Rank of free group

Question

Let $G$ be a free group with a basis $S$.

Let $G'$ be the commutator subgroup of $G$.

Define $S'=\{gG'\in G/G': g\in S\}$

Then, $S'$ is a basis for the free abelian group $G/G'$.

Following the above argument, How do I prove that $|S|=|S'|$?

Answer 1

Free group. Let $G$ be the free group on $S$. It consists of words ("strings") over the alphabet $S$. For instance $S = \{a,b\}$. Then if $x \in$ the commutator subgroup $G'$, then $xy = yx \ \forall y \in G$. But take $y = aaa$, $x = b$. Then $xy \neq yx$. Thus if $|S| \gt 1$, $G' = \{1\}$, and if $|S| = 1, \ G' = G$. Is that correct? Then $G/G' \approx G$ in the first case and $G/ G' \approx \{1\}$ in the second case.