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.

  • 1
    $\begingroup$ What you are trying to prove is false, this quotient is isomorphic to $Z^n$. $\endgroup$ – Moishe Kohan Nov 24 '13 at 21:22
  • $\begingroup$ ok, let me look at my notes. $\endgroup$ – user95747 Nov 24 '13 at 21:24
  • $\begingroup$ @studiosus. I asked my teacher. You are right. Sorry. $\endgroup$ – user95747 Nov 24 '13 at 21:32
  • $\begingroup$ @studiosus. I change it, because I think I have the same problem. $\endgroup$ – user95747 Nov 24 '13 at 21:39


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)

  • $\begingroup$ When you say "universal property" you only mean to the fact that the map in my post can be extended uniquely to $F_n$, or there are other implication that you use. $\endgroup$ – user95747 Nov 24 '13 at 22:38
  • $\begingroup$ No, exactly that...and with the hint given in the 2nd-3rd lines above you already have that $\;F_n'=\ker\phi\;$ (I'm calling the map and the group homomorphism it yields thesame name $\;\phi\;$ ) $\endgroup$ – DonAntonio Nov 24 '13 at 22:40
  • $\begingroup$ It is strictly necessary show that $F_n/F′_n$ is abelian? I think it is sufficiently show that their set of generators have the same number of elements. $\endgroup$ – user95747 Nov 24 '13 at 23:48
  • $\begingroup$ Well, it may depend on what you know. If you already I wrote in lines 2-3 then obviously it is not necessary, but if you don't know that then you may want to prove it first. $\endgroup$ – DonAntonio Nov 25 '13 at 0:14

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