# If $F$ is free with basis $X$ and if $F'$ is the commutator subgroup of $F$, then $F/F'$ is free abelian with basis all cosets $xF'$ with $x \in X$

In response to the duplicate claim: The question I've asked is specifically about showing the isomorphism I've chosen is well defined. I see nothing in the linked question that answers this.

From Rotman's algebraic topology:

If $F$ is free with basis $X$ and if $F'$ is the commutator subgroup of $F$, then $F/F'$ is free abelian with basis all cosets $xF'$ with $x \in X$.

I see that I would have to show that for each $x \in X$, $\langle xF' \rangle \approx (\Bbb Z, +)$ and that $F/F' = \oplus_{x \in X} \langle xF' \rangle$

Showing $F/F' = \sum_{x \in X}\langle xF'\rangle$ is simple, as:

$z \in F/F' \Rightarrow z = wF'$, where $w$ is a reduced word in $F$ and $q \in \oplus_{x \in X} \langle xF' \rangle \Rightarrow q = \oplus x_i^{n_i}F'$ which is the definition of a word.

Next I try to show $\langle xF' \rangle \approx (\Bbb Z, +)$:

$x^nF' \mapsto n$ should be the isomorphism, but I don't understand how to show it's well defined.

I know I need to show that $x^nF' = x^mF' \Rightarrow n = m$, but I don't see how I would continue from here.

Anyone have any ideas?