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Given a free group of rank $n$ -> $F_n$ .

Is it easy to see what is the index of the subgroup $ [F_n, F_n ] \subseteq F_n $ ?

Hope someone will be able to help me understand this

Thanks !

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I think it is infinite index. The commutator subgroup is an infinitely generated (free) group. If it were a finite index subgroup of a finitely generated free group, it too would be finitely generated. –  Geoff Robinson Jul 30 '12 at 8:03
The quotient is free abelian of rank $n$, and hence infinite (assuming $n\geq 1$). This is equivalent to saying that the index of the commutator subgroup is infinite. –  Shane O Rourke Jul 30 '12 at 8:05
Great! Thanks a lot ! ! –  joshua Jul 30 '12 at 8:15
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2 Answers

Lemma. Given a normal subgroup $N$ of a group $G$, $G/N$ is abelian if and only if $G'\leqslant N$.

Proof. $$aNbN=bNaN\Leftrightarrow (a^{-1}b^{-1}ab)N=N \Leftrightarrow a^{-1}b^{-1}ab\in N$$ Letting $a,b$ be arbitrary in the above calculation yields the result.

With the above in mind we see that taking the quotient of the free group on $n$ generators by its commutator subgroup is equivalent to making the largest possible abelian quotient. It's easy to see that such a quotient is given by imposing the relations $[x,y]=1$ for each $x,y$ in the set of $n$ generators. Thus $F_n/F_n^\prime$ is the free abelian group of rank $n$, which for $n>0$ is countably infinite.

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$F_n / F_n'$ is the free abelian group on the free group on the set $\{1,\dotsc,n\}$, hence (by composing adjunctions) is the free abelian group on this set. –  Martin Brandenburg Mar 28 '13 at 18:43
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On an intuitive level we should automatically expect the group generated by $n$ letters modulo the commutativity relations should simply be the free abelian group on $n$ letters.

For a formal justification, one thing we should always look to depend on are the isomorphism theorems, as they are rather fundamental. In particular, there is a homomorphism from $F_n$ to the free abelian group ${\bf Z}^n$; the $k$th coordinate of the image of a word under this homomorphism is the sum of all the exponents of the $k$the letter in the word, e.g. $a^{\color{Blue}{3}}b^{\color{Green}{2}}a^{\color{Blue}{1}}b^{\color{Green}{-1}}\mapsto(\color{Blue}{3+1},\color{Green}{2-1})$.

It is straightforward to verify this is a homomorphism and that the commutator subgroup is contained in the kernel. The lemma Alexander mentions tells us that since the image is abelian the kernel must be contained in the commutator subgroup. Since inclusion holds both ways, there must be equality between the kernel and commutator subgroups.

Note that the quotient $G/[G,G]$ is the definition of the abelianization $G^{\rm ab}$ of $G$; the mentioned lemma states that every abelian homomorphic image (equivalently, quotient) of $G$ is isomorphic to a subgroup of the abelianization, so it is in this sense "universal."

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