Unique homomorphism, commutator subgroup.

Question

Let $F = F(A)$ be a free group, and let $f: A \to G$ be a set-function from the set $A$ to an abelian group $G$. What is the easiest way to see that $f$ induces a unique homomorphism $F/[F, F] \to G$, where $[F, F]$ is the commutator subgroup of $F$? Can we conclude that $F/[F, F] \cong F^{ab}(A)$?

Answer 1

This is true of any map to an abelian group. The comment just posted is not correct, because the kernel does not have to be precisely the group $[F,F]$, but since $f(a)f(b)f(a^{-1})f(b^{-1})$ does equal $1$ $[F,F]$ must be a subgroup of the kernel. So, by the second isomorphism theorem, we get $F/Ker(f) = (F/[F,F])/(Ker(f)/[F,F])$. This is how you know that it induces a homomorphism from $F/[F,F]$ to $G$. Since the group $F/[F,F]$ is the largest abelian group in this sense, in that every map to an abelian group factors through it, and since $F^{ab}(A) = \mathbb{Z}^{|A|}$ is abelian and there is a natural surjective map, you know that $F/[F,F]$ surjects onto $F^{ab}(A)$. To prove this is an isomorphism, you need to take the map $(x) \in F^{ab}(A) \rightarrow [x]$, where $[x]$ is the isomorphism class of $(x) \in F$.