator subgroups.

Now, $$\{aC\mid a\in A\}$$ is a basis (linearly independent) for $$F[A]/C$$ follows immediately from this lemma. For if $$a_1 C, a_2 C, ..., a_n C$$ are distinct (this is a requirement in the definition of a basis for a free abelian group) and $$(a_1 C)^{k_1}(a_2 C)^{k_2}\cdots (a_n C)^{k_n}=C$$, then $$(a_1^{k_1}a_2^{k_2}\cdots a_n^{k_n})C=C$$ and $$a_1^{k_1}a_2^{k_2}\cdots a_n^{k_n}\in C$$. It follows that $$k_1=k_2=\cdots=k_n=0$$ by the lemma.

It also can be proved by using the universal property (see Theorem II.1.1.(iv) in Hungerford's Algebra). However, this method also uses the lemma when proving the homomorphism is one-to-one.

Universal Property for Free Abelian Groups. If $$H$$ is the free abelian group on $$A$$, then given an abelian group $$G$$ and function $$f:A\to G$$, there exists a unique homomorphism of groups $$\overline{f}:H\to G$$ such that $$\overline{f}\iota=f$$.

Let $$F[A]/C$$ be the abelian group and define $$f:A\to F[A]/C$$ by $$f(a)=aC$$ in the theorem. Then there exists a homomorphism $$\overline{f}:H \to F[A]/C$$ such that $$\overline{f}\iota=f$$. For $$k_1 a_1+k_2 a_2+\cdots +k_n a_n\in H$$, since $$H$$ is abelian, we can assume $$a_1, a_2, ..., a_n$$ are distinct. $$\begin{eqnarray*} &\text{If}& \overline{f}(k_1 a_1+k_2 a_2+\cdots +k_n a_n)=C \\ &\Rightarrow& \overline{f}(a_1)^{k_1}\overline{f}(a_2)^{k_2}\cdots \overline{f}(a_n)^{k_n}=C \\ &\Rightarrow& \overline{f}(\iota(a_1))^{k_1}\overline{f}(\iota(a_2))^{k_2}\cdots \overline{f}(\iota(a_n))^{k_n}=C \\ &\Rightarrow& f(a_1)^{k_1}f(a_2)^{k_2}\cdots f(a_n)^{k_n}=C\\ &\Rightarrow& (a_1 C)^{k_1} (a_2 C)^{k_2}\cdots (a_n C)^{k_n}=C \\ &\Rightarrow& a_1^{k_1}a_2^{k_2}\cdots a_n^{k_n}\in C\\ &\stackrel{\text{Lemma}}{\Rightarrow}& k_1=k_2=\cdots =k_n=0. \end{eqnarray*}$$ That is, $$\ker{\overline{f}}=0$$ and $$\overline{f}$$ is one-to-one.