I am reading Fraleigh's Abstract Albebra recently, and I cannot prove a statement about free abelian group:
- 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 $\{aC:a\in A\}$.
- 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
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.