# Factor group of a free group

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?

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.