Let $F$ be a free group of rank $n$. Let $G$ be the commutator subgroup of $F$. I need prove that $F/G\cong\mathbb{Z}^n$.
I have tried with the isomorphism theorem: With $\varphi$, I send the $x_i$ element in the basis of $F$ to the vector with $1$ in the i-th entree and $0$ in the others, in $\mathbb{Z}^n$. But I can't prove that $ker\varphi\subseteq G$. Any help.
Hints:
In any group $\;G\;$ and for any normal subgroup $\;H\lhd G\;$ , we have that the quotient $\;G/H\;$ is abelian iff $\;G'\le H\;$
If $\;\{x_1,...,x_n\}\;$ is a set of free generators of the free group $\;F_n\;$, then $\;\{xF_n',\ldots,x_nF_n'\}\;$ is a basis for the abelianization $\;F_n/F_n'\;$ , and thus this last group is a free abelian group of rank $\;n\;$
Since $\;\Bbb Z^n:=\underbrace{\Bbb Z\times\ldots\times\Bbb Z}_{n\;\text{times}}\;$ is a free abelian group of rank $\;n\;$ , we have that $\;F_n/F_n'\cong\Bbb Z^n\;$ .(You can also get this by means of the universal property of $\;F_n\;$ and the map you described in your post)