Factor group of a free group

Question

Let $F[A]$ be the free group on the generating set $A$. Let $C$ be the commutator subgroup of $F[A]$, then show that $F[A]/C$ is a free abelian group with basis $\{aC \mid a \in A\}$. It is trivial that $F[A]/C$ is abelian and generated by $\{aC \mid a \in A\}$, but how can I prove that $\{aC \mid a \in A\}$ is linearly independent?

Answer 1

You can avoid having to prove that by simply showing that $F[A]/C$ and $\{aC\mid a\in A\}$ have the relevant universal property... That is done in egreg’s answer in the question you link to.

To show they are linearly independent, suppose that $a_1,\ldots,a_n\in A$ be pairwise distinct and $\beta_1,\ldots,\beta_n\in\mathbb{Z}$ are such that $$a_1^{\beta_1}\cdots a_n^{\beta^n}C=eC.$$ Then $a_1^{\beta_1}\cdots a_n^{\beta^n}\in C$.

But in elements of $C$, the sum of the exponents of each $a\in A$ is $0$. This follows by noting that the map from $F[A]$ to $\mathbb{Z}$ obtained by sending $a$ to $1$ and all other elements of $A$ to $0$ must have $C$ in the kernel (since $\mathbb{Z}$ is abelian), and so if $g\in C$ then $g\mapsto 0$. In particular, the sum of the exponents of $a$ in $g$ add up to $0$. This holds for each $a\in A$.

Since $a_1^{\beta_1}\cdots a_n^{\beta_n}\in C$, then $\beta_i=0$ for each $i$. Thus, the set $\{aC\mid a\in A\}$ is linearly independent in $F[A]/C$, as claimed.